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Difficulty: HardElectrochemical Series and Reaction Spontaneity

The standard reduction potentials for four half-cell reactions at 25∘C25^\circ\text{C} are given below:

1. Cd2+(aq)+2e−→Cd(s)E∘=−0.40 V\text{Cd}^{2+}(aq) + 2e^- \rightarrow \text{Cd}(s) \quad E^\circ = -0.40\text{ V}
2. Pb2+(aq)+2e−→Pb(s)E∘=−0.13 V\text{Pb}^{2+}(aq) + 2e^- \rightarrow \text{Pb}(s) \quad E^\circ = -0.13\text{ V}
3. Fe3+(aq)+e−→Fe2+(aq)E∘=+0.77 V\text{Fe}^{3+}(aq) + e^- \rightarrow \text{Fe}^{2+}(aq) \quad E^\circ = +0.77\text{ V}
4. Ag+(aq)+e−→Ag(s)E∘=+0.80 V\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s) \quad E^\circ = +0.80\text{ V}

Which of the following reaction processes is non-spontaneous under standard conditions?

  1. Oxidation of lead metal by aqueous cadmium ionsAnswer
  2. B
    Oxidation of cadmium metal by aqueous lead ions
  3. C
    Oxidation of iron(II) ions by aqueous silver ions
  4. D
    Reduction of iron(III) ions by cadmium metal

Answer

Oxidation of lead metal by aqueous cadmium ions is non-spontaneous because its standard cell potential is negative (Ecell∘=−0.27 VE^\circ_{\text{cell}} = -0.27\text{ V}).
The reaction process involving the oxidation of lead metal by aqueous cadmium ions requires cadmium ions to act as the oxidizing agent (reduced at the cathode) and lead metal to act as the reducing agent (oxidized at the anode). Calculating the standard cell potential gives Ecell∘=Ecathode∘−Eanode∘=−0.40 V−(−0.13 V)=−0.27 VE^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = -0.40\text{ V} - (-0.13\text{ V}) = -0.27\text{ V}. Because Ecell∘E^\circ_{\text{cell}} is negative, the reaction is non-spontaneous under standard conditions.

Step-by-Step Solution

1
Identify the reduction and oxidation half-reactions for the proposed reaction.
For the oxidation of lead metal by cadmium ions: Cathode (reduction): Cd2+(aq)+2e−→Cd(s)\text{Cd}^{2+}(aq) + 2e^- \rightarrow \text{Cd}(s), Anode (oxidation): Pb(s)→Pb2+(aq)+2e−\text{Pb}(s) \rightarrow \text{Pb}^{2+}(aq) + 2e^-.
Reaction spontaneity depends on the overall standard cell potential, calculated as Ecell∘=Ecathode∘−Eanode∘E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}.
2
Substitute the standard reduction potentials into the cell potential formula.
Ecell∘=E∘(Cd2+/Cd)−E∘(Pb2+/Pb)=−0.40 V−(−0.13 V)=−0.27 VE^\circ_{\text{cell}} = E^\circ(\text{Cd}^{2+}/\text{Cd}) - E^\circ(\text{Pb}^{2+}/\text{Pb}) = -0.40\text{ V} - (-0.13\text{ V}) = -0.27\text{ V}.
The cathode potential is the reduction potential of the species being reduced, and the anode potential is the reduction potential of the species being oxidized.
3
Evaluate reaction spontaneity based on the sign of Ecell∘E^\circ_{\text{cell}}.
Since Ecell∘=−0.27 V<0E^\circ_{\text{cell}} = -0.27\text{ V} < 0, this reaction cannot occur spontaneously under standard conditions.
A chemical reaction is spontaneous under standard conditions if and only if Ecell∘>0E^\circ_{\text{cell}} > 0.

Key Concept

Standard Cell Potential and Reaction Spontaneity
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Topics

Electrochemical Series and Reaction Spontaneity

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