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© Mr. E. Therrien Chemistry 12 Notes ©

1 Thermochemistry

1. Thermochemistry Thermochemistry is the study of heat changes that occur during chemical reactions. Fuels such as coal, oil and gasoline have high chemical potential energy. Their combustion products, carbon dioxide and water, have very low chemical potential energy. Energy is the capacity for doing work or supplying heat. We can also state that the energy is the capacity to produce changes. It is found in two forms: Energy Form Symbol Definition Chemistry Example Potential Ep Stored energy Energy in chemical bonds (Chemical Energy) Kinetic Ek Energy of motion Movement of particles (Thermal Energy) Kinetic Molecular Theory: The particles in a substance are constantly moving and colliding. Adding heat energy to an object increases its energy content (Ek + Ep).

Increases Ek → increases particle motion Adding Heat Energy Increases Ep → increases bond energies Furthermore, kinetic energy is continually converted to potential energy, and vice versa. Consider a head-on collision between two gas molecules with the same initial velocity – energy must be conserved.

Event Diagram Energy Description Approach Ek is high

Ep is low Collision Ek is zero

Ep is high Rebound Ek is high again

Ep is low again Heat is a form of energy. Heat and temperature are not the same. Heat involves a transfer of energy between two objects because of a difference in temperature. Temperature is the measure of the average kinetic energy of molecules. Any temperature change (∆T) depends on the amount of heat (q) flowing into or out of the substance. The SI unit for measuring energy, including heat, is the Joule (J). Only temperature can be measured. Heat is not measurable but can be calculated from measuring temperatures changes. The temperature change produced by an amount of heat varies directly with three factors:

q ∞ mass (m) of the object q ∞ specific heat (c) of the object q ∞ temperature change (ΔT) in the object

Therefore, q = mcΔT (i) Specific Heat (c) = heat required to raise the temperature of 1 g of a substance by 1 oC. = has the unit J / g · oC


2 (ii) Some specific heat values have been tabled for you. Be careful. They depend on the physical state of the substance. (iii) ∆T = Tfinal - Tinitial If ∆T > 0, the q will become a positive number (heat is gained). If ∆T < 0, the q will become a negative number (heat is lost). Try This: Calculate the missing values. Substance Heat

Exchanged (J)

Mass (g)

Specific heat (J / g · oC)

T (oC)

Initial Temperature (oC)

Final Temperature (oC)

Water +1530 20.0 23.2 Aluminum 14 95 26 +3600 20 14 48 Steam -1500 -25 116 Mercury +825 152 42.0 It is useful in thermochemistry to define a system as the part of the universe on which we focus our attention. For example, a system might be the chemicals that will undergo a reaction and the surroundings could be the water bath that surrounds the flask that contains the chemicals. Together, the system and surroundings constitute the universe. The changes that are of interest in thermochemistry are the flow of heat from the system to the surroundings or to the system from the surroundings. The first law of thermodynamics is that the energy of the universe remains constant.

No energy is created or destroyed as a result of the transfer of heat. 2. Internal Energy, Heat, and Work All of the energy in a system is called the system's internal energy. It is impossible to measure the internal energy of a system, but we can measure the heat changes that accompany a chemical or physical change. ∆E = E2 - E1 Internal energy is a state property and is independent of how the substance is prepared. Other state properties include pressure, temperature and volume. The internal energy of a system is changed by the flow of heat or work or both during a chemical or physical process. This change in internal energy is expressed as follows. ΔE = q = w q = heat w= work If energy is exchanged by mechanical means we call it work. If it is exchanged between the system and its surroundings due to a difference in temperature, we call it heat. A process that releases heat that flows to the surroundings is called an exothermic process. (q < 0) A process that absorbs heat from the surroundings is called an endothermic process. (q > 0) Work done by the system is a negative quantity (w < 0) Work done on the system is a positive quantity (w > 0)


3 3. Calorimetry A calorimeter is used to determine the heat released by a reaction. In calorimetry, the heat evolves by the system is equal to the heat absorbed by the surroundings. Calorimeters are devices used to measure the absorptions or release of heat in physical or chemical processes. In a closed system, the mass of the system remains constant throughout a process. A simple calorimeter is a Styrofoam cup used as an open (constant pressure) system. In a chemical reaction, system = chemical components surroundings = everything outside the system Calorimeter = system + surroundings dissolved chemicals water Two events may happen in a calorimeter:

Analysis Analysis Endothermic reaction Exothermic reaction Bonds in the system reform at a higher energy level.

Bonds in the system reform at a lower energy level.

Temperature of surroundings decreased (measured)

Temperature of surroundings increased (measured)

Enthalpy of system increased (has to be calculated)

Enthalpy of system decreases (has to be calculated)

∆H > 0 (positive values) ∆H < 0 (negative values) Energy Level Diagram

Energy Level Diagram

In a constant-pressure calorimeter we measure the change in enthalpy of the system. Enthalpy (H) is the amount of heat that a substance has at a given temperature and pressure.

*** Try to remember Pexo and Rendo The energy values of foods cans be determined by burning them in a bomb calorimeter. 4. Enthalpy Heat and enthalpy are the same for reactions at constant pressure. Heat change at constant pressure is the enthalpy change.

surroundings

energy system

surroundings

energy system

reactants

products

∆H = +

reaction path

reactants

products ∆H = -

reaction path


4 System Pressure Volume Energy Example Open Constant Varies Exchanged Beaker Closed Varies Constant Exchanged Gas cylinder Most changes in chemistry happen in open systems. SATP = Standard Ambient Temperature and Pressure = 101 kPa and 25oC The total energy amount (Ek + Ep) in an open system is called its heat content or enthalpy (H). The enthalpy of a system cannot be known. However, enthalpy changes (∆H) can be tracked because they will occur in the form of a heat transfer (q).

∆H = H (products) – H (reactants) = q This change in enthalpy is mostly due to the energy required for breaking and remaking bonds. We will study three events that involve enthalpy (heat) changes: Event Symbol ∆T Kinetic Molecular Theory Flow of heat q Yes Change to the movement of molecules Phase change ∆H No Change to the intermolecular bonds between

particles Chemical reaction ∆H No (at SATP) Change to the intramolecular bonds within the

particles Note that temperature changes cannot always be expected with a change in the amount of heat (enthalpy). Physical Changes Substances exchange heat with their surroundings as they undergo physical changes such as melting/freezing, vaporization,/condensation and dissolving. All pure substances show similar curves as they are heated and cooled.


5 All phase changes involve an exchange (in or out) of a unique amount of heat. H2O(s) + 6.01 kJ → H2O(l) C6H6(l) – 9.87 kJ → C6H6(s) What is wrong with this equation???? The energy exchange in kJ needed to produce these phase changes have been measured for most chemicals and tabled (see reference sheet).

(i) heat of fusion = - heat of freezing heat of vaporization = - heat of condensation

(ii) The measurement is referred to as the molar enthalpy constant. Its unit is kJ/mol. q = ∆H = nH Thus, the amount of heat involved as a substance is heated or cooled can be calculated using two formulas:

1. q = mc∆T (relates heat to events with temperature changes) 2. q = nH (relates heat to events that produce no temperature change)

Molar Enthalpies for phase changes Molar Enthalpy of Fusion (Hfus) – is the heat absorbed by one mole of a substance when melting from a solid to a liquid at the same temperature. Molar enthalpy of freezing (Hfr) is the same amount of heat released when the substance solidifies from a liquid to as solid.

1 mol (s) + H(fus) → 1 mol (l) 1 mol (l) + H(fr) → 1 mol (s)

(i) most molar enthalpies have been measure and tabled (see reference sheet) (ii) Hfus is also called Hmelt. (iii) The unit for all H is kJ/mol

Molar Enthalpy of Vaporization (Hvap) – This is the heat absorbed by one mole of a substance while vaporizing from a liquid to a gas at the same temperature.

1 mol (l) + H(vap) → 1 mol (g) 1 mol (g) + H(cond) → 1 mol (l)

Try This: 1. How much heat is required to melt 1.5 mol C6H6? 2. How much heat is released by freezing 1.5 mol CH3OH? 3. How much heat is needed to vaporize 2.6 mol H2O? 4. How much heat is released by the condensation of 25.0 g H2O? 5. What mass of ice can be melted by 152 kJ of heat? 6. What is the Hcond of a substance if 25 kJ of heat is released by the condensation of 2.3

mol?

Hfus positive number takes in heat

Hfr negative number gives off heat

Hvap positive number takes in heat

Hcond negative number gives off heat


6 Putting it All Together Example: Calculate the total energy change (∆Etotal) involved in raising the temperature of a 25.7 g block of ice from freezing temperature (-18 oC) to room temperature (22 oC). Heat solid → q1 = mc∆T q1 = 25.7 g (2.1 J/g · oC)(18 oC) q1 = 970 J = 0.97 kJ Melt solid → q2 = nHfus

q2 = (25.7 x 1 mol / 18.0g)(6.01 kJ/mol) q2 = 8.59 kJ

Heat liquid → q3 = mc∆T q3 = (25.7 g)(4.19 J/g · oC)(22 oC) q3 = 2369 J = 2.4 kJ

Total → ∆Etotal = q1 + q2 + q3 = (0.97 kJ) + (8.59 kJ) + (2.4 kJ) = 12.0 kJ Note:

(i) Specific heat capacities will vary between phases (ii) Pay close attention to units!

