Question

If \(E^\circ (Cd^{+2}|Cd_{(s)}) = -0.40\) V. What is potential for \(Cd_{(s)} \rightarrow Cd^{+2}_{(aq)}(0.01M) + 2e^{-}\) at 298 K?

• A. +0.4592 V
• B. -0.4592 V
• C. +0.3408 V
• D. -0.3408 V

Hint:

Reverse the sign of the standard reduction potential for oxidation and apply the Nernst equation.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
For \( Cd \rightarrow Cd^{2+} + 2e^{-} \), the standard oxidation potential is \( E^\circ = +0.40 \) V.
Nernst equation: \( E = E^\circ - \frac{0.0592}{n}\log Q \), with \( Q = [Cd^{2+}] \).

Step 2: Calculation
\( E = 0.40 - \frac{0.0592}{2}\log(0.01) \)
\( E = 0.40 - 0.0296\times(-2) = 0.40 + 0.0592 \).

Step 3: Result
\( E = +0.4592 \) V, option (A).

Final Answer:
The potential is +0.4592 V. \[ \boxed{\text{(A) }+0.4592\ \text{V}} \]