C/P 09 · Electrochemistry
Galvanic Cells and Electroplating
Use reduction potentials, ΔG° = −nFE°, the Nernst equation and Faraday’s law.
Passage
In a galvanic (voltaic) cell, a spontaneous redox reaction produces an electric current. Oxidation occurs at the anode and reduction occurs at the cathode, and electrons flow through an external wire from the anode to the cathode. The standard cell potential is E°cell = E°(cathode) − E°(anode), using standard reduction potentials for both half-reactions. A positive E°cell means that the reaction is spontaneous under standard conditions. The standard free energy change is related to the cell potential by ΔG° = −nFE°, where n is the number of moles of electrons transferred and F is the Faraday constant, about 96,500 C per mole of electrons.
When the concentrations are not standard, the cell potential is given by the Nernst equation. At 298 K, E = E° − (0.06/n) log₁₀ Q, where Q is the reaction quotient. The Nernst equation shows that the cell potential changes with the concentrations of the ions in solution, and it reaches zero when the cell comes to equilibrium.
Part 1. A student has four metal electrodes, each in a 1.0 M solution of its own ions. Table 1 shows the standard reduction potentials for the half-reactions.
Table 1. Standard reduction potentials.
| Half-reaction | E° (V) |
|---|---|
| Cu²⁺ + 2e⁻ → Cu | +0.34 |
| Zn²⁺ + 2e⁻ → Zn | −0.76 |
| Ag⁺ + e⁻ → Ag | +0.80 |
| Fe²⁺ + 2e⁻ → Fe | −0.44 |
Part 2. The student builds a cell with a zinc electrode in a Zn²⁺ solution and a copper electrode in a 1.0 M Cu²⁺ solution, connected by a wire and a salt bridge. She keeps [Cu²⁺] at 1.0 M and changes [Zn²⁺]. Table 2 shows the measured cell potential at 298 K.
Table 2. Cell potential of the Zn–Cu cell at different [Zn²⁺].
| [Zn²⁺] (M) | [Cu²⁺] (M) | Cell potential (V) |
|---|---|---|
| 1.0 | 1.0 | 1.10 |
| 0.10 | 1.0 | 1.13 |
| 0.010 | 1.0 | 1.16 |
The salt bridge contains an inert electrolyte. Ions move through it to keep each half-cell electrically neutral, so that the circuit is completed without the two solutions mixing.
Part 3. In a separate experiment, the student connects a battery to two copper electrodes in a solution of Cu²⁺, so that copper metal is deposited on the electrode connected to the negative terminal. This is an electrolytic cell. A constant current of 2.0 A flows for 965 s. The molar mass of copper is 64 g/mol. Charge, current and time are related by q = It.
In the body, a similar flow of electrons occurs in the electron transport chain, in which electrons move from NADH to oxygen through a series of carriers with increasingly positive reduction potentials. The large difference in potential between NADH and oxygen is the reason that the free energy released is enough to pump protons across the inner mitochondrial membrane. The student wants to use the same relationships to compare the zinc–copper cell with this biological example.
Questions
Question 1 of 6
What is the standard cell potential of the Zn–Cu cell, and which electrode is the anode?
Copper has the more positive reduction potential, so Cu²⁺ is reduced at the cathode. Zinc is oxidized at the anode.
The cell potential is E°(cathode) − E°(anode) = 0.34 − (−0.76), which is positive.
This is 0.34 + (−0.76), a sum of potentials. The cell potential is the difference.
Why the answer is what it is
Zn²⁺ has the lower reduction potential, so Zn is oxidized (anode) and Cu²⁺ is reduced (cathode). E°cell = +0.34 − (−0.76) = +1.10 V.
Review: Electrochemistry · Redox Chemistry & Electron Carriers · Gibbs Free Energy & Spontaneity · Chemical Equilibrium & Le Châtelier’s Principle · Equations: Equilibrium, Acids & Solutions
Question 2 of 6
What is the approximate standard free energy change for the Zn–Cu reaction, per mole of Zn?
The sign is wrong. A positive E° means that ΔG° is negative.
This would be the result for n = 1. Two electrons are transferred, so n = 2.