Using a Calorimeter Enthalpy changes in a system can only be found indirectly by measuring temperature changes in its surroundings. This idea can be used to calculate any enthalpy change in the calorimeter. Here are two examples:

1. Chemicals exchange heat as they dissolve in water. The Molar Enthalpy of Solution (Hs) is the amount of heat absorbed or released per mole of a substance as it dissolves to form a solution. KNO3(s) → K+

(aq) + NO3-(aq) where Hs = 15 kJ/mol

2. The process of a fuel burning in oxygen is referred to as combustion. The Molar Enthalpy

of Combustion (Hcomb) is the amount of energy released per mole of fuel as it is burned. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) where Hcomb= -2220 kJ/mol

We know that energy must be conserved: q(system) = - q (surroundings) Therefore mc∆T (system) = - mc∆T (surroundings) Or nH (system) = - mc∆T (surroundings) Note:

(i) do not confuse system data with surroundings data. (ii) The heat units may not be balanced. (1kJ = 1000 J).

You may wish to use nH = - mC∆T/1000 (iii) There are many tables of different enthalpy terms – Hs, Hcomb, Hfr, …

5. Thermochemical Equations A thermochemical equation includes the heat change. The heat of reaction is the heat change for the equation, exactly as it is written.


7 CaO(s) + H2O(l) Ca(OH)2(s) + 65.2 kJ In this reaction the heat is released so it’s an

exothermic reaction and the heat of reaction is 65.2 kJ

Another way to show this is… CaO(s) + H2O(l) Ca(OH)2(s) Ho = -65.2 kJ In this reaction the heat is absorbed so it’s an endothermic reaction and the heat of reaction is 129 kJ 2NaHCO3(s) + 129 kJ Na2CO3(s) + H2O(g) + CO2(g) or 2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g) Ho = 64.5 kJ The heat of reaction for the burning of 1 mol of a substance completely is often given the specific name, heat of combustion Ho

comb at SATP. The physical state of the reactants and products in a thermochemical reaction must be stated. Example: H2O(g) H2(g) + ½ O2(g) Ho = 241.8 kJ H2O(l) H2(g) + ½ O2(g) Ho = 285.8 kJ 6. Heats of physical processes The heat absorbed by one mole of a substance when melting from a solid to a liquid at the same temperature is the molar heat of melting ∆Hmelt or the molar heat of ∆Hfus. The heat lost when one mole of a liquid changes to a solid at the same temperature is the molar heat of freezing, ∆Hfreez, or the molar heat of solidification ∆Hsolid. Temperature remains constant as heat is transferred during a phase change. Melting is an endothermic process; freezing is an exothermic process. Fusion is another name for melting; solidification is another name for freezing. 7. Chemical Changes (Ho) Molar enthalpies can be used to calculate the enthalpy change that occurs during a chemical reaction. The molar enthalpy that is determined when the initial and final condition of the chemical system are at SATP is called the Standard Molar Enthalpy Change (Ho) or the Enthalpy of Reaction. Example: When hydrogen gas and chlorine gas at SATP combine to form two moles of hydrogen chloride gas at SATP, 184 kJ of heat is released. H2(g) + Cl2(g) → 2 HCl(g) Ho = -184 kJ

Reactants at SATP

heating Reaction cooling Products at SATP

Standard Molar Enthalpy Change (Ho)


8 This is an exothermic reaction. Heat was released during the reaction. The enthalpy content of the system decreased. The enthalpy content of the surroundings would therefore increase and we would expect its temperature to rise.

Endothermic Reactions Exothermic Reactions reactants + heat → products reactants → products + heat Ho > 0 Ho < 0 Hprod > Hreact Hreact > Hprod Enthalpy content of the system has increased Enthalpy content of the system has decreased Reporting Ho The standard molar enthalpy change is reported for each reaction exactly as it is written. This can be done several ways:

(i) HCl(g) + ½ I2(g) → HI(g) + ½ Cl2(g) Ho = 118 kJ (Since the equations refer to the change of one mole of one of the chemicals, fractional coefficients are permitted.)

(ii) It may also be reported as HCl(g) + ½ I2(g) + 118 kJ → HI(g) + ½ Cl2(g)

(iii) It may be doubled if desired to 2 HCl(g) + I2(g) → 2 HI(g) + Cl2(g) H = 236 kJ

(iv) Since the equation is reversible, it can be rewritten as HI(g) + ½ Cl2(g) → HCl(g) + ½ I2(g) Ho = -118 kJ

Or HI(g) + ½ Cl2(g) → HCl(g) + ½ I2(g) + 118 kJ

Try This: For the reaction A + ½ B → C + 2D Ho = 100 kJ

a) Rewrite the same information three more ways. b) Rewrite the equation to show the heat involved in reacting (R)

(i) 1 mol B (ii) 1 mol D (iii) 2 mol A (iv) 5 mol C

c) Rewrite the equation to show the heat involved in forming (P) (i) 1 mol B (ii) 1 mol D (iii) 2 mol A (iv) 5 mol C

Predicting Ho Molar Enthalpies of Reaction (Ho) can be measured directly in a calorimeter or they can be calculated indirectly using three methods:

a) Using Molar Heats of Formation (Hof)

This is the change in heat that accompanies the formation of one mole of a compound from its elements (which have a defined Ho

f = 0).


9 Example: H2(g) + ½ O2(g) → H2O(l) Ho

f = -286 kJ (This means that 286 kJ of energy was released in the formation of one mole of the compound water from the elements hydrogen and oxygen.) Theses Ho

f values have been measured for most compounds and collected into tables (see sheet standard molar enthalpies of formation). We can use these Ho

f values to calculate the overall Ho value for any chemical change. Example: What is the enthalpy change that occurs in this reaction? 2 SO2(g) + O2(g) → 2 SO3(g) Ho = ? 2 mol (-297 kJ/mol) + 1 mol (kJ/mol) → 2 mol (-396 kJ/mol) -594 kJ → -792 kJ

H = -198 kJ (reaction is exothermic)

Try this: Find the molar enthalpy of reaction for the following: 1. 2 CO2(g) → 2 CO(g) + O2(g) 2. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) 3. 4 CO2 + 2 H2O(g) → 2 C2H2(g) + 5 O2(g) 4. 2 HI(g) → H2(g) + I2(g) 5. P4O6(s) + 2 O2(g) → P4O10(s) 6. 2 Fe(s) + O2(g) → 2 FeO(s) 7. 2 Fe2O3(s) → 4 FeO(s) + O2(g) 8. SO2(g) + H2O(l) → H2SO3(aq)

b) Using Hess’s Law

The enthalpy change for a reaction is the sum of all of the enthalpy changes for all of the reactions that add up to the overall reaction = Hess’s Law If you can add the equations, then you can add the H’s. Example: Find the molar heat of combustion for carbon monoxide burned in excess oxygen:

CO(g) + ½ O2(g) → CO2(g) Ho = ? Given 2 C(s) + O2(g) → 2 CO(g) H = -222 kJ 1 And C(s) + O2(g) → CO2(g) H = -394 kJ 2 (arrange equations) CO(g) → C(s) + ½ O2(g) Ho = 111 kJ 1

C(s) + O2(g) → CO2(g) Ho = -394 kJ 2 (cancel, then add) CO(g) + ½ O2(g) → CO2(g) Ho = -283 kJ Remember that an equation can be reversed and/or multiplied by a constant (which can be a whole number or a fraction). Opposite formulas in any equation can be cancelled before combining. Try this:

1. Find P4O10(s) → 4 P(s) + 5 O2(g) Ho = ? Given 4 P(s) + 3 O2(g) → P4O6(s) H = -1640 kJ

And P4O6(s) + 2 O2(g) → P4O10(s) H = -1344 kJ


10 2. Find 3 C2H2(g) → C6H6(l) Ho = ? If C2H2(g) + 5/2 O2(g) → 2 CO2(g) + H2O(l) H = -1299 kJ And C6H6(l) + 15/2 O2(g) → 6 CO2(g) + 3 H2O(l) H = -3267 kJ

c) Using Bond Energies During a chemical reaction, bonds are broken and reformed into new bonds. Energy (usually in the form of heat) must be supplied to break bonds (a positive number). When the bonds are reformed, though, the heat energy is released (a negative number). H2(g) + ½ O2(g) → H2O(g) Ho = -242 kJ

Break 1 mol break ½ mol form 2 mol product bonds H-H bonds O=O bonds O-H bonds are stronger

(+) (+) (-) The heat energy absorbed at constant pressure when a chemical bond is broken is called bond energy. The values for some bond energies are given on your reference sheet. The numbers are usually the average value for breaking bonds of gaseous molecules. Bond energies can be used to estimate molar enthalpies of reaction. Example: Determine the Ho for the reaction: ½ H2(g) + ½ Cl2(g) → HCl(g)

Reactants (+) Products (-) ½ mol H-H x 436 kJ/mol = +218 kJ ½ mol Cl-Cl x 243 kJ/mol = 122 kJ

1 mol H-Cl x 432 kJ/mol = -432 kJ

Ho = 218 kJ + 122 kJ – 432 kJ = - 92 kJ Try this

1. Estimate the Ho for these reactions using average bond energy. Use structural diagrams to determine the bonds in the molecules.

a) H2(g) + F2(g) → 2 HF(g) b) N2(g) + 3 H2(g) → 2 NH3(g) c) N2(g) + 2 H2(g) → N2H4(g) d) C2H4(g) + H2(g) → C2H6(g) e) CS2(g) + 2 H2O(g) → CO2(g) + 2 H2S(g)

2. Given XeF2(g) + H2(g) → 2 HF(g) + Xe(g) Ho = -430 kJ

Find the bond energy of the Xe-F bond.


