Question

The standard potential of electrode $Cu^{++ (0.02 M)} | Cu_{(s)}$ is 0.337 volt. Calculate its potential in volt.
 

Hint:

Note that when ion concentration is less than 1M, the reduction potential decreases for metal electrodes like Cu/Cu2+.

Answer 1

Step 1: Understanding the Concept:
The potential of an electrode depends on the concentration of ions in the solution. This relationship is given by the Nernst equation.
Step 2: Key Formula or Approach:
The reduction half-reaction is: \( Cu^{2+} + 2e^- \rightarrow Cu(s) \).
Number of electrons transferred, \( n = 2 \).
Nernst Equation:
\[ E = E^\circ - \frac{0.0592}{n} \log \frac{1}{[Cu^{2+}]} \]
Step 3: Detailed Explanation:
Given:
\( E^\circ = 0.337 \text{ V} \)
\( [Cu^{2+}] = 0.02 \text{ M} = 2 \times 10^{-2} \text{ M} \)
Substituting the values:
\[ E = 0.337 - \frac{0.0592}{2} \log \frac{1}{0.02} \]
\[ E = 0.337 - 0.0296 \log(50) \]
Since \( \log(50) = \log(100/2) = \log(100) - \log(2) = 2 - 0.3010 = 1.6990 \).
\[ E = 0.337 - (0.0296 \times 1.6990) \]
\[ E = 0.337 - 0.0503 \]
\[ E = 0.2867 \text{ V} \approx 0.287 \text{ V} \]
Step 4: Final Answer:
The electrode potential is 0.287 volt.