Question

If $E^{\circ}(\text{Cu}^{2+}/\text{Cu})=+0.34\text{ V}$. What is potential for $\text{Cu}(s)\rightarrow \text{Cu}^{2+}(aq)(0.1\text{M})+2e^{-}$ at 298 K?

• A. +0.3696 V
• B. -0.3696 V
• C. +0.3104 V
• D. -0.3104 V

Hint:

Read the requested reaction carefully! If it shows electrons on the right side ($X \rightarrow X^+ + e^-$), it's asking for the Oxidation Potential. You must flip the sign of the standard reduction potential before plugging it into Nernst!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given the standard reduction potential of copper. We are asked to find the non-standard oxidation potential for the reaction $\text{Cu}(s) \rightarrow \text{Cu}^{2+}(0.1\text{M}) + 2e^-$ using the Nernst equation.

Step 2: Detailed Explanation:
1. Determine the Standard Oxidation Potential:
The standard reduction potential is given as $E^\circ_{\text{red}} = +0.34$ V.
Because the required reaction is an oxidation (loss of electrons), we flip the sign to find the standard oxidation potential:
$E^\circ_{\text{ox}} = -0.34$ V.
2. Apply the Nernst Equation for the half-cell:
The Nernst equation for an oxidation half-reaction at 298 K is:
$E_{\text{ox}} = E^\circ_{\text{ox}} - \frac{0.0592}{n} \log(Q)$
For the reaction $\text{Cu}(s) \rightarrow \text{Cu}^{2+}(aq) + 2e^-$:
- Number of electrons transferred, $n = 2$.
- Reaction quotient, $Q = [\text{Cu}^{2+}] = 0.1 \text{ M} = 10^{-1} \text{ M}$. (Solids like Cu are omitted).
Substitute the values:
$E_{\text{ox}} = -0.34 - \frac{0.0592}{2} \log(10^{-1})$
$E_{\text{ox}} = -0.34 - (0.0296) \times (-1)$
$E_{\text{ox}} = -0.34 + 0.0296$
$E_{\text{ox}} = -0.3104$ V.

Step 3: Final Answer:
The potential is -0.3104 V, matching option (d).