11

Dissolving

Crystallization

+ heat

+ solute

- heat + seed crystal

- solute

Amount of solute

Temperature

Supersaturated region

Unsaturated region

Saturated line

From Solutions to Kinetics to Equilibrium 1. Concentrations

1a) Solutions Solute + Solvent = Solutions All solutions are homogeneous mixtures that cannot be separated by filtration. They are found in all physical states. Ionic substances dissolve most easily in water. These compounds can dissociate into individual aqueous ions. The polar water molecules attract the positive sodium ions and the negative chlorine ions in the salt crystal. The ions are drawn out of the crystal, surrounded by water molecules and carried away. There is a limit to the amount of solute that can be dissolved is a solvent. At some point an equilibrium is reached and the solution is saturated. In a saturated solution - rate of dissolving = rate of crystallization.

- dissolved particles are in equilibrium with undissolved particles. The solubility of most substances increases as the temperature of the solvent increases. This allows us to make a supersaturated solution: When the solubilities of different solutes are graphed at different temperatures, a solubility curve is obtained.

A solution is a homogeneous mixture of two or more substances, the composition of which can be varied within definite limits. As well, the components do not settle out upon standing and cannot be separated by filtration. Terminology Solvent = the substance doing the dissolving. Water is the universal solvent. Solute = the substance being dissolved. Electrolytes = water solutions that conduct electricity (salt water, acid solutions) Non-electrolytes = solutions that do not conduct electricity (sugar water)

Salt crystal

Dissolved particles

saturated solution

saturated solution

supersaturated solution

saturated solution


12 Miscible = two liquids soluble in each other (alcohol/water, ether/alcohol) Immiscible = two liquids insoluble in each other (oil/water, gasoline/water) Dilute = more solvent than solute in solution (opposite is concentrated) Dissociation = the separation of ions from an ionic crystal (salt crystals in water dissociate into sodium and chloride ions) Saturated solution = a solution in which no more solute will dissolve in a given amount of solvent at that temperature. (opposite is unsaturated) Precipitation = the separation of a solid from a solution – particles settle upon standing. Super saturation solution = a solution containing more solute than if saturated under given conditions (a hot solution cooled very slowly so that excess solute will not precipitate). Solution equilibrium = when two opposing processes of dissolving and crystallizing occur at equal rates. Types of Solutions: 1b) Molarity Concentration is a measure of the amount of solute that is dissolved in a given amount of solution. The most important measure of concentrations is molarity (M). Molarity is the number of moles of a solute dissolved in 1 L of the solution. Molar solutions are prepared in volumetric flasks. To make a 1 M solution: Add some add 1 mol fill to the Solvent of solute 1 L mark. concentration (c) = molarity (M) = n/V = moles of solute / liters of solution. (SI term) (non-SI term) Lab solutions are usually sold in a concentrated form called a stock solution. Dilution is the process of decreasing the concentration of the stock solution by adding more solvent (usually water)

MsVs = MdVd stock solution diluted solution

SOLUTE SOLVENT EXAMPLE gas solid Gas bubbles in ice gas liquid Carbonated beverages gas gas Carbon dioxide in air liquid solid Mercury/copper, mercury/silver liquid liquid Alcohol/water (antifreeze) liquid gas Water vapor/air, perfume/air solid solid Alloys (copper + tin = brass) solid liquid Sugar/water, salt/water

solid gas Sulfur vapor/air


13 Try This: Use the formulas or factor-label to complete the tables. 1. Molarity

Substance Mass (g) Moles (mol) Volume (L) Molarity (mol/L) CaCl2 12 1.5 NH4NO3 1.61 0.652 NaCl 450 mL 1.5 MgF2 0.53 0.050 Ca(OH)2 51.0 0.25 2. Dilutions Substance Mole Ms Vs Water

added (L) Md Vd

NaNO3 1.64 2.51 1.25 BaCl2 0.582 2.54 0.0800 Mg(OH)2 0.45 0.85 0.15 2. Solvation

2 a) Aqueous Solutions Aqueous solutions (aq) have water as a solvent. Ionic Compounds: Water causes ionic compounds to become separated:

NaCl(s) → Na+(aq) + Cl-(aq)

This separation of ions from their crystal is called dissociation. The presence of ions in a liquid permits the solution to conduct electricity; the more ions present, the better the conduction. When a crystal contains proportionately more than one ion of a particular type, all of the ions dissociate: Al2(SO4)3(s) → 2 Al3+

(aq) + 3 SO42-

(aq) Molecular Compounds: these substances may also be soluble: (i) Many molecules create ions when dissolved in water in a process called ionization.

HCl(g) → H+(aq) + Cl-(aq)

Note that since H+ ions are always attached to a water molecule (H+ • H2O), this equation is also written as

HCl(g) + H2O(l) → H3O+(aq) + Cl-(aq)

(ii) Some molecular compounds, such as glucose, dissolve without breaking up. They form hydrogen bonds with the water molecules. They are drawn apart and carried away by the water molecules.

C6H12O6(s) → C6H12O6(aq) Precipitation is the reverse of dissociation.

Ag+(aq) + Cl-(aq) → AgCl(s)

2 b) Ionic Equations

A reaction between two chemicals dissolved in water is shown three ways:

(i) Molecular Equation - the reaction is written as if all of the chemicals consist of molecules:

BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2 NaCl(aq)


14

Reactants Products

Activated complex

Activation Energy (Ea)

Energy of reaction (∆Ho)

(ii) Total Ionic Equation - the equation is written to show that the dissolved chemicals are actually in the forms of ions. Ba2+

(aq) + 2 Cl-(aq) + 2 Na+

(aq) + SO42-

(aq) → BaSO4(s) + 2 Na+(aq) + 2 Cl-

(aq)

(iii) Net Ionic Equation - the equation is written to show only the ions or molecules that are actively involved in the reaction. Ba2+

(aq) + SO42-

(aq) → BaSO4(s) The Na+ and Cl- ions were eliminated since they did not take part in the reaction. Ions like this that occur in the same form on both sides of the equation are called spectator ions. Note: (i) Ionic equations should be written from molecular equations.

(ii) All solution equations must contain the physical state symbols. (iii) Ionic equations are balanced in terms of mass and charge.

Try This: Write the total and net ionic equations for the following:

a) AgNO3 + K2CO3 → b) Na3PO4 + MgSO4 → c) SnSO4 + Na2S → d) FeBr2 + KOH → e) SnCl2 + K3PO4 →

3. Particle Theory

3 a) Collision Theory Reacting particles must collide, and with enough energy for bond breaking and bond making to occur. A reshuffling of the bonds leads to the formation of new products. Some collisions have enough energy in the proper orientation to cause an interpenetration of the electron clouds. The distance between the reactant particles is now small enough for new bonds to form. The minimum amount of energy needed for these effective collisions is called activation energy. In the brief moment between bond disruption and bond formation, some sort of partial bonding exists - call the activated complex. Its lifetime is short. An activated complex is equally likely to return to reactants as proceed to products. For the reaction H2(g) + I2(g) →2 HI(g) Energy

Reaction path This particle theory applies to all reactions:


15

-10

+10

-10

+10 A + B → C

C → A + B

Ea = +10 Ea = +15 ∆Ho = -5 ∆Ho = +5 Exothermic (reaction is favored) Endothermic

3 b) Reaction Rates Rate of Reaction is measured by tracking the change in concentration of a chemical over a unit of time. Rate of reaction varies directly with the number of particle collisions that occur in a measured amount of time. Changing the conditions of a reaction may affect either the frequency of collisions or the collision efficiency. Five variables will affect reaction rates:

1. Nature of the Reactants weak bond → increases collision efficiency → faster reaction

2. Surface Area (solids only) crushing → increases surface area → increases collision frequency → faster reaction

3. Concentration (liquid and gases only) increase pressure (gases only) → increases concentration→ more particles → increases collision frequency → faster reaction

4. Temperature (endothermic and exothermic) increases temperature → higher Ek → faster particles → increases collision frequency → faster reaction

5. Catalysts - these are substances that increase a reaction rate without being permanently changed. They provide the reactants with an alternative reaction path of lower activation energy.

catalyst → lower energy barrier → increases collision efficiency → faster reaction

3 c) Calculating Reaction Rates Reaction Rate is a measure of how fast a reactant is used up or how fast a product is formed. By convention, both rates are assumed to be positive. Rreact = ∆ [reactant] = ∆ [product] ∆t ∆t

Normal energy barrier

Energy barrier with catalyst


16 Where Rreact = average rate of reaction. It depends on when the measurement was taken. The

unit is mol/L time unit [ ] = molar concentration (mol/L) t = time elapsed (s, min, h)

Only one substance in any reaction needs to be measured in order to calculate rates for the other substances. Note that changes in concentration can be found by measuring change in color, volume, temperature, pressure, etc. Try This:

1. In the decomposition reaction 2N2O5(g) → 4 NO2(g) + O2(g) that takes place in 2.5 L of pure carbon tetrachloride solution, 0.14 mole of O2 is produced over 6.5 min. Calculate the rate of (a) production of O2 gas (b) loss of N2O5(g).

2. During the reaction CO(g) + NO2(g) → CO2(g) + NO(g), the concentration of CO was 0.019 M at 27 min and 0.013 M at 45 min. Calculate the average rate for (a) the loss of CO(g) (b) the gain of CO2(g).

3. Given: 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) Measured: [O2] = 0.022 M at 12.0 s and 0.018 M at 23.0 s Calculate: the average rate for (a) loss of NH3(g) (b) gain of H2O(g)

4. Equilibrium Systems In some chemical reactions, all of the reactants are converted to products: 2 Cu(s) + O2(g) → 2 CuO(s) This reaction is said to go to completion. In other processes, however, the final mixture consists of both products and reactants

Reactants → Products Reactants ← Products Reactants products

This reaction is at equilibrium. The reactants are colliding to form products while the products are also colliding to reform reactants. This is known as a dynamic equilibrium = the rate of the forward reaction is equal to the rate of the reverse reaction. Equilibrium is reached in a closed system when all properties of the system are constant. The forward and reverse reactions are happening at the same rate.

