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: Use Nernst equation.
The Nernst equation gives the potential of a half-reaction at non-standard conditions:
\[ E = E^\circ - \frac{0.0591}{n} \log \frac{[red]}{[ox]} \]
where:
- \( E^\circ \) is the standard electrode potential,
- \( n \) is the number of electrons involved in the reaction (here \(n = 1\)),
- \([red]\) is the concentration of the reduced form (solid silver),
- \([ox]\) is the concentration of the oxidized form (\(Ag^+\)).

Step 2: Substitute the given values.
We are given:
\[ E^\circ (Ag^+ | Ag) = +0.80 \, V, \, [Ag^+] = 0.01 \, M, \, [Ag] = 1 \, M \]
Substitute into the Nernst equation:
\[ E = 0.80 - \frac{0.0591}{1} \log \frac{1}{0.01} \]

Step 3: Simplify the equation.
\[ E = 0.80 - 0.0591 \log 100 \] \[ \log 100 = 2 \quad \Rightarrow \quad E = 0.80 - 0.0591 \times 2 = 0.80 - 0.1182 = 0.9184 \, V \]

Step 4: Final answer.
Thus, the potential developed is \(+0.9184 \, V\), corresponding to option \((1)\).