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?
If \( E^\circ (Ag^+ | Ag) = +0.80 \, V \), What is the potential developed for \( Ag (s) \rightarrow Ag^+ (0.01M) + e^- \) at 298 K?
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Find the reduction potential from the Nernst equation, then reverse its sign because the reaction is an oxidation.
Updated On: Nov 13, 2026
- +0.9184 V
- -0.9184 V
- +0.6816 V
- -0.6816 V
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
For \(\text{Ag}^+ + e^-\rightarrow \text{Ag}\), \(E=E^\circ+0.0591\log[\text{Ag}^+]\) at 298 K.
Step 2: Reduction potential
\(E_{red}=0.80+0.0591\log(0.01)=0.80-0.1182=0.6818\ \text{V}\)
Step 3: Oxidation potential
The given reaction \(\text{Ag}\rightarrow\text{Ag}^++e^-\) is the reverse, so \(E_{ox}=-0.6818\ \text{V}\approx-0.6816\ \text{V}\).
Step 4: Compare the options
The fourth option, \(-0.6816\ \text{V}\), matches.
Final Answer:
The potential developed for the oxidation is about -0.6816 V. \[ \boxed{\text{(D) }-0.6816\ \text{V}} \]
For \(\text{Ag}^+ + e^-\rightarrow \text{Ag}\), \(E=E^\circ+0.0591\log[\text{Ag}^+]\) at 298 K.
Step 2: Reduction potential
\(E_{red}=0.80+0.0591\log(0.01)=0.80-0.1182=0.6818\ \text{V}\)
Step 3: Oxidation potential
The given reaction \(\text{Ag}\rightarrow\text{Ag}^++e^-\) is the reverse, so \(E_{ox}=-0.6818\ \text{V}\approx-0.6816\ \text{V}\).
Step 4: Compare the options
The fourth option, \(-0.6816\ \text{V}\), matches.
Final Answer:
The potential developed for the oxidation is about -0.6816 V. \[ \boxed{\text{(D) }-0.6816\ \text{V}} \]
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