Unit 9 · Topic 9.9 Beta

Cell Potential and Free Energy

Standard reduction potentials rank how readily species gain electrons.

Practice 5: Mathematical RoutinesPractice 6: Argumentation

Question set for this topic

Part 1 · Hook

Why this matters

Stack two different metals with a damp salty paper between them and you can measure a voltage, the way the first battery was made in 1800. Which pair gives the most voltage? A table of standard reduction potentials answers that for any pair, and the voltage tells you ΔG° and K as well.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. In a galvanic cell, where does reduction happen?

  1. At the cathode
  2. At the anode
  3. In the salt bridge
  4. In the wire
Show the answer

Reduction at the cathode, oxidation at the anode.

  • Correct: At the cathode:
  • At the anode:
  • In the salt bridge:
  • In the wire:

2. If ΔG° < 0, what is true of K?

  1. K > 1
  2. K < 1
  3. K = 1
  4. K = 0
Show the answer

ΔG° = −RT ln K: negative ΔG° gives K > 1.

  • Correct: K > 1:
  • K < 1:
  • K = 1:
  • K = 0:

Part 4 · See it

See it first

A triangle linking ΔG° at the top to K at the bottom left by ΔG° = −RT ln K and to E°cell at the bottom right by ΔG° = −nFE°cell. A table below: favored reactions have ΔG° below zero, K above 1 and E°cell above zero; unfavored ones have the opposite signs.
ΔG°, K and E°cell all describe how far a reaction goes, so each can be found from the others. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Each half-reaction has a tendency to gain electrons, measured by its standard reduction potential E°the half-reaction with the more positive E° is reduced at the cathode
  2. The other half-reaction runs in reverse as the oxidation at the anodeE°cell = E°(cathode) − E°(anode), positive for a galvanic cell
  3. E° is energy per unit charge, an intensive propertybalancing electrons never multiplies an E°
  4. Total electrical work is charge × voltageΔG° = −nFE°, so a positive E°cell means ΔG° < 0 and K > 1

Part 6 · Key ideas

Key ideas

  • A standard reduction potential E° measures how readily a species gains electrons. More positive: stronger oxidizing agent.
  • E°cell = E°(cathode) − E°(anode), using both values as reduction potentials. Positive E°cell: favored (galvanic).
  • E° is intensive: never multiply it by a coefficient when you balance electrons.
  • ΔG° = −nFE°, with n the mol e⁻ transferred in the balanced equation and F = 96,485 C/mol e⁻ (1 J = 1 C·V).
  • E° > 0 ⇔ ΔG° < 0 ⇔ K > 1. A negative E°cell means the reaction needs an electrolytic cell.

Part 7 · Misconception

A common mistake

The wrong idea: When a half-reaction is multiplied by 2 to balance electrons, its E° is multiplied by 2 too.

What actually happens: E° is energy per coulomb, an intensive property, like temperature. Multiplying a half-reaction changes the number of electrons (n), which changes ΔG° = −nFE°, but E° stays the same.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Data table

Standard reduction potentials

Standard reduction potentials at 25 °C. F = 96,485 C/mol e⁻.

Standard reduction potentials, E°, at 25 °C
Half-reactionE° (V)
Ag⁺(aq) + e⁻ → Ag(s)+0.80
Cu²⁺(aq) + 2e⁻ → Cu(s)+0.34
Pb²⁺(aq) + 2e⁻ → Pb(s)−0.13
Ni²⁺(aq) + 2e⁻ → Ni(s)−0.25
Fe²⁺(aq) + 2e⁻ → Fe(s)−0.44
Zn²⁺(aq) + 2e⁻ → Zn(s)−0.76
Al³⁺(aq) + 3e⁻ → Al(s)−1.66
Mg²⁺(aq) + 2e⁻ → Mg(s)−2.37

1. A galvanic cell is built from a Ni/Ni²⁺ half-cell and an Ag/Ag⁺ half-cell under standard conditions. Calculate E°cell.

Type a number and its unit.

Show the answer

Silver has the more positive E°, so Ag⁺ is reduced (cathode) and Ni is oxidized (anode). E°cell = E°(cathode) − E°(anode) = 0.80 V − (−0.25 V) = 1.05 V.

  • Answer: 1.05 V

2. Which is the balanced overall reaction for the Ni/Ag galvanic cell?

  1. Ni²⁺(aq) + 2 Ag(s) → Ni(s) + 2 Ag⁺(aq)
  2. Ni(s) + 2 Ag⁺(aq) → Ni²⁺(aq) + 2 Ag(s)
  3. Ni(s) + Ag⁺(aq) → Ni²⁺(aq) + Ag(s)
  4. Ni(s) + 2 Ag(s) → Ni²⁺(aq) + 2 Ag⁺(aq)
Show the answer

Oxidation Ni → Ni²⁺ + 2e⁻; reduction 2 Ag⁺ + 2e⁻ → 2 Ag; overall Ni + 2 Ag⁺ → Ni²⁺ + 2 Ag.