R(forwared) = R(reverse)

concentration time At equilibrium, the overall reaction has reached a balanced state in which there is no net reaction. The concentrations of products and reactants may be different, but they are no longer changing. 5. Describing Equilibriums


17

t t

5 a) Law of Mass Action For any equilibrium in a closed system, the amount of reactants and products can be calculated using the Law of Mass Action:

If mA + nB pC + qD Then [C]p • [D]q = Keq

[A]m • [B]n Note:

(i) At any chosen temperature, if the concentrations of the products or reactants are changed, they will yield the constant Keq value after the equilibrium has re-established itself.

(ii) Each reaction temperature will have a unique Keq value. (iii) The unit for Keq varies with each expression so they are usually omitted. (iv) The [ ] indicate the Molar concentration only at equilibrium. (v) The expression can only include substances which can actually vary in concentration:

Included Not included Gases (g) Solids (s)

Solutions (aq) Pure liquids (l) Liquid mixtures (l + l)

Try This: Write the equilibrium expressions for

a) 3 A(g) + 2 B(g) 2 C(g) b) A(aq) + 2 B(g) 2 C(aq) + D(aq) c) A(g) + B(l) 3 C(g) d) 2 A(g) B(s) e) B(s) 2 A(g)

5 b) Analyzing Equilibriums

The concentration may be changed in any equilibrium but the Keq value will not change, provided the temperature is held constant. Thus, the Keq values for many equilibrium reactions have been measured at recommended temperatures and tabled for reference. The Keq values may be very large or small but never equal to zero. Keq relates the amount of product or to the amount of reactant in a closed chemical system.

Forward Reaction Preferred Reverse Reaction Preferred

(i) [products] > [reactants] (i) [reactants] > [products] (ii) [products] / [reactants] > 1 (ii) [products] / [reactants] < 1 (iii) Keq > 1 (iii) Keq < 1 (iv) products are favored (iv) reactants are favored

c [products]

[reactants] [products]

c [reactants]

[reactants]


18 Try This:

1. At 460oC, the following system is at equilibrium 2 NO(g) + O2(g) 2 NO2(g) The concentrations were then measured at 0.100 M NO(g), 0.014 M O2(g) and 0.100 M NO2(g). Are the products or reactants preferred?

2. For the equilibrium F2(g) + I2(g) 2 FI(g) the concentrations were [FI(g)] = 5.00 M, [F2(g)] = 2.50 M and [I2] = 1.00 M at a fixed temperature.

a. Prove that products are favored in this reaction. b. Rewrite the equation so that reactants would be favored at this temperature.

6. Equilibrium Calculations Even though the concentrations of chemicals may vary widely at equilibrium, the Keq ratio remains the same at a constant temperature. Equilibrium problems involve three sources of information:

(1) a balanced equation – for mole ratios (2) Keq values – known or unknown (3) Equilibrium concentrations – known or unknown – [ ] means Molarity at equilibrium

only. Note:

(i) Increasing Keq values indicate an increasing concentration of products. (ii) There is no unit with Keq value. (iii) Unit Alert = mol vs. M

6 a) Given Equilibrium Concentrations Try This:

1. Consider the reaction 2 SO2(g) + O2(g) 2 SO3(g) Find the equilibrium constant (Keq) if the concentrations at equilibrium are as follows: [SO3(g)] = 0.10 M, [SO2(g)] = 0.052 M, [O2(g)] = 0.048 M

2. At a different temperature, Keq = 85 for the reaction in question 1. Find [SO3(g)] at this new temperature knowing that [SO2(g)] = [O2(g)] = 0.050 M

6 b) Given Starting Concentrations Only (Find Keq) We are often required to calculate the change in concentration during the course of a reaction. These problems will involve measured starting concentrations, which will have to be changed to equilibrium concentrations before they can be used in an equilibrium expression. Our strategy will be starting concentrations ( )table equilibrium concentrations ( exp )Keq ression Keq value Example: When 1.0 mol of N2O4(g) was introduced into a 1.5 L bulb, only 0.80 mol remained after equilibrium was reached. The reaction is described by N2O4(g) 2 NO2(g). What is Keq for this reaction? Molarity N2O4(g) 2 NO2(g)

Start = Initially I 1.0 mol / 1.5 L = 0.67 M 0 M ∆ = Change C -0.14 M +0.28 Equilibrium E 0.80 mol / 1.5 L = 0.53 M +0.28

Keq = [NO2(g)]2 = (0.28)2 = 0.15 [N2O4(g)] (0.53)


19 Try This: A 10.0 L bulb is filled with 4.0 mol of SO2(g), 2.2 mol of O2(g) and 5.6 mol of SO3(g). The gases then reached equilibrium according to the equation 2 SO2(g) + O2(g) 2 SO3(g). At equilibrium, the bulb contained only 2.6 mol of SO2(g). Calculate Keq for this reaction. 7. Manipulating Equilibrium Systems Keq values are constant only for closed systems (constant mass) at fixed temperatures. If the temperature of the system or the concentration of the chemicals is disturbed, the equilibrium is upset and will have to react to the change. Le Chatelier’s Principle – When a chemical system at equilibrium is disturbed by a change in temperature or concentration (including pressure of gases), the system adjusts in a way to minimize the change. Note that catalysts do not affect the final position of the equilibrium: they can only make it occur faster.

Variable Type of Change Response of System Increase Shifts to consume some added material. Concentration Decrease Shifts to replace some removed material. Increase Shifts to consume some added heat. Temperature Decrease Shifts to replace some removed heat. Increase Shifts to reduce pressure (less moles of gas) Pressure (gases) Decrease Shifts to add pressure (more moles of gas)

We predict the effect of these disturbances by doing a “stress analysis”. Example: What is the effect of removing some NO(g) from this system: 2 NO2(g) 2 NO(g) + O2(g) + heat

1. Disturbance Reduce concentration 2 NO2(g) 2 NO(g) + O2(g) + heat

2. Shift → 3. Result ↓ ↑ ↑ ↑

Now identify the effects of (a) decreasing the container size (b) cooling the system.


20 Acids and Bases

1. Acids and Bases 1 a) Properties We can classify aqueous solutions as being an acidic, basic or a salt solution.

1. An acid produces H+ ions when dissolved in water. The presence of the H+ ions gives the substance its acidic properties. HNO3(aq) → H+

(aq) + NO3-(aq)

2. A base produces OH- ions when dissolved in water: Ca(OH)2(aq) → Ca2+

(aq) + 2 OH-(aq)

3. A salt is composed of a positive ion and a negative ion (not H+ or OH-) when dissolved in water. NaCl(aq) → Na+

(aq) + Cl-(aq) These categories are based on the neutralization reaction: Acid + base → salt + water Molecular: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Total Ionic Equation: H+

(aq) + Cl-(aq) + Na+(aq) + OH-

(aq) → Na+(aq) + Cl-(aq) + H2O(l)

Net Ionic Equation: H+(aq) + OH-

(aq) → H2O(l) These solutions can be identified by their properties:

Properties Acids Bases Salts Taste Sour Bitter Feel Soapy Electrolytic Yes Yes Yes Composition H+ OH- No H+ or OH- Identifying reaction

Acid + metal → H2

Neutralizes acids

Litmus Red Blue No effect Phenolphthalein Colorless Pink No effect 1 b) Water Equilibrium Water molecules are highly polar. Sometimes the collisions between two water molecules are energetic enough that a hydrogen ion is transferred from one molecule to another. This is called the self-ionization of water. + + Water molecules hydronium ion hydroxide ion or

H2O(l) H+(aq) + OH-

(aq) hydrogen ion hydroxide ion

O H

H

O H

H

O H

H

H +

O H -


21 Note:

1. A hydrogen ion (H+) is always joined to a water molecule to form a hydronium ion (H3O+). Therefore, either formula will actually represent the hydrogen ion in aqueous solution.

2. The hydrogen ion is also referred to as a proton. This self-ionization process is at equilibrium. The ionization constant for water, Kw, quantifies this equilibrium:

Kw = [H+][OH-] = 1.0 x 10-14 For pure water at SATP Since [H+] = [OH-] Then [H+] = 1.0 x 10-7 M And [OH-] = 1.0 x 10-7 M

Note that 1.0 x 10-7 = 10-7

2. Arrhenius Acids - An acid was originally defined as a substance that ionizes in water to produce hydrogen ions: HCl(g) H+

(aq) + Cl-(aq) - The additional H+ provided by the acid increases the [H+] in water. Thus, [H+] > 10-7 and the [H+] > [OH-] and the solution is considered to be acidic. Note:

(i) Polyprotic acid contributes a greater proportion of H+ ions. H2SO4(l) 2 H+

(aq) + SO42-

(aq) (ii) A basic solution has [OH-] > 10-7 and [OH-] > [H+]. They are also known as alkaline

solutions. - This is an equilibrium situation. If [H+] is increased in water, then [OH-] must decrease so that the constant Kw remains at 10-14 (Le Châtelier’s Principle). - Since Kw applies to all aqueous solutions, we can use it to find unknown concentrations:

[H+] [OH-] = Kw = 10-14 Example: If [H+] = 1.0 x 10-6 M, then find [OH-]. [OH-] = Kw / [H+] = 1.0 x 10-14 / 1.0 x 10-6 = 1.0 x 10-8 M (acidic) Try This: Complete the table. Show work!

