CHEM1612 - PharmacyWeek 9: Galvanic Cells

Dr. Siegbert SchmidSchool of Chemistry, Rm 223Phone: 9351 4196E-mail: siegbert.schmid@sydney.edu.au

Unless otherwise stated, all images in this file have been reproduced from:

Blackman, Bottle, Schmid, Mocerino and Wille,     Chemistry, John Wiley & Sons Australia, Ltd. 2008

     ISBN: 9 78047081 0866

Lecture 25-3

Electrochemistry Blackman, Bottle, Schmid, Mocerino & Wille:

Chapter 12, Sections 4.8 and 4.9

Key chemical concepts: Redox and half reactions Cell potential Voltaic and electrolytic cells Concentration cells

Key Calculations: Calculating cell potential Calculating amount of product for given current Using the Nernst equation for concentration cells

Lecture 25-4

The measured voltage across the cell is called cell potential (Ecell). This driving force for the reaction is also called ELECTROMOTIVE

FORCE (emf) of the cell.

Every galvanic cell has a different cell potential. How can we measure the cell potential relative to each species? How can we tabulate cell potentials?

Electromotive Force

Zn(s) + Cu2+(aq) Zn2+

(aq) + Cu(s)

Lecture 25-5

The electromotive force cannot be measured absolutely for one element (we cannot measure EZn and ECu in isolation).

Only differences in potential have meaning (ΔE).

We assign a “half-cell” potential to each “half-reaction”, and then add up two half-reactions to get the full reaction and full potential.

We choose as a standard the reaction of hydrogen :

2H+ (aq) + 2e– H2 (g) E0 = 0.00 V

(1 atm H2, [H+] = 1 M, at all temperatures)

Standard Reduction Potential

Indicates standardconditions

Lecture 25-6

Determining E0cell

We measure half-cell potentials relative to this reference hydrogen (H+/H2) electrode, which has E0

ref = 0.00 V. We build voltaic cells that have this reference half-cell and another

half-cell whose potential is unknown, E0unknown.

We define: E0cell = E0

cathode – E0anode

When H2 is oxidised, the reference electrode is the anode, so:E0

cell = E0cathode – E0

ref = E0unknown – 0.00 = E0

unknown

When H+ is reduced, the reference electrode is the cathode, so:E0

cell = E0ref – E0

anode = 0.00 - E0unknown = -E0

unknown

Lecture 25-7

Standard

Hydrogen

Electrode

Experiment 1: Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)

0.76 VPotential(V)

all reagents at standard concentration of 1.0 M

Demo: Potential of cell Zn/H2

Zn(s)|Zn2+||H+(aq)|H2(g)|Pt

We measure E0cell = 0.76 V;

Since E0cell = 0 - E0

unknown Then

E0Zn = -0.76 V

Lecture 25-8

Cu2+(s) + H2(g) 2H+(aq) + Cu(s) Experiment 2:

+0.34 VPotential(V)

Demo: Potential of cell Cu/H2

H2(g)| H+(aq)|| Cu2+(s)|Cu(s)| Pt

all reagents at standard concentration of 1.0 M

We measure E0cell =0.34 V

Since E0cell = E0

unknown Then

E0Cu = 0.34 V

Lecture 25-9

E0Cu

0.34 V

E0Zn -0.76 V

1.10 V

Potential(V)Experiment 3: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

Demo: Potential of Zn/Cu cell

We measure a potential

E0cell = 1.1 V.

Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s)

E0cell = E0

cathode – E0anode = 0.34 – (–0.76)= 0.34 + 0.76 = 1.10 V

Lecture 25-10

Cell Potential

all reagents at standard concentration of 1.0 M

- Experiment 1: Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)

Experiment 3: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

Cu2+(aq) + H2(g) Cu(s) + 2H+(aq)Experiment 2:

-(-0.76 V)

0.34 V

= 1.10 V

Potential(V)

Conclusion 2: You can add cell potentials as you add chemical reactions.

