Electrochemistry

• “Marriage” of redox and thermo

• Spontaneous electron-transfer reactions can result in spontaneous electric current if the reactants are separated by a wire– Voltaic (Galvanic) cells [Experiment 32!]

– We can use the “spontaneity” of the reaction to do electrical work

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(Continued)

• We can “push” electrons through a cell in order to make a nonspontaneous redox reaction occur.– Electrolysis cell [Experiment 32!]

– Doing work to “force” chemical reaction to occur [opposite of voltaic cell]

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Balancing Redox Equations

• Deferred until later

• For now just know that:– A half reaction has electrons written as a reactant

or a product• Oxidation half reaction: A reactant gets oxidized

(loses electrons); electrons appear as a “product”• Reduction half reaction: A reactant gets reduced

(gains electrons); electrons appear as a “reactant”

– A balanced redox equation does not show electrons explicitly. #e-’s lost = #e-’s gained (called “n”)

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Voltaic Cells

• Recall lab…

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• The spontaneous rxn occurs in the cell

• e-’s flow from – to + (“get to go where they want to go”)

• Anode = where ox occurs• Cathode = where red occurs• Salt bridge prevents charge

buildup (which would stop flow)

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You used graphite in place of Pt in lab for Fe2+/Fe3+ and I2/I-cells. A cheaper “inert” electrode.

Used when neither redox species in a half reaction (or electrode) is a neutral metal.

Neither is a neutral metal

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Standard Reduction Potentials (E°red)

• Recall lab– Make a bunch of different cells, get different

Ecell values (Eºcell if at standard state).

– Clearly some reductions are more favorable than others

• How do you know? [Which direction did e-’s flow?]• By how much?• Rank them? (Must pick a zero as reference.)

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Quick quiz

• NOTE: Every electrode compartment has one oxidizing agent and one reducing agent (this pair is called the redox “couple”)

Ox agent is ___

Red agent is ___ Ni (b/c it’s “more negative”; it has an electron to give)

Ni2+ (b/c it’s “more positive”; it has “room for an electron”

• If an electrode has Ni(s) and Ni2+ ions in it, which species is the oxidizing agent and which the reducing agent (of the pair)?

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Revisit Earlier Cell—Look at this as a “Competition for the electrons”. Which oxidizing agent “wants

them more”?

Who is the (possible)

oxidizing agent on the right? _____

Hint: The two “players” are Cu and Cu2+.

Cu2+

Who is the (possible)

oxidizing agent on the left? _____

Hint: The two “players” are Zn and Zn2+

Zn2+

Who wins? (Which one “got” the electrons?) ____Cu2+

Cu2+ “pulled harder”

So…which of the half reactions shown at the right is more favorable (greater tendency to happen)?

Cu2+ + 2 e- Cu(s)

Zn2+ + 2 e- Zn(s)

By how much?.....11

Reducing Cu2+ is more favorable than reducing

Zn2+ …by 1.10 V! (Measure it w/voltmeter!)

We define a “standard reduction potential”, E°red, for every reduction half reaction such that:

E°cell = E°red(cathode) - E°red(anode)

The more positive the “E” (Ecell, Ered, or Eox), the more favorable the process

Where reduction takes place

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Reducing Cu2+ is more favorable than reducing

Zn2+ …by 1.10 V! (Measure it w/voltmeter!)

E°cell = E°red(cathode) - E°red(anode)

If E°red(Zn2+/Zn) were 0 V, E°red(Cu2+/Cu) would be +1.10 V

If E°red(Zn2+/Zn) were -1.0 V, E°red(Cu2+/Cu) would be +0.10 V

NOTE:

1.10 V = E°red(Cu2+/Cu) - E°red(Zn2+/Zn)

If E°red(Zn2+/Zn) were +1.0 V, E°red(Cu2+/Cu) would be +2.10 V

The “zero” is arbitrary, but must be chosen / agreed upon!13

This electrode (SHE) was ultimately the one chosen by the scientific community to be the “zero” of potential.

2 H+ + 2 e- H2 (g); E°red = 0.0 V

Upshot:One can determine any E°red experimentally by just setting up a cell where one of the half cells is SHE!(next slide → )

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Ecell Ecathode - EanodeBoth as reductions

0.76 V = E°red(SHE) - E°red(Zn2+/Zn)

0.76 V = 0 - E°red(Zn2+/Zn)

E°red(Zn2+/Zn) = - 0.76 V

The reduction of H+ is more favorable than the reduction of Zn2+…. by 0.76 V! H+ (not Zn2+) gets reduced

Determining a Standard Reduction Potential using the SHE

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2 H+(aq) + 2 e- H2(g) 0

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

Could use this info to predict that this direct reaction would occur:

E°cell E°red(cat) - E°red(an)

Use these values to:

predict which reactions are spontaneous at standard state

and to

find any E°cell!

E°cell 0 – (-0.76) = +0.76 V

H+ gets reduced; Zn2+ does not (Zn gets oxidized):

2 H+ + Zn → H2(g) + Zn2+

is spontaneous: E°cell > 0

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Refers only to species on the left side of the arrow. E.g., F2, is a better ox agent than H2O2 which is better ox agent than Au3+ (but all of these species are very good oxidizing agents relative to most!)

Refers only to species on the right side of the arrow. E.g., F-, is a poorer red agent than H2O which is poorer red agent than Au(s) (but all of these are very poor reducing agents relative to most!)