[H+] [OH-] Type of solution

1 1.0 x 10-3 2 1.0 x 10-3 3 1.0 x 10-7 4 2.3 x 10-9 5 6.4 x 10-6 6 1.7 x 10-5


22 3. pH A Math Review Logarithms are used to find exponents: 32 = 9 or log3 9 = 2 exponent Although logarithms can be used with any number, the most useful one is “log 10”. This is called the common log and is usually just written as “log”. log 7 = log10 7 Only common logs can be found on your calculator: log antilog 0.845 The hydrogen ion concentration of a solution is more easily expressed as a pH value. This is the negative logarithm of the H+ concentration.

pH = - log [H+] In any sample of pure water at 25oC, [H+] = 1.0 x 10-7 M Therefore, pH = - log [H+] = - log (l.0 x 10-7) = -(-7) = 7 This pH = 7 value is the starting point on the pH scale for a solution that is neither acidic nor basic (neutral). As the [H+] increases, the neutral solution becomes more acidic. The pH values move from 7 towards 0. If the [OH-] increases, however, the neutral solution becomes more basic and pH values move from 7 towards 14.

The pH Scale 0 7 14 increasing acidity increasing basicity The definitions of pOH follow the same patterns: pOH = - log [OH-] Finally, it is possible to prove that pH + pOH = 14 (at SATP) Note:

(i) For numbers with two sig. fig., the fractional part of the logarithm of the number should also contain two sig. fig.

If [H+] = 4.7 x 10-11, then pH = 10.33 (rounded to 2 sig. fig.) (ii) You may prefer to remember the definition of pH as: [H+] = 10-pH

4. Strengths of Acids 4 a) Percent Ionization All acids share the same properties but not to the same degree: 0.10 M HCl has a pH = 1.00 but 0.10 M HC2H3O2 has a pH = 2.89 Strong acids ionize completely (100%) in aqueous solution. HCl(aq) H+

(aq) + Cl-(aq) 1 M 1 M 1 M

7 7


23 Weak acids ionize incompletely in water. For example, only 1% of acetic acid ionizes when dissolved. HC2H3O2(aq) H+

(aq) + C2H3O2-(aq)

1 M 0.01 M 0.01 M Monoprotic acid – contains one ionisable hydrogen HCl(aq) + HOH(l) H3O+

(aq) + Cl-(aq) Polyprotic acid- Diprotic acid – contains 2 ionizable hydrogens H2SO4(aq) + HOH(l) H3O+

(aq) + HSO4-(aq)

HSO4-(aq) + HOH(l) H3O+

(aq) + SO42-

(aq) H2SO4(aq) + 2 HOH(l) 2 H3O+

(aq) + SO42-

(aq) Triprotic acid – contains 3 ionizable hydrogens H3PO4(aq) + HOH(l) H3O+

(aq) + H2PO4-(aq)

H2PO4-(aq) + HOH(l) H3O+

(aq) + HPO42-

(aq) HPO4

2-(aq) + HOH(l) H3O+

(aq) + PO43-

(aq) H3PO4(aq) + 3 HOH(l) 3 H3O+

(aq) + PO43-

(aq) Weak acids, then, have weaker properties because they make fewer H+ ions. Most acids are weak. Note that only the weak acids (which ionize less than 100%) will establish an equilibrium. Strong acids go to completion. Acids have been ranked according to their strengths by comparing the pH values of 0.10 M solutions of the acids (see handout). These values are only valid for the concentration and temperature as stated (Le Châtelier’s of course) Let (i) p = percent ionization (%), (ii) [H+] = concentration of H+ ions at equilibrium, (iii) M(acid) = measured (starting) concentration of the acid. P = [H+] / M(acid) x 100 Note that polyprotic acids ionize successively with each step having a unique % ionization number. We will work only with monoprotic acids. H2SO4(aq) H+

(aq) + HSO4-(aq) (100%)

HSO4-(aq) H+

(aq) + SO42-

(aq) (27%) Try This:

1. Find the percent ionization of a 0.100 M solution of formic acid if its [H+] = 4.22 x 10-3 M. 2. The pH of a 0.10 M methanoic acid solution is 2.15. What percent of this acid ionizes? 3. What is the concentration of an acid that ionizes 15% and has a [H+] = 0.052 M?

4 b) Ionization Constant (Ka) Weak acids, which only partially ionize in water are said to have a percent ionization < 100%. Percent ionization numbers, though, have a limited usefulness because they vary with the temperature and the concentration of the solution. To calculate the amount of ionization over a range of concentrations (but at a fixed temperature) it is more useful to use the equilibrium constant:


24 Weak acid aqueous equilibrium concentrations are constant described by Keq For any weak acid HX(aq) H+

(aq) + X-(aq)

Therefore Ka = [H+][X-] / [HX] = Ka (acid ionization constant). Ka reflects the amount of the acid that is in ion form. Higher Ka more product more H+ ions stronger acid.

Note (i) Most Ka values have been determined at SATP (see handout). (ii) For bases, if XOH(aq) X+

(aq) + OH-(aq)

Therefore Kb = [X+][OH-] / [XOH] = Kb (base ionization constant). The concentration (Molarity) of an acid is a measured (initial) quantity (think table!) and cannot be used to calculate Ka values. However, if we can prove that only a few molecules ionize (1:1000), then M(init.) = M(equilib.) Try This: 1. Find Ka for these acids:

a) Acetylsalicyclic acid (ASA) is a weak, monoprotic acid that we can write as HAsa. A 0.100 M solution of the acid has a [H+] = 0.0058 M.

b) For a 0.100 M aqueous solution of butanoic acid (HBut), the [H+] = 1.24 x 10-3 M

2. Find [H+] for 0.100 M solutions of these acids: a) Ascorbic acid (Vitamin C) has a Ka = 8.0 x 10-5 b) Hydrocyanic acid (HCN) has a Ka = 6.20 x 10-10

3. Find the pH of a 0.200 M solution of methanoic acid (Ka = 1.78 x 10-4) 4. The pH of a 0.16 M aqueous solution of Benzoic acid is 2.50. Find the Ka for this acid.

5. Bronsted-Lowry Acids

5 a) Hydrogen Ion Exchanges Acid-base reaction can be described more accurately as hydrogen ion exchange reactions: Acid = hydrogen ion (proton) donor Base = hydrogen ion acceptor HCl(aq) + H2O(l) H3O+

(aq) + Cl-(aq) where HCl = acid (donor) H2O = bases (acceptor) Note that an acid must be in the presence of a base willing to accept the hydrogen ion: A base is any chemical capable of bonding with the hydrogen ion: NH3(aq) + H2O(l) NH4

+(aq) + OH-

(aq) (where the OH- makes the solution basic.) When an acid loses a hydrogen ion it becomes a base; when the base accepts a hydrogen ion, it becomes an acid.

HNO3(aq) + H2O(l) H3O+(aq) + NO3

-(aq)

Acid 1 Base 1 Acid 2 Base 2


25

Therefore, ACID + BASE CONJUGATED ACID + CONJUGATE BASE A conjugated acid-base pair is two substances that are related by the loss/gain of a single hydrogen ion. base conjugate acid

NH3(aq) + H2O(l) NH4

+(aq) + OH-

(aq)

Acid conjugate base Some substances, such as water, are amphoteric – they can act as an acid or as a base. NH3(aq) + H2O(l) NH4

+(aq) + OH-

(aq) (where H2O is the acid) HCl(aq) + H2O(l) H3O+

(aq) + Cl-(aq) (where H2O is the base) The stronger the acid, the weaker will be its conjugate base, and vice-versa. 5 b) Colour Indicators Substances that change colour when reacted with acids and bases are called indicators. They are usually organic compounds that establish an equilibrium between their molecular form (colour 1) and their ionic form (colour 2) in solution:

Hin(aq) H+(aq) + In-

(aq) colour 1 colour 2

increase [H+] (add acid) decrease [H+] (add base) The pH at which the colour change occurs is called the end point. It refers to the colour transition when the two colours are present in equal amounts.

[Hin] = [In-] Indicators are useful only in colourless solutions and over a narrow pH range. Therefore, they must be chosen to match the pH of the solution being formed (see handout). Example: phenolphthalein (HPh) 8.2 10.0 colourless (HPh) pink (Ph-) Try This:

1. Bromothymol blue (HBb) forms the anion Bb-. a) What colour is it?

b) At what pH range will it start to appear? 2. Two drops of chlorophenol red is added to a 0.001 M HBr(aq) solution. What colour will it

be? 3. A solution is tested: methyl red = yellow and bromothymol blue = yellow. What is the

concentration of the sample? 6. Neutralizations


26 6 a) Reactions Neutralization is a double replacement reaction: Acid(aq) + base(aq) salt(aq) + water(l) Assuming the acid = strong (100% ionized) base = strong (100% ionized) salt = exists only as ions. Example: HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) H+

(aq) + Cl-(aq) + Na+(aq) + OH-

(aq) Na+(aq) + Cl-(aq) + H2O(l)

H+(aq) + OH-

(aq) H2O(l) Essentially neutralization is the production of water. If a weak acid/base is used (< 100% ionized), neutralization will still go to completion even though the weak acid/base exists in equilibrium. Let’s analyze a weak acid/strong base situation:

1. Stress scarce acid ions removed (during neutralization) Acid molecules acid ions

2. Shift 3. Result

As the acid ions continue to be used up the stress continues. The weak acid eventually will ionize and react completely. (It will react like a strong acid). At least one of the reactants must be strong in a neutralization reaction. Then we will assume there is enough stress placed on the equilibrium of the weak chemical for the reaction to go to completion. Polyprotic acid will donate all of their acidic H+ ions for the same reason. H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + 2 H2O(l) Try This:

1. Write a balanced molecular equation for the reaction of these chemicals: a) HNO2 + NaOH b) HCl + Ba(OH)2 c) H2SO4 + KOH

2. How many moles of NaOH is needed to neutralize 0.50 mol of: a) HNO3 b) H2CO3 c) H3PO4

6 b) Titrations A titration is a technique for finding an unknown concentration of one chemical from the known concentration of a second chemical. Standard stoichiometry steps can now be used to calculate the unknown Ma.