Conclusion 1: Reverse the reaction – reverse the sign of the potential.

Lecture 25-11

We measured for the reaction

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

A cell potential Ecell = ΔE = 1.1 V

Measured in volts, V = J·C–1

Moving one Coulomb of charge from Zn to Cu2+ releases 1.1 J of energy

Electromotive Force ΔE

Lecture 25-12

We define the standard reduction potential E0 as the potential of each redox couple when all reagents are in the standard state, i.e. 1 M concentration, 298 K, and gases at 1 atm pressure.The standard reduction potentials are tabulated.

In data tables all half reactions are written as reductions.

Fe3+ + e– Fe2+ E0 = 0.77 V Cu2+ + 2e– Cu E0 = 0.33 V2H+ + 2e– H2 (g) E0 = 0.00 V Zn2+ + 2e– Zn E0 = –0.76 V

The higher E (more positive), the greater the tendency to acquire electrons (be reduced).

Standard Reduction Potentials

Lecture 25-13

Reduction potential tableHalf-reaction Half-cell potential (V)

Au3+(aq) + 3e- Au(s) +1.50

Cl2(g) + 2e- 2Cl- (aq) +1.36

O2(g) + 4H+(aq) + 4e- 2H2O(l) +1.23

Ag+(aq) + e- Ag(s) +0.80

Cu2+(aq) + 2e- Cu(s) +0.34

2H+(aq) + 2e- H2(g) 0.00*

Sn2+(aq) + 2e- Sn(s) -0.14

Fe2+(aq) + 2e- Fe(s) -0.44

Zn2+(aq) + 2e- Zn(s) -0.76

2H2O(l) + 2e- H2(g) + 2OH-(aq) -0.83

Mg2+(aq) + 2e- Mg(s) -2.37

* by definition

Strong oxidising

agent

Weak oxidising

agent

Strong reducing

agent

Weak reducing

agent

No/slow oxidation by H+ due to anover-potential

Lecture 25-14

Using the redox potential tables

1. Write down the two half-reactions.2. Work out which is the oxidation and which is the reduction half-

reaction.3. Balance the electrons.4. Add up the half-reactions to get full reaction.5. Add up half-cell potentials to get E 0.

6. A spontaneous (working) voltaic cell, ALWAYS has a positive cell potential, E 0.

Lecture 25-15

ApproachCalculate the standard cell potential for a galvanic cell formed by Mg2+(aq) | Mg(s) in one half-cell and Sn2+(aq) | Sn(s) in the other.

Which reaction is the oxidation and which is the reduction?

Which half reaction is turned around?

Sn2+ + 2e- Sn E0 = -0.14V

Mg2+ + 2e- Mg E0 = -2.37V

In general, you reverse the reaction that appears lower in the table of standard reduction potentials.

Lecture 25-16

Example 1Calculate the standard cell potential for a galvanic cell formed by Mg2+(aq) | Mg(s) in one half-cell and Sn2+(aq) | Sn(s) in the other.

We saw that Mg is a stronger reducing agent than Sn, i.e. it likes to be oxidised. Turn the Mg reaction around to get an oxidation reaction.

Sn2+ + 2e- Sn E0 = -0.14V

Mg2+ + 2e- Mg E0 = -2.37V

Sn2+ + 2e- Sn E 0 = -0.14V

Mg Mg2+ + 2e- E 0 = +2.37V

Mg(s) + Sn2+(aq) Mg2+(aq) + Sn(s) E 0 = +2. 23V

Lecture 25-17

Example 2Calculate the standard cell potential for a galvanic cell with Ag+(aq) | Ag(s) in one half-cell and Cr3+(aq) | Cr(s) in the other. E0 (Ag+(aq)|Ag(s)) = +0.80 V; E0(Cr3+(aq)|Cr(s)) = -0.74V

Cr half reaction is lower (more negative), turn it around…

Cr(s) + 3Ag+(aq) Cr3+(aq) + 3Ag(s) E 0 = +1.54V

Balance the electrons… Note!

