Question

What is the approximate \(E_{\text{cell}}\) (in V) for the following cell at \(298\ \mathrm{K}\)? \[ \mathrm{Sn(s)\,|\,Sn^{2+}(0.05\,M)\,||\,H^+(0.02\,M)\,|\,H_2(1\,bar)\,|\,Pt(s)} \] Given: \[ E^\circ_{\mathrm{Sn^{2+}/Sn}}=-0.14\ \mathrm{V},\qquad E^\circ_{\mathrm{H^+/H_2}}=0.00\ \mathrm{V} \] \[ \log(125)=2.097 \]

• A. \(0.218\)
• B. \(0.078\)
• C. \(0.020\)
• D. \(0.040\)

Hint:

For a cell, \[ \boxed{ E=E^\circ-\frac{0.0591}{n}\log Q } \] at \(298\ \mathrm{K}\).

Answer 1

The Correct Option is B

Step 1: Calculate standard cell potential. Cathode: \[ \mathrm{2H^+ +2e^- \rightarrow H_2} \] Anode: \[ \mathrm{Sn \rightarrow Sn^{2+}+2e^-} \] \[ E^\circ_{\text{cell}} =0-(-0.14) =0.14\ \mathrm{V} \]

Step 2: Reaction quotient. Overall reaction: \[ \mathrm{Sn+2H^+\rightarrow Sn^{2+}+H_2} \] \[ Q=\frac{[\mathrm{Sn^{2+}}]P_{H_2}}{[\mathrm{H^+}]^2} =\frac{0.05}{(0.02)^2} =125 \]

Step 3: Apply Nernst equation. \[ E = E^\circ-\frac{0.0591}{2}\log Q \] \[ =0.14-\frac{0.0591}{2}(2.097) \] \[ =0.14-0.062 \approx0.078\ \mathrm{V} \] Thus, \[ \boxed{0.078\ \mathrm{V}} \] Hence, \[ \boxed{(B)} \] is the correct answer.