Stop at Unknown Chemical A Known Va Unknown Ma

Known Chemical B Known Vb Known Mb

End point (colour change)

Add


27 Note: (i) M = n/V

(ii) This will be out three-step stoichiometry solution: 1. Va x Ma → mol A 2. mol A → mol B 3. mol B → Mb (n = MV) (mole ratio) (M = n/V) (iii) At least one reactant must be strong. (iv) Mixture problems (known acid + known base) will have a reactant in excess- either [H+] or [OH-]. Remember that their concentrations can cancel (neutralize) each other. (v) Some problems involve findings % by mass from known concentrations: M (mol/L) g / L % (g / 100 g) Molar mass divide by 100 (assuming 1 L of solution = 1000 g)

Try This:

1. 25 mL of HNO3 is neutralized by 18 mL of 1.2 M NaOH. Find the concentration of the HNO3 knowing HNO3 + NaOH → NaNO3 + H2O

2. How many L of 0.050 M Ca(OH)2 will be neutralized by 125 mL of 2.0 M HCl? 3. Suppose 50.0 mL of 1.60 M NaOH exactly neutralized 100.0 mL of vinegar, which is an

acetic acid (HC2H3O2) solution. Calculate a) the molarity of the acid in the vinegar. b) Mass of the acid in the vinegar. c) % by mass of acetic acid in vinegar (assume 1 L = 1000 g)

4. Calculate the pH of a solution that was made from mixing 125.4 mL of 0.24 M HCl with 97.7 mL of 0.12 M KOH.


28 Electrochemistry

1. Tracking Electrons 1 a) Electrons in Formulas - All atoms are electrically neutral (total p+ charge + total e- charge = 0) - Metals atoms tend to lose electrons easily. They then form a positively charged ion, called a cation: Ca → Ca2+ + 2 e-. This is an oxidation reaction. The electrical state of the atom has increased from 0 → +2.

Family Number No. of e- lost Electrical State

IA 1 +1 IIA 2 +2

METALS IIIA 3 +3

- Non-metals atoms tend to gain electrons easily. They then form a negatively charged ion, called an anion: O + 2 e- → O2-. This is called a reduction reaction. The electrical state of the particle has been reduced from 0 → -2.

Family Number No. of e- gain Electrical State VA 3 -3 VIA 2 -2

NON-METALS VIIA 1 -1

Remember: ‘LEO the lion says GER’ - Molecular compounds also show electron changes. Each atom has a different attraction for electrons that is rated in an electronegativity (En) scale (see handout). As the atoms combine, they compete for the bonding electrons. The electrons will be shifted toward one atom and away from the other atom. The more electronegative atom (with a higher En value) becomes more negative and the atom with the lower En value becomes more positive. Example: A water molecule O H H has an En = 2.1

H O has an En = 3.5 Note the electron shift away from H and towards O.

The oxygen has been reduced and the hydrogen has been oxidized. Try This: Locate the elements oxidized and reduced. a) MgO b) NH3 c) O2 d) CaF2 e) CCl4 f) OCl2 g) Na2O 1 b) Electrons in Equations Consider these two reactions: Double Replacement Molecular Equation: BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2 NaCl(aq) Net Ionic Equation: Ba2+

(aq) + SO42-

(aq) → BaSO4(s) Change in Charges: Ba2+ → Ba2+

SO4

2- → SO42- (no change in charges)


29

Single Replacement Molecular Equation: Cu(s) + 2 AgNO3(aq) → Cu(NO3)2(aq) + 2 Ag(s) Net Ionic Equation: Cu(s) + 2 Ag+

(aq) → Cu2+(aq) + 2 Ag(s)

Change in Charges: Cu → Cu2+

2 Ag+ → 2 Ag (charges have changed)

- Composition, decomposition, and single replacement reaction involve electron changes and are called an oxidation-reduction reaction (or a redox reaction). Redox reactions are always coupled – oxidation and reduction must occur together. Electrons lost must equal to electrons gained. - Nearly every reaction (except double replacements) is a redox reaction. Even non-ionic reactions involve changes to electrons. These changes will involve electron shifts, though, rather than electron exchanges. Example: 2 H2 + O2 → 2 H2O Atom Initial Charge Final Charge Type of change Hydrogen 0 2(+1) Oxidized Oxygen 0 -2 Reduced Try This: Identify which reactant is oxidized and which is reduced in the following reactions.

a) H2 + Cl2 → 2 HCl b) 2 Na + S → Na2S c) N2 + 3 H2 → 2 NH3 d) 2 K + Br2 → 2 KBr e) S + O2 → SO2 f) 4 Al + 3 O2 → 2 Al2O3

2. Oxidation Numbers 2 a) Determining Oxidation Numbers - A precise way of describing oxidation and reduction is with oxidation number (or oxidation state). It is a number assigned to an atom according to a set of arbitrary rules. It is the charge an atom would have if all of the atoms in a molecule were 100% ionic. - The calculation of an unknown oxidation number is routine algebra:

Assume that - elemental atoms = 0 - monoatomic ions = ionic charge - oxygen = -2 - hydrogen = +1 If necessary, then assume (in order) - Family I atoms = +1 - Family II atoms = +2 - Family VII atoms = -1


30 The oxidation number of any one other atom can be calculated. Rule: The sum of the positive changes plus the negative changes must equal the total charge of the molecule (zero) or of the Polyatomic ion. Note:

(i) Oxidation numbers may be fraction or even zero. (ii) Oxidation numbers are always per atom. They will be recorded above the atom in the

formula. (iii) The oxidation number of an element may vary form molecule to molecule.

Try This: Find the oxidation number of each element in the following formulas:

1. H2SO4 2. C2H8 3. HCO3

- 4. KClO3

2 b) Using Oxidation Numbers - A reaction in which the oxidation number of any two elements changes is a redox reaction. - The reactant that contains the atom being oxidized is the reducing agent. The reactant that contains the atom being reduced is the oxidizing agent.

Reduction has occurred in an element when - its oxidation number has been algebraically lowered. - its electrical state has been decreased. - it has acted as an oxidizing agent. Oxidation has simultaneously occurred in an element when - its oxidation number has been algebraically raised. - its electrical state has been increased. - it has acted as an reducing agent.

Example: In the following reaction, identify which elements are oxidized and reduced, and then identify the two agents:

0 +1 -1 +2 -1 0 Mg + 2 HCl → MgCl2 + H2

Mg: oxidation number has increased (0 → +2) so it has been oxidized. Magnesium metal has acted as the reducing agent. H: oxidation number has decreased (+1 → 0) so it has been reduced. Hydrochloric acid has acted as the oxidizing agent. Note that the coefficients must be considered in the calculation of oxidation numbers in redox equations. Try This: Use oxidation numbers to determine if the reaction is a redox one. If so, identify the chemical units acting as oxidizing and reducing agents.

1. Ca(NO3)2(aq) + 2 Na(s) → 2 NaNO3(aq) + Ca(s) 2. Cl2(g) + 2 HBr(aq) → 2 HCl(aq) + Br2(s) 3. C(s) + O2(g) → CO2(g) 4. BaCl2(aq) + Na2SO4(aq) → 2 NaCl(aq) + BaSO4(s) 5. C(s) + H2O(g) → CO(g) + H2(s) 6. 2 HNO3(aq) + 6 HI(g) → 2 NO(g) + 3 I2(s) + 4 H2O(l) 7. CO(g) + PbO(s) → CO2(g) + Pb(s) 8. Br2(aq) + SO2(g) + 2 H2O(l) → 2 HBr(aq) + H2SO4(aq)


31 3. Half-Reactions 3 a) Writing Half-Reactions - Redox reaction can be separated into two half-reactions. These reactions are written to include the electrons produced during oxidation (added to the product side) and the electrons reacted during reduction (added to the reactant side). Complete Method: Step 1. Write the reaction in ionic form. This mainly affects aqueous solutions. Step 2. Assign oxidation number (ON). Step 3 Write separate half-reactions for the oxidation and reductions processes. Include the electrons necessary to account of the changes in ON. Example: S(s) + HNO3(aq) → SO2(g) + NO(g) + H2O(l) Step 2: 0 +1 +5-2 +4-2 +2-2 +1-2 Step 1: S+ H+ + NO3

- → SO2 + NO + H2O Step 3: Oxidation: S(s) → SO2(g) + 4 e- Reduction: NO3

-(aq) + 3 e- → NO(g)

Note: (i) The remainder of the species will be treated as spectator ions. (ii) Include state symbols in the half-reactions.