Note: E0 does not depend on stoichiometry!

Ag+ + e- Ag E 0 = +0.80V

Cr Cr3+ + 3e- E 0 = +0.74V

3Ag+ + 3e- 3Ag E 0 = +0.80V

Cr Cr3+ + 3e- E 0 = +0.74V

Lecture 25-18

Al + Fe2+ Al3+ + Fe

ONs: Al Al3+

Fe2+ Fe

Oxidation half cell: Al Al3+ + 3e-

Reduction half cell: 2e- + Fe2+ Fe

Combine: 2 x (Al Al3+ + 3e-) 3 x (2e- + Fe2+ Fe) 2Al + 3Fe2+ 2Al3+ + 3Fe

Balancing Redox Equations – Example 1

(oxidation)(reduction)

balance charge with e-

balance charge with e-

Lecture 25-19

Cr2O72- + I- Cr3+ + I2 (in acidic solution)

O.N.s: Cr2O72- (+6) Cr3+ (+3)

I- (-1) I2 (0)

Oxidation half cell: 2I- I2 + 2e-

Reduction half cell: 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O

Combine: 3 x (2I- I2 + 2e-) 6e- + 14H+ + Cr2O7

2- 2Cr3+ + 7H2O

6 I- + 14H+ + Cr2O72- 3I2 + 2Cr3+ + 7H2O

(oxidation)(reduction)

Balancing Redox Equations – Example 2

Lecture 25-20

HgO + Zn Hg + Zn(OH)2 (in basic solution)

ONs: HgO (+2) Hg (0) Zn (0) Zn2+ (+2)

Oxidation half cell: Zn Zn2+ + 2e-

Reduction half cell: 2e- + H2O + HgO Hg + 2OH-

(Add OH- and water to neutralise the charge and balance O and H)

Combine: Zn Zn2+ + 2e-

2e- + H2O + HgO Hg + 2OH-

Zn + H2O + HgO Zn2+ + Hg + 2OH-

Balancing Redox Equations – Example 3

(oxidation)(reduction)

Lecture 25-21

Step 1 Assign O.N. to all elements. Step 2 Identify the species being reduced/oxidised.Step 3 Calculate and balance the gain/loss electrons.Step 4 Balance the number of all the atoms other than O and H.Step 5 Balance O by adding H2O to either side as required.

Balance H and charges by adding H+ to either side as required.

Step 6 If basic conditions are specified, add OH– as required. (H+ and OH- cannot be present at the same time, they will

convert into H2O).

Finally, check that the whole reaction is balanced, i.ethat the charges and the moles of reactants and products are balanced.

Balancing Redox Equations – oxidation number method

Lecture 25-22

Example 1

Balance the equation for the oxidation of HCl by MnO4 –

MnO4– + HCl Mn2+

+ Cl2

Answer:

2 MnO4–(aq) + 10 HCl(aq) + 6 H+(aq) 2 Mn2+(aq) + 5 Cl2(aq) + 8 H2O(l)

Lecture 25-23

Example 2

Balance the reaction of oxidation of H2SO3 by Cr2O72-:

Cr2O72- + H2SO3 Cr3+ + SO4

2-

Answer: Cr2O72- + 3 H2SO3 + 2 H+ → 2 Cr3+ + 3 SO4

2- + 4H2O

Lecture 25-24

Example 3

Balance the reaction between NaMnO4 and Na2C2O4:

MnO4- + C2O4

2- MnO2 + CO32-

Answer: 2 MnO4- + 3 C2O4

2-+ 2 H2O → 2 MnO2 +6 CO32- + 4 H+

Lecture 25-25

SummaryCONCEPTS

Galvanic/Voltaic cells Cell notation Standard Reduction Potentials Balancing Redox Equations

CALCULATIONS Work out cell potential from reduction potentials

Lecture 25-26

Half-cell standard reduction potentials