  • Ni²⁺(aq) + 2 Ag(s) → Ni(s) + 2 Ag⁺(aq): This is the reverse, with E° = −1.05 V: not favored, so it is not what the galvanic cell runs.
  • Correct: Ni(s) + 2 Ag⁺(aq) → Ni²⁺(aq) + 2 Ag(s): Right: nickel is oxidized, silver ions are reduced, and two Ag⁺ take the two electrons from each Ni.
  • Ni(s) + Ag⁺(aq) → Ni²⁺(aq) + Ag(s): Atoms balance but charge does not (+1 vs +2). Two electrons need two Ag⁺.
  • Ni(s) + 2 Ag(s) → Ni²⁺(aq) + 2 Ag⁺(aq): Both metals are oxidized here; a redox reaction needs one species reduced.

3. Calculate ΔG° for the reaction in the Ni/Ag cell, Ni(s) + 2 Ag⁺(aq) → Ni²⁺(aq) + 2 Ag(s).

Type a number and its unit.

Show the answer

ΔG° = −nFE° = −(2 mol e⁻)(96,485 C/mol e⁻)(1.05 V) = −2.026 × 10⁵ J/mol = −203 kJ/mol (1 J = 1 C·V).

  • Answer: -203 kJ/mol

Model

An aluminum-copper cell

A galvanic cell is built under standard conditions from an Al strip in 1.0 M Al(NO₃)₃ and a Cu strip in 1.0 M Cu(NO₃)₂. Standard reduction potentials: Al³⁺(aq) + 3e⁻ → Al(s), E° = −1.66 V; Cu²⁺(aq) + 2e⁻ → Cu(s), E° = +0.34 V. F = 96,485 C/mol e⁻.

Vsalt bridge, KNO₃(aq)Al(s) in 1.0 M Al(NO₃)₃Cu(s) in 1.0 M Cu(NO₃)₂Beaker 1Beaker 2

4. Calculate E°cell for the cell.

Type a number and its unit.

Show the answer

Cu²⁺ has the more positive E°, so it is reduced; Al is oxidized. E°cell = 0.34 V − (−1.66 V) = 2.00 V. The coefficients 2 and 3 used to balance electrons do not change any E°.

  • Answer: 2.00 V

5. How many moles of electrons, n, are transferred in the balanced equation 2 Al(s) + 3 Cu²⁺(aq) → 2 Al³⁺(aq) + 3 Cu(s)?

  1. 3
  2. 2
  3. 6
  4. 5
Show the answer

n is the least common multiple of the electrons in the two half-reactions: 2 × 3 = 6.

  • 3: That is the electrons for one Al atom. The equation has two Al atoms.
  • 2: That is the electrons for one Cu²⁺ ion. The equation has three of them.
  • Correct: 6: Right: 2 Al each lose 3 electrons and 3 Cu²⁺ each gain 2: six electrons.
  • 5: Adding 3 and 2 does not count electrons. Multiply electrons per atom by atoms: 2 × 3 = 3 × 2 = 6.

6. A student writes E°cell = 3(+0.34) − 2(−1.66) = 4.34 V "because the equation has 3 Cu²⁺ and 2 Al." Which correction is best?

  1. The coefficients should be divided, not multiplied, giving 0.34/3 + 1.66/2.
  2. The student is right; the coefficients belong in E° and in n as well.
  3. E° is intensive, like temperature, so coefficients never change it: E°cell = 2.00 V.
  4. The anode’s E° alone should be multiplied, because electrons are released there.
Show the answer

E° is energy per coulomb. Balancing electrons changes the total energy (through n in ΔG° = −nFE°) but not the energy per coulomb.

  • The coefficients should be divided, not multiplied, giving 0.34/3 + 1.66/2.: Coefficients do not enter E° at all, whether multiplied or divided.
  • The student is right; the coefficients belong in E° and in n as well.: The coefficients affect n (and so ΔG°), but never E°.
  • Correct: E° is intensive, like temperature, so coefficients never change it: E°cell = 2.00 V.: Right: potential is energy per unit charge; doubling the reaction doubles both, leaving E° unchanged.
  • The anode’s E° alone should be multiplied, because electrons are released there.: Neither E° is multiplied.

7. For Cu(s) + Zn²⁺(aq) → Cu²⁺(aq) + Zn(s), E° = −1.10 V. What does this tell you?

  1. The reaction is favored but slow, so it needs a catalyst.
  2. The cell will run with electrons flowing from Zn to Cu, so it is still galvanic.
  3. The reaction is at equilibrium, because E° is negative.
  4. It is not favored; it needs an electrolytic cell with a power supply.
Show the answer

A negative E°cell means ΔG° = −nFE° > 0. The reaction can be driven by a power supply, which is an electrolytic cell.

  • The reaction is favored but slow, so it needs a catalyst.: A negative E° means not favored; a catalyst cannot change that.
  • The cell will run with electrons flowing from Zn to Cu, so it is still galvanic.: For this reaction as written, copper would be oxidized; with E° < 0 that does not happen on its own.
  • The reaction is at equilibrium, because E° is negative.: At equilibrium the cell potential E is zero, not E°.
  • Correct: It is not favored; it needs an electrolytic cell with a power supply.: Right: a negative E°cell means ΔG° > 0, so the reaction must be driven.

Part 9 · Summary

Summary

Standard reduction potentials rank how readily species gain electrons. In a galvanic cell the half-reaction with the more positive E° is the cathode, and E°cell = E°(cathode) − E°(anode). E° is intensive and is never multiplied by coefficients. ΔG° = −nFE°, so a positive E°cell means a favored reaction with K > 1.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections