Question

Determine the electrode potential of \(\text{Sn}^{2+}(0.01\text{M}) | \text{Sn(s)}\) at \(25^{\circ}\text{C}\) if \(E_{\text{Sn}}^{\circ}\) is \(-0.136\) V.

• A. \(0.186\) V
• B. \(-0.186\) V
• C. \(-0.195\) V
• D. \(0.195\) V

Hint:

Use the Nernst equation for a metal in its own ion; the reaction is Sn2+ + 2e- going to Sn.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The electrode potential changes with ion concentration according to the Nernst equation. The electrode reaction is a reduction:
\[ \text{Sn}^{2+}+2e^-\to\text{Sn(s)},\quad n=2 \]

Step 2: Key Formula or Approach
\[ E=E^\circ-\frac{0.0591}{n}\log\frac{1}{[\text{Sn}^{2+}]} \]

Step 3: Substitute
\[ E=-0.136-\frac{0.0591}{2}\log\frac{1}{0.01} \]
\[ \log\frac{1}{0.01}=\log100=2 \]
\[ E=-0.136-0.0591=-0.1951\ \text{V} \]

Step 4: Check the options
The dilute solution makes the reduction less favourable, so the potential is lower than the standard value of \(-0.136\) V. Options (A) and (D) are positive and so cannot follow. Option (B) \(-0.186\) V is not given by the equation. The answer is \(-0.195\) V, option (C).

Final Answer:
The electrode potential is \(-0.195\) V, option (C). \[ \boxed{-0.195\ \text{V}} \]