- Half-reactions are especially useful for analysing ionic reactions in solutions. It shows which species are actively involved in the redox process. Try This: Write half reactions for the following unbalanced redox reactions:

a) H2(g) + S(s) → H2S(g) b) NO2(g) → O2(g) + N2(g) c) CuO(s) + H2(g) → Cu(s) + H2O(l) d) HNO3(aq) + H3PO3(aq) → NO(g) + H3PO4(g) + H2O(l) e) HNO3(aq) + I2(s) → HIO3(aq) + NO2(g) + H2O(l) f) Cl2(g) + H2O(l) → HCl(aq) + HClO(aq) g) PbO2(aq) + HI(aq) → I2(g) + PbI2(aq) + H2O(l)

3 b) Using Half-Reactions - Let’s analyse this single-replacement reaction between a metal and a solution: Molecular equation: Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s) Net ionic equation: Zn(s) + Cu2+

(aq) → Zn2+(aq) + Cu(s)

This is a spontaneous redox reaction: Oxidation reaction: Zn(s) → Zn2+

(aq) + 2 e- (the metal is ionized) Reduction reaction: Cu2+

(aq) + 2 e- → Cu(s) (the solute is precipitated) If copper is placed in a ZnSO4(aq), no reaction will occur (the reaction is non-spontaneous). Cu(s) does not react with Zn2+

(aq) ions. Substances vary in their ability to lose and gain electrons. Therefore, they have been organized into an activity series (see handout). The table ranks substances in decreasing order of their ability to gain electrons (to reduce). Each of these reduction equations must be paired with an oxidation equation found by reading the equations in reverse (from right to left).


32 Spontaneity Rule: In a reduction table, a reaction will only occur if the equilibrium favors products. If RR then the reaction is spontaneous. OR Where RR = reduction reaction (aq → s) and OR = oxidation reaction (aq ← s) Try This:

1. Rank these four substances in decreasing order in a reduction table (by their attraction for electrons) given the following:

a) R2+(aq) + X(s) → X2+

(aq) + R(s) b) Q2+

(aq) + X(s) → no reaction c) X2+

(aq) + M(s) → M2+(aq) + X(s)

d) Q2+(aq) + M(s) → no reaction

2. Determine which of the following reaction will be spontaneous. Then write the oxidation and reduction reactions for each redox reaction.

a) solid iron in lead(II) nitrate solution b) a copper wire in zinc sulfate solution c) calcium wire in zinc sulfate solution d) a magnesium strip in iron(II) nitrate solution

4. Balancing Redox Equations 4 a) Oxidation Number Method - A balanced redox equation means (i) mass and (ii) charge is conserved. - The oxidation number method is used for balancing complex redox reactions. It involves balancing the changes in oxidation number (∆ON) that occur during every oxidation-reduction reaction. This method works well for molecular equations or when a complete ionic equation is needed. RULE: Increase in ON = Decrease in ON. - Oxidation Number Method

1. Start with a complete but unbalanced equation. Assign oxidation numbers above the symbols.

2. Identify the atoms being oxidized and reduced. Record the number of electrons being transferred per particle below its symbol.

3. Below that (2nd line), record the number of electrons being transferred per molecule. 4. Below that (3rd line), balance the redox participants. Record the coefficients in your

equation. 5. Balance the other atoms by the inspection method. 6. Check particles and charge.

Example: Balance using the ON method: HNO3 + H2S → NO + S + H2O +1+5-2 +1–2 +2 -2 0 +1-2

HNO3 + H2S → NO + S + H2O +3e-/N -2e-/S +3e-/HNO3 -2e-/H2S (+3)x2 (-2)x3 +6 -6


33 (Redox Answer: charges are balanced) 2 HNO3 + 3 H2S → NO + S + H2O (Final Answer: mass is also balanced) 2 HNO3 + 3 H2S → 2 NO + 3 S + 4 H2O Check particles: 19=19; charges 0=0 Try This: Balance these reactions by the oxidation number method:

a) I2 + HNO3 → HIO3 + NO2 + H2O b) MnO4

- + H+ + Cl- → Mn2+ + Cl2 + H2O c) CrO2

- + ClO- + OH- → CrO42- + Cl- + H2O

d) MnO4- + H2SO3 + H+ → Mn2+ + HSO4

- + H2O e) HBr + HBrO3 → Br2 + H2O

4 b) Reduction Table Method (Half-Reactions) The half-reaction method of balancing redox equations involves splitting the reaction into its two half-reactions. This allows us to track actual electron movements. After balancing the electron movements, the two half-reaction are then combined back into an overall reaction. If the reduction table lists the half-reaction, we can solve using these steps:

1. Determine the half-reaction in the unbalanced equation. 2. Write the two half-reactions from the reduction table. 3. Balance the electron exchange. 4. Add the two half-reactions (like terms cancel) to get a net ionic equation. 5. Check atoms and charge.

Example: Balance this redox equation: Cr2O7

2-(aq) + Sn2+

(aq) → Sn4+(aq) + Cr3+

(aq) Step 1. Cr2O7

2- → Cr3+ Sn2+ → Sn4+ Step 2. Locate the two half reactions. Balance the electron exchange. Cr2O7

2-(aq) + 14 H+

(aq) + 6 e- 2 Cr3+(aq) + 7 H2O(l)… reduction

(Sn2+(aq) Sn4+

(aq) + 2 e-) x 3…oxidation) Step 3. Combine

Cr2O72-

(aq) + 14 H+(aq) + 6 e- 2 Cr3+

(aq) + 7 H2O(l) + 3 Sn2+

(aq) 3 Sn4+(aq) + 6 e-

Cr2O72-

(aq) + 14 H+(aq) + 3 Sn2+

(aq) 3 Sn4+(aq) + 2 Cr3+

(aq) + 7 H2O(l) Step 4. Atoms are balanced (26=26); charges are balanced (+18=+18) Try this: Balance these redox equations:

1. ClO4-(aq) + H2O2(l) → Cl-(aq) + O2(g)

2. Cr2O72-

(aq) + I-(aq) → I2(s) + Cr3+

(aq) 3. HNO2(aq) + Cl-(aq) → N2O(g) + ClO-

(aq) 4. MnO4

-(aq) + H2O2(l) → Mn2+

(aq) + O2(aq) 5. MnO4

-(aq) + Cu(s) → Cu2+

(aq) + Mn2+(aq)

6. NO3-(aq) + H2S(aq) → NO2(g) + S(s)

4 c) Acidic Mediums Many redox reactions occur in aqueous solutions or suspensions. In an acidic medium many of the reactants and products exist as ions and the reaction is affected by the pH of the medium. This means that H2O molecules and H+ ions are available to interact with the dissolved chemicals.


34 Method

1. Write the equation in total ionic form, if necessary. 2. Write the half-reaction for mass only. 3. Balance (i) the atoms in this order – most atoms, then O by adding H2O,

then H by adding H+ ions. (ii) the charges by introducing electrons where necessary. (iii) the electron exchanges by introducing coefficients.

4. Add the half-reactions (electrons must cancel). 5. Answer is a net ionic equation. Check particles and charges.

Example: Balance the redox equation for adding copper metal to nitric acid (an acidic medium): Cu(s) + HNO3(aq) → Cu(NO3)2(aq) + NO2(g) 0 +1 +5-2 +2 +5-2 +4-2 Step 1 Cu + H+ + NO3

- → Cu2+ + 2 NO3- + NO2 total ionic equation

Step 2 Cu → Cu2+ half-reactions NO3

- → NO2 balance - mass Step 3 (Cu → Cu2+ + 2 e-) x 1 - charge (NO3

- + 2 H+ + e- → NO2 + H2O) x 2 - electrons Step 4 Cu → Cu2+ + 2 e- balance half-reactions + 2 NO3

- + 4 H+ + 2 e- → 2 NO2 + 2 H2O Step 5 Cu(s) + 2 NO3

-(aq) + 4 H+

(aq) → Cu2+(aq) + 2 NO2(g) + 2 H2O(l) (answer)

Try This: Balance these reactions that are occurring in acidic mediums:

1. MnO4-(aq) + Cu(s) → Cu2+

(aq) + Mn2+(aq)

2. BrO3-(aq) + SO2(g) → SO4

2-(aq) + Br2(l)

3. Co2+(aq) + H2SO3(aq) → Co3+

(aq) + S2O32-

(aq) 4. H2C2O4(aq) + MnO4

-(aq) → Mn2+

(aq) + CO2(g) 5. Cr2O7

2-(aq) + S(s) → SO2(g) + Cr2O3(s)

4 d) Basic Mediums When a redox reaction occurs in a basic medium, H2O and OH- particles will be introduced into the reaction to help balance masses. The method remains the same except that when mass is balanced

- balance all atoms except O and H - balance each O by adding H2O

(as in an acidic solution.) - balance each H by adding a H2O on the side that it is needed and an equal number of

OH- on the other side (H2O – OH- = H+ !) Example: Consider the oxidation of the oxalate ion to the carbonate ion by the permanganate ion, which in turn is reduced to manganese (IV) oxide. The reaction takes place in basic solution. +3-2 +7 –2 +4-2 +4-2

C2O42-

(aq) + MnO4-(aq) → MnO2(s) + CO3

2-(aq)

C2O42- → CO3

2- MnO4

- → MnO2 (C2O4

2- + 2 H2O + 4 OH- → 2 CO32- + 4 H2O + 2 e-) x 3

(MnO4- + 4 H2O + 3 e- → MnO2 + 2 H2O + 4 OH-) x 2


35 3 C2O4

2- + 6 H2O + 12 OH- → 6 CO32- + 12 H2O + 6 e-

2 MnO4- + 8 H2O + 6 e- → 2 MnO2 + 4 H2O + 8 OH-

3 C2O42-

(aq) + 4 OH-(aq) + 2 MnO4

-(aq) → 6 CO3

2-(aq) + 2 MnO2(s) + 2 H2O(l)

(36 = 36, -12 = -12) Try This:

1. Balance the following equation in a basic solution: a) Cr2O7

2-(aq) + AsO2

-(aq) → Cr3+

(aq) + AsO43-

(aq) b) P4(s) → H2PO2

-(aq) + P2H4(g)

c) CrO42-

(aq) + Sn2+(aq) → Sn4+

(aq) + CrO2-(aq)

d) BrO3-(aq) + SO2(aq) → Br2(l) + SO4

2-(aq)

2. Balance this reaction in acidic and in a basic solution: Ni(s) + NO3

-(aq) → NH4

+(aq) + Ni2+

(aq) 5. Electrochemical Processes All electrochemical processes involve redox reactions. Cells Chemical Energy → Electrical Energy (spontaneous) Electrolysis Electrical Energy → Chemical Energy (non-spontaneous) We know that oxidation and reduction occurs in the interaction of a metal with aqueous ions.

