Question

If \( E^\circ (Ag^+ | Ag) = +0.80 \, V \), What is the potential developed for \( Ag (s) \rightarrow Ag^+ (0.01M) + e^- \) at 298 K?

• A. +0.9184 V
• B. -0.9184 V
• C. +0.6816 V
• D. -0.6816 V

Hint:

The Nernst equation is essential for calculating the potential of electrochemical reactions at non-standard conditions, such as varying ion concentrations.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given the standard reduction potential for the silver electrode and need to calculate the non-standard oxidation potential for a specific concentration.
Step 2: Key Formula or Approach:
Use the Nernst equation for the oxidation half-reaction:
\[ \text{E}_{\text{ox}} = \text{E}^\circ_{\text{ox}} - \frac{0.0592}{n} \log_{10} [\text{Ag}^+] \]
Step 3: Detailed Explanation:
Given standard reduction potential \( \text{E}^\circ_{\text{red}} = +0.80 \text{ V} \).
Standard oxidation potential \( \text{E}^\circ_{\text{ox}} = -\text{E}^\circ_{\text{red}} = -0.80 \text{ V} \).
Reaction: \( \text{Ag}_{(s)} \longrightarrow \text{Ag}^+(0.01\text{M}) + \text{e}^- \), where \( n = 1 \) and \( [\text{Ag}^+] = 10^{-2} \text{ M} \).
Substitute into Nernst equation:
\[ \text{E}_{\text{ox}} = -0.80 - \frac{0.0592}{1} \log_{10}(10^{-2}) \] \[ \text{E}_{\text{ox}} = -0.80 - (0.0592) \times (-2) \] \[ \text{E}_{\text{ox}} = -0.80 + 0.1184 \] \[ \text{E}_{\text{ox}} = -0.6816 \text{ V} \] Step 4: Final Answer:
The potential developed is \( -0.6816 \text{ V} \).