This ignores n and F. ΔG° = −nFE° is in the hundreds of kilojoules.
Why the answer is what it is
ΔG° = −nFE° = −(2)(96,500 C/mol)(1.10 V) ≈ −212,000 J/mol, or about −212 kJ/mol. A negative ΔG° agrees with a spontaneous reaction.
Review: Electrochemistry · Redox Chemistry & Electron Carriers · Gibbs Free Energy & Spontaneity · Chemical Equilibrium & Le Châtelier’s Principle · Equations: Equilibrium, Acids & Solutions
Question 3 of 6
In Table 2, the cell potential rises as [Zn²⁺] falls. What explains this?
E° is a constant for the half-reaction at standard conditions. The potential changes because of Q.
The number of electrons transferred per reaction is unchanged. The driving force changes.
At equilibrium the potential is zero. A rising potential means the cell is moving away from equilibrium.
Why the answer is what it is
Q = [Zn²⁺]/[Cu²⁺], so lowering [Zn²⁺] lowers Q. In E = E° − (0.06/2) log Q, a smaller Q makes log Q more negative and E larger. For [Zn²⁺] = 0.010 M, E = 1.10 − 0.03 × (−2) = 1.16 V.
Review: Electrochemistry · Redox Chemistry & Electron Carriers · Gibbs Free Energy & Spontaneity · Chemical Equilibrium & Le Châtelier’s Principle · Equations: Equilibrium, Acids & Solutions
Question 4 of 6
Using Table 1, which pairing of two of these half-cells gives the largest standard cell potential?
E° = −0.44 − (−0.76) = 0.32 V. This is the smallest of the pairings listed.
E° = 0.80 − 0.34 = 0.46 V. Other pairings have larger differences.
E° = 0.80 − (−0.44) = 1.24 V, which is large but smaller than the Zn–Ag pairing.
Why the answer is what it is
The cell potential is the difference between the two reduction potentials, so the largest value comes from the highest and lowest potentials: Ag (+0.80) and Zn (−0.76), giving 1.56 V.
Review: Electrochemistry · Redox Chemistry & Electron Carriers · Gibbs Free Energy & Spontaneity · Chemical Equilibrium & Le Châtelier’s Principle · Equations: Equilibrium, Acids & Solutions
Question 5 of 6
How much copper is deposited in Part 3?
This is half of the correct mass. Recheck the charge (2.0 A × 965 s = 1930 C) and the moles of copper (one mole of Cu per 2 moles of electrons).
This would result from assuming that 1 mole of electrons deposits 1 mole of copper. Each Cu²⁺ ion needs 2 electrons.
This is the charge in kilocoulombs (1.93 kC) and not a mass.
Why the answer is what it is
q = It = 2.0 A × 965 s = 1930 C. Moles of electrons = 1930 ÷ 96,500 = 0.020 mol. Cu²⁺ + 2e⁻ → Cu uses 2 mol of electrons per mole of Cu, so 0.010 mol of Cu is deposited, which is 0.010 × 64 = 0.64 g.
Review: Electrochemistry · Redox Chemistry & Electron Carriers · Gibbs Free Energy & Spontaneity · Chemical Equilibrium & Le Châtelier’s Principle · Equations: Equilibrium, Acids & Solutions
Question 6 of 6
Which statement about the electron transport chain is consistent with the relationships in this passage?
The chain is a spontaneous process, like a galvanic cell. It is the electrolytic cell in Part 3 that needs an external power source.
Electrons flow toward the carrier with the more positive reduction potential, so they flow from NADH to oxygen.
ΔG° = −nFE°, so the free energy released depends directly on the potential difference.
Why the answer is what it is
As in a galvanic cell, electrons move from the half-reaction with the lower reduction potential (NADH) to the one with the higher potential (O₂). The positive E° corresponds to a negative ΔG°, and that energy is used to pump protons.
Review: Electrochemistry · Redox Chemistry & Electron Carriers · Gibbs Free Energy & Spontaneity · Chemical Equilibrium & Le Châtelier’s Principle · Equations: Equilibrium, Acids & Solutions