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Step-by-Step Solution
Step 1: Identify the Known Half-Reactions and Their Standard Potentials
We are given two standard electrode (half-cell) reactions and their standard electrode potentials:
1. $ \text{Cu}^{2+} + 2e^- \to \text{Cu} \quad E^\circ = 0.337\,\text{V} $
2. $ \text{Cu}^{2+} + e^- \to \text{Cu}^+ \quad E^\circ = 0.153\,\text{V} $
We want to find the standard electrode potential for the reaction:
$ \text{Cu}^+ + e^- \to \text{Cu} $
Step 2: Express the Relationship Using Gibbs Free Energy Changes
Recall that the standard Gibbs free energy change $ \Delta G^\circ $ for a half-cell reaction is related to its standard electrode potential $ E^\circ $ by:
$ \Delta G^\circ = -nFE^\circ $
where
- $ n $ is the number of moles of electrons transferred in the reaction,
- $ F $ is the Faraday constant, and
- $ E^\circ $ is the standard electrode potential for the reaction.
Step 3: Write the Gibbs Free Energy for Each Given Reaction
Reaction (i): $ \text{Cu}^{2+} + 2e^- \to \text{Cu} $
Number of electrons, $ n = 2 $.
$ E^\circ = 0.337\,\text{V} $
So,
$ \Delta G^\circ_{(i)} = -2F \times 0.337 = -0.674\,F $
Reaction (ii): $ \text{Cu}^{2+} + e^- \to \text{Cu}^+ $
Here, $ n = 1 $.
$ E^\circ = 0.153\,\text{V} $
So,
$ \Delta G^\circ_{(ii)} = -1F \times 0.153 = -0.153\,F $
Step 4: Combine the Reactions to Obtain the Desired Reaction
Observe that if we add reaction (ii) “($ \text{Cu}^{2+} + e^- \to \text{Cu}^+ $)” to the unknown reaction “($ \text{Cu}^+ + e^- \to \text{Cu} $),” we should get reaction (i) “($ \text{Cu}^{2+} + 2e^- \to \text{Cu} $).”
Hence, in terms of Gibbs free energy,
$ \Delta G^\circ_{(i)} = \Delta G^\circ_{(ii)} + \Delta G^\circ_{\text{(desired)}} $
Let $ \Delta G^\circ_{\text{(desired)}} $ be the standard Gibbs free energy change for the reaction
$ \text{Cu}^+ + e^- \to \text{Cu}. $
Then,
$ \Delta G^\circ_{\text{(desired)}} = \Delta G^\circ_{(i)} - \Delta G^\circ_{(ii)}. $
Substitute the values we have:
$ \Delta G^\circ_{\text{(desired)}} = (-0.674\,F) - (-0.153\,F) = -0.674\,F + 0.153\,F = -0.521\,F. $
Step 5: Relate $ \Delta G^\circ_{\text{(desired)}} $ to $ E^\circ $ for $ \text{Cu}^+ + e^- \to \text{Cu} $
For the desired reaction, $ n = 1 $ (one electron transfer). So,
$ \Delta G^\circ_{\text{(desired)}} = -1 \times F \times E^\circ_{\text{(desired)}}. $
We have found:
$ \Delta G^\circ_{\text{(desired)}} = -0.521\,F. $
Thus,
$ -0.521\,F = -F \times E^\circ_{\text{(desired)}}. $
Dividing by $ -F $ on both sides,
$ E^\circ_{\text{(desired)}} = 0.521\,\text{V} \approx 0.52\,\text{V}. $
Final Answer
The standard electrode potential for the reaction $ \text{Cu}^+ + e^- \to \text{Cu} $ is 0.52 V.
