Question

The standard potential of \(\text{Cu}^{+2}/\text{Cu}\) is \(0.34\) V at \(298\) K. Calculate its potential at the same temperature when the \(\text{Cu}^{+2}\) ion concentration is \(0.1\) M.

• A. \(0.64\) V
• B. \(0.34\) V
• C. \(0.31\) V
• D. \(0.37\) V

Hint:

Use the Nernst equation with n = 2 and log(0.1) = -1.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The electrode potential changes with the concentration of the ion. The Nernst equation gives this dependence.

Step 2: Key Formula:
For \(\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}\), at 298 K:
\[ E = E^{\circ} - \dfrac{0.0591}{n}\log\dfrac{1}{[\text{Cu}^{2+}]} \]

Step 3: Substitute:
\(E^{\circ} = 0.34\) V, \(n = 2\), \([\text{Cu}^{2+}] = 0.1\) M.
\[ E = 0.34 - \dfrac{0.0591}{2}\log\dfrac{1}{0.1} = 0.34 - 0.02955 \times 1 \]
\[ E = 0.3105 \text{ V} \approx 0.31 \text{ V} \]

Step 4: Why the other options are wrong.
0.64 V adds 0.30 V, which is an error of sign and size. 0.34 V ignores the dilution. 0.37 V has the wrong sign for the correction: lower ion concentration lowers the reduction potential.

Final Answer:
The electrode potential is about 0.31 V. \[ \boxed{\text{(C) }0.31\ \text{V}} \]