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 Nernst equation allows the calculation of electrode potential under non-standard conditions.
Step 2: Key Formula or Approach:
The Nernst equation for the reduction \(Cu^{2+} + 2e^- \rightarrow Cu(s)\) is:
\[ E = E^\circ - \frac{0.0592}{n} \log \frac{1}{[Cu^{2+}]} \]
Given: \(E^\circ = 0.337 \text{ V}\), \([Cu^{2+}] = 0.02 \text{ M}\), \(n = 2\).
Step 3: Detailed Explanation:
\[ E = 0.337 - \frac{0.0592}{2} \log \frac{1}{0.02} \]
\[ E = 0.337 - 0.0296 \log(50) \]
Since \(\log(50) = \log(100/2) = \log 100 - \log 2 = 2 - 0.301 = 1.699\):
\[ E = 0.337 - (0.0296 \times 1.699) \]
\[ E = 0.337 - 0.0503 \]
\[ E = 0.2867 \approx 0.287 \text{ V} \]
Step 4: Final Answer: The potential of the copper electrode is \(0.287 \text{ volt}\).