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Difficulty: MediumElectrochemical Series and Reaction Spontaneity
The standard reduction potentials for two half-cell reactions at 25∘C25^\circ\text{C} are given below:
Mn2+(aq)+2e−→Mn(s)E∘=−1.18 V\text{Mn}^{2+}(aq) + 2e^- \rightarrow \text{Mn}(s) \quad E^\circ = -1.18\text{ V}
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}

Calculate the standard cell potential (Ecell∘E^\circ_{\text{cell}}), in volts, for the spontaneous electrochemical reaction between these two half-cells.

Answer: 1.05 V

Answer

The standard cell potential for the spontaneous reaction is +1.05 V+1.05\text{ V}.
For a spontaneous redox reaction in a galvanic cell, the half-cell with the higher standard reduction potential acts as the cathode, and the one with the lower standard reduction potential acts as the anode. Lead(II) ions (Pb2+\text{Pb}^{2+}, E∘=−0.13 VE^\circ = -0.13\text{ V}) have a higher reduction potential than manganese(II) ions (Mn2+\text{Mn}^{2+}, E∘=−1.18 VE^\circ = -1.18\text{ V}). Substituting these values into Ecell∘=Ecathode∘−Eanode∘E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} yields Ecell∘=−0.13 V−(−1.18 V)=+1.05 VE^\circ_{\text{cell}} = -0.13\text{ V} - (-1.18\text{ V}) = +1.05\text{ V}.

Step-by-Step Solution

1
Identify cathode and anode roles for a spontaneous galvanic cell.
Cathode (reduction): Pb2+(aq)+2e−→Pb(s)\text{Pb}^{2+}(aq) + 2e^- \rightarrow \text{Pb}(s) with E∘=−0.13 VE^\circ = -0.13\text{ V}. Anode (oxidation): Mn(s)→Mn2+(aq)+2e−\text{Mn}(s) \rightarrow \text{Mn}^{2+}(aq) + 2e^- with E∘=−1.18 VE^\circ = -1.18\text{ V}.
For a spontaneous process (Ecell∘>0E^\circ_{\text{cell}} > 0), the species with the more positive reduction potential acts as the oxidizing agent and undergoes reduction at the cathode.
2
Use the cell potential formula Ecell∘=Ecathode∘−Eanode∘E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}.
Ecell∘=−0.13 V−(−1.18 V)E^\circ_{\text{cell}} = -0.13\text{ V} - (-1.18\text{ V})
Subtracting the standard reduction potential of the anode from that of the cathode gives the overall electromotive force of the cell.
3
Perform the subtraction to find the final numerical answer.
Ecell∘=+1.05 VE^\circ_{\text{cell}} = +1.05\text{ V}
−0.13+1.18=1.05-0.13 + 1.18 = 1.05, confirming a positive standard cell potential for the spontaneous reaction.

Key Concept

Standard Cell Potential and Reaction Spontaneity
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Topics

Electrochemical Series and Reaction Spontaneity

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