Question

Difficulty: MediumElectrochemical Series and Reaction Spontaneity
The standard reduction potentials for iron and copper half-cells are given as follows:
Fe2+(aq)+2eFe(s)E=0.44 V\text{Fe}^{2+}(aq) + 2e^- \rightarrow \text{Fe}(s) \quad E^\circ = -0.44\text{ V}
Cu2+(aq)+2eCu(s)E=+0.34 V\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s) \quad E^\circ = +0.34\text{ V}

What is the standard electromotive force (EcellE^\circ_{\text{cell}}) for the overall reaction Fe(s)+Cu2+(aq)Fe2+(aq)+Cu(s)\text{Fe}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Fe}^{2+}(aq) + \text{Cu}(s), and is the reaction spontaneous under standard conditions?

  1. +0.78 V+0.78\text{ V}, spontaneousAnswer
  2. B
    0.78 V-0.78\text{ V}, non-spontaneous
  3. C
    0.10 V-0.10\text{ V}, non-spontaneous
  4. D
    +0.10 V+0.10\text{ V}, spontaneous

Answer

The standard electromotive force (EcellE^\circ_{\text{cell}}) is +0.78 V+0.78\text{ V}, and the reaction is spontaneous.
In the given reaction, Cu2+\text{Cu}^{2+} ions are reduced to copper metal at the cathode (E=+0.34 VE^\circ = +0.34\text{ V}), while Fe\text{Fe} metal is oxidized to Fe2+\text{Fe}^{2+} ions at the anode (E=0.44 VE^\circ = -0.44\text{ V}). Using the standard formula Ecell=EcathodeEanodeE^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}, we calculate Ecell=+0.34 V(0.44 V)=+0.78 VE^\circ_{\text{cell}} = +0.34\text{ V} - (-0.44\text{ V}) = +0.78\text{ V}. Because the cell potential is positive, the reaction is spontaneous under standard conditions.

Step-by-Step Solution

1
Identify the reduction and oxidation half-reactions
Cathode (reduction): Cu2+(aq)+2eCu(s)\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s) with Ecathode=+0.34 VE^\circ_{\text{cathode}} = +0.34\text{ V}. Anode (oxidation): Fe(s)Fe2+(aq)+2e\text{Fe}(s) \rightarrow \text{Fe}^{2+}(aq) + 2e^- with Eanode=0.44 VE^\circ_{\text{anode}} = -0.44\text{ V}.
Copper ions gain electrons (reduction at cathode) while iron metal loses electrons (oxidation at anode).
2
Calculate the standard cell potential (EcellE^\circ_{\text{cell}})
Ecell=EcathodeEanode=+0.34 V(0.44 V)=+0.78 VE^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = +0.34\text{ V} - (-0.44\text{ V}) = +0.78\text{ V}.
The cell potential is the difference between the reduction potential of the cathode species and that of the anode species.
3
Determine reaction spontaneity
Since Ecell=+0.78 V>0E^\circ_{\text{cell}} = +0.78\text{ V} > 0, the reaction is spontaneous.
A positive standard cell potential indicates a thermodynamically feasible (spontaneous) redox reaction under standard conditions.

Key Concept

Standard Cell Potential and Reaction Spontaneity
Estimated Time:1m 15s
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