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Excerpt from Voltaic Cell lab reading

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Table 18.1 (continued)

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Recall earlier slide

We define a “standard reduction potential”, E°red, for every reduction half reaction such that:

E°cell = E°red(cathode) - E°red(anode)

Ecell Ered + Eox

OR (could also write Ecell as…

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Lab interlude

• See overhead / board

• The lab manual initially asks you to pretend that the Ag+/Ag reduction potential is 0.0 V just to show you the “arbitrary-ness” of this.

• Then it tells you that in reality, Ag+/Ag reduction potential is 0.80 V if the H+/H2 potential (SHE) is 0.0 V

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Cu/Cu2+ & Zn/Zn2+

Cu/Cu2+ & Fe3+/Fe2+

Zn/Zn2+ & Ag/Ag+

**Circle the species that is the better oxidizing agent**

1.50 V Zn/Zn2+

1.05 V Zn/Zn2+

0.32 V Cu/Cu2+ Cu Cu2+ + 2e-

Cu2+ + 2e- Cu

Zn Zn2+ + 2e-

Zn Zn2+ + 2e-

Fe3+ + e- Fe2+

Ag+ + e- Ag 0.0 V1.50 V

1.50 V -0.45 V

+0.45 V -0.13 V

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Ag+ + e- Ag0.0 V

Fe3+ + e- Fe2+-0.13 V

Cu2+ + 2e- Cu-0.45 V

Zn2+ + 2e- Zn-1.50 V

0.80 V

0.67 V

0.35 V

-0.70 V

0.80 V

0.77 V

0.34 V

-0.76 V

From Text Table

(See next slide)

Zn2+ + 2e- Zn -1.50 VFlip

Cu2+ + 2e- Cu

Zn Zn2+ + 2e-

Fe3+ + e- Fe2+

Ag+ + e- Ag 0.0 V

1.50 V

-0.45 V

-0.13 V

From Table A.1:

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Table 18.1 (continued)

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Determine the cell reaction, calculate Ecell , identify the cathode and anode, label the (+) and (-) electrodes, and show electron flow (see board)

Ni2+(aq) + 2 e- Ni(s) -0.23 V

Mn2+(aq) + 2 e- Mn(s) -1.18 V

The better oxidizing agent is:___

So ___ actually gets reduced, and thus electrons flow to the _______ side, which must be the ____ode.

So the Ni electrode must be ______ive

E°cell = _____ - _____

= _______

Ni2+

Ni2+

rightcath

posit

-0.23 V -1.18 V

+0.95 V 26

(Because its Ered is more positive)

Determine the cell reaction, calculate Ecell , identify the cathode and anode, label the (+) and (-) electrodes, and show electron flow (see board)

Fe2+(aq) + 2 e- Fe(s) -0.45 V

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

The better oxidizing agent is:___

So ___ actually gets reduced, and thus electrons flow to the _______ side, which must be the ____ode.

So the Fe electrode must be ______ive

E°cell = _____ - _____

= _______

Fe2+

Fe2+

rightcath

posit

-0.45 V -2.37 V

+1.92 V 27

Saltbridge

1 M Fe2+

1 M Pb2+

e

Fe2+(aq) + 2 e- Fe(s) -0.45 V

Pb2+(aq) + 2 e- Pb(s) -0.13 V

The better oxidizing agent is:___

So ___ actually gets reduced, and thus electrons flow to the _______ side, which must be the ____ode.

So the Pb electrode must be ______ive

E°cell = _____ - _____

= _______

Pb2+

Pb2+

leftcath

posit

-0.13 V -0.45 V

+0.32 V

Pb(s) Fe(s)

Determine the cell reaction, calculate Ecell , identify the cathode and anode, label the (+) and (-) electrodes, and show electron flow (see board)

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• See last week’s pink lab handout (Voltaic Cells), board, and below– Start with G = G + RTlnQ and substitute in

G = -nFEcell and G = -nFEcell

After some algebra (and substituting in values for R, assuming T = 298 K, and converting to a base 10 log):

Nernst Equation

(T = 25C)

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cell cell

0.0592 Vlog E E Q

no o

Standard vs. Nonstandard Cell

cell cell

0.0592 Vlog E E Q

no o

30**Always write out the balanced redox equation before using the Nernst Equation or predicting whether Ecell should increase or decrease.**

Zn + Cu2+ Zn2+ + Cu; Q = ??

Recall lab—adding NH3 to Cu2+ side!

(T = 25C)

Explain in detail an in a conceptual way why the cell potential goes up when the NH3 is added. Is the driving force for the cell rxn greater or smaller after the NH3 is added?

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Relationship between variables (at any conditions; Mines Fig)

G G RT ln Q

cell cell

0.0592 Vlog E E Q

no o

Q

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(T = 25C)

Relationship between variables (at standard state conditions; From Tro)

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(T = 25C)

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EXAMPLE 18.8 Calculating Ecell Under Nonstandard ConditionsDetermine the cell potential for an electrochemical cell based on the following two half-reactions:

Oxidation:

Reduction:

E°cell = E°red(cat) - E°red(an)

ORE°cell = E°red(cat) + E°ox(an)

cell cell

0.0592 Vlog E E Q

no o**Need to write balanced equation before

using Nernst! What will “n” be here?**

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Fig. 18.22

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What mass of gold is plated in 25 minutes if the current is 5.5 A?

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

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