Oxidation: Cu(s) → Cu2+(aq) + 2 e-

Reduction: Cu2+(aq) + 2 e- → Cu(s)

If the reactants are in contact, the energy released during the electron transfer is in the form of heat. If redox is to be used as a source of electrical energy the two half-reactions must be physically separated. An electric current (a flow of electrons) will then occur in a wire that connects the two electrodes (metal strips) in the half-cells. Electrical Quantities:

Quantity Symbol Meter Unit Analogy Current I Ammeter Ampere (A) Amount of water in a pipe Potential Difference V Voltmeter Volt (V) Force of water in a pipe 6. Cells a) Voltaic Cells - These are electrochemical cells that convert chemical energy into electrical energy by a spontaneous redox reaction. It contains separate half-cells for the two half-reactions. - The two half-cells are separated with a porous partition or a salt bridge. They prevent the solutions from mixing but allow for the passage of ions.


36

Half-cell Type of metal

Electrode Charge Electron activity

Ion activity

oxidation - All parts of the cell remain balanced in charge at all times. We can show this by using a standard notation for cells.

Electrons

anode(-) │electrolyte ║electrolyte │ cathode(+) (oxidation) (reduction)

anions cations 6 b) Dry Cells

- This “dry-cell” battery is made up of a metal electrode or graphite rod surrounded by a moist electrolytic paste enclosed in a metal cylinder. - Cathode = positive electrode = surface in which reduction occurs. Sometimes the reduction occurs in the paste if it is a strong oxidizing agent. Then the cathode will be an inert material (such as graphite) to provide a location to complete the circuit and a surface on which the half-reaction occurs. - The thin zinc cylinder is the anode that undergoes oxidation at its surface. Zn(s) → Zn2+ In an acidic dry cell, reduction occurs in the paste (NH4Cl + MnO2):

NH4+ + MnO2 → Mn2O3 + NH3

Try This: 1. Did you notice that the half-reaction are not balanced?

a) Balance each half-reaction. b) Find the overall net reaction.

(-)

Anode (usually zinc metal)

(+)

Cathode

Paper or cardboard Salt bridge

Moist Electrolyte paste


37 2. An alkaline battery replaces NH4Cl with a strong hydroxide. They last longer because the

zinc corrodes less rapidly under basic conditions. Balance these half-reactions and find the overall net reaction of a) + b)

a) Zn → ZnO b) MnO2 → Mn2O3

6 c) Batteries - A battery is a group of cells that that are connected together. The lead storage battery consists of six voltaic cells connected together.

- Each cell contains two lead electrodes or grids immersed in sulphuric acid (5M) and separated by a steel plate. The anode is packed with lead and the cathode via lead(IV) oxide. - Oxidation: Pb(s) + SO4

2- → PbSO4(s) - Reduction: PbO2(s) + SO4

2- → PbSO4(s)

- Lead(II) sulphate and lead(IV) oxide are insoluble in sulphuric acid so a salt bridge is unnecessary. - During discharge, lead sulphate builds up on the plate, and the concentration of the sulphuric acid decreases. When it is recharged, the reaction is reversed and the reactants are restored. This reaction is non-spontaneous, though, and a current must be passed through the cell in the opposite direction. Try This:

1. a) Balance the two half-reactions. b) Find the overall equation.

6 d) Fuel Cells


38 - Fuel cells are voltaic cells in which a fuel substance undergoes oxidation and from which electrical energy is obtained continuously. Its wasted material is water. - The hydrogen-oxygen fuel cell has three compartments separated from one another by two carbon electrodes. Oxygen (the oxidizer) is fed into the cathode compartment. Hydrogen (the fuel) is fed into the anode compartment. The central compartment contains a hot, concentrated hydroxide. - The hydrogen gas is oxidized at the anode and the electrons are passed through a circuit before entering the cathode compartment to be reduced.

Oxidation: H2(s) → H2O(l) Reduction: O2(g) → OH-

(aq) Try This: You probably noticed the redox equations were not balanced.

a) Balance each half-reaction. b) Find the net reaction.

7. Half-Cells 7 a) Standard Reduction Potential (Eo Cell) - Half-cells may be described by their ability to acquire a flow of electrons. The tendency of electrons to flow in an electrochemical reaction:

- is called the voltage. - is called the electrical pressure. - is called the electrical potential to do work. - is given the symbol E.

- The voltage of a half-cell cannot be measured, only the difference in voltage between two half-cells. To establish a definite scale, we define a “zero-point” voltage:

2 H+(aq) + 2 e- → H2(g) Eo = 0.00 V

Note: (i) Eo = standard reduction potential (ii) Standard conditions = 25oC

= freshly made 1 M solutions = 1 atm pressure

(iii) Any half-cell can now be connected with the standard half-cell and the voltage produced recorded in a table (see Redox half-reactions sheet).

- Reduction potential of a half-cell is a measure of its tendency to occur as reduction. The higher numbers have greater ability to gather electrons.

Consider a zinc and copper half-cell: Zn2+ + 2 e- → Zn Eo = -0.76 V Cu2+ + 2 e- → Cu Eo = 0.34 V

- Copper ions have a greater tendency to acquire electrons than zinc ions. Copper ions will reduce and the zinc will be forced to give up electrons. It will oxidize. Zn → Zn2+ + 2 e-

Cu2+ + 2 e- → Cu Zn + Cu2+ → Zn2+ + Cu

Note: (i) The half-cell with the smaller tendency to reduce will be forced to become the

oxidation half-cell. The greater the difference in the half-cell potentials, the greater the reaction tendency.


39 7 b) Calculating Eo

cell - A table of reduction potentials enables us to predict direction, maximum voltage and products of an electrolytic cell. This series is a list arranged in order of ability to attract electrons in competition with other oxidizing agents. - Electron Flow Review

From To Electron donor Electron acceptor Oxidation half-reaction Reduction half-reaction Reducing agent Oxidizing agent Anode Cathode

- Standard cell potential (Eocell) is the sum of the voltages produced in the two half-cells, where

the ions are at SATP and are 1.00 M. Example: For the cell Fe | Fe2+ || Br2 | Br- find:

a) the two half-cell reactions. b) the cell reaction and its standard cell potential.

a) Fe2+ + 2 e- Fe Eo = -0.44 V Br2 + 2 e- 2 Br- Eo = +1.07 V

b) Br2 + 2 e- 2 Br- Eo = +1.07 V Fe Fe2+ + 2 e- Eo = +0.44 V Br2 + Fe 2 Br- + Fe2+ Eo

cell = 1.51 V Note:

(i) Electrons must cancel when the half-reactions are combined. (ii) Eo of a half-cell is not changed if the equation must be multiplied by a coefficient. Eo

depends only on molar concentration, not moles. Try This: Find the cell reaction and Eo

cell for the following: a) Zn | Zn2+ || Fe2+

| Fe b) H2 | H+ || Br2 | Br- c) Cu | Cu2+ || Ag+

| Ag d) Mg | Mg2+ || Br2 | Br- e) Ni | Ni2+ || Hg2+

| Hg f) Pb | Pb2+ || Cl2 | Cl

- If the final redox reaction, as written, ends up with a positive cell potential (Eo

cell > 0) then this reaction is spontaneous. A negative cell potential means the reaction is non-spontaneous. A non-spontaneous reaction will be spontaneous in the reverse direction and the Eo

cell value will become equally positive. Example: Is the following redox reaction spontaneous as written?

Ni(s) + Fe2+(aq) → Ni2+

(aq) + Fe(s) Ni Ni2+ + 2 e- Eo = +0.26 V Fe2+ + 2 e- Fe Eo = -0.45 V

Ni + Fe2+ Ni2+ + Fe Eocell = -0.19 V

This reaction is non-spontaneous. Energy must be supplied to make it occur.


40 Try This: Determine if the following cells will be spontaneous:

a) Co2+(aq) + Fe(s) → Fe2+

(aq) + Co(s) b) Cu(s) + 2 H+

(aq) → Cu2+(aq) + H2(g)

c) 2 Ag(s) + Fe2+(aq) → 2 Ag+

(aq) + Fe(s) d) 3 Zn2+

(aq) + 2 Cr(s) → 3 Zn(s) + 2 Cr3+(aq)

8. Electrolytic Cells Electrolysis is the process by which electrical energy is passed through a solution to cause a non-spontaneous reaction to occur. This occurs in an electrolytic cell. Electrolytic Cell Electrochemical Cell

Diagram

Energy Applied to cell Generated by cell Partition No Yes Anode Sign + - Anode Electron flow Out Out Another Half-reaction Oxidation Oxidation 8 a) Electrolysis of Water - Electrons are forced into an aqueous salt solution at the cathode and removed from the solution at the cathode. The electrolyte acts like a salt bridge by ensuring that ions are able to migrate.

Anode: H2O(l) → ½ O2(g) + H+

(aq) Cathode: H2O(l) → H2(g) + OH-

(aq) Try This: 1. a) Balance each half-reaction b) Now find the overall reaction and its standard cell potential.

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