Question

The emf of the cell :
$Sn\left(\right.s\left.\right)\left|\left(Sn\right)^{+ 2} \left(\right. 0 . 01 M \left.\right) \parallel H^{+} \left(\right. 0 . 01 M \left.\right)\right|H_{2}\left(\right.g\left.\right)$ Pt is
(Given : $E_{Sn^{+ 2} / Sn}^\circ =-0.14V$ ):-

• A. $-0.08V$
• B. $0.8V$
• C. $0.22V$
• D. $0.08V$

Answer 1

Answer: (D)

$Sn\left(\right.s\left.\right)+2H^{+}\left(\right.0.03M\left.\right)\overset{n = 2}{ \rightarrow }\left(Sn\right)^{+ 2}\left(\right.0.06M\left.\right)+H_{2}\left(\right.g\left.\right)$
$E_{\text{cell }}^\circ =\left(\right.SRP\left(\left.\right)_{\text{Cathode }}-\left(\right.SRP\left(\left.\right)_{\text{anode }}$
$E_{\text{cell }}^\circ =0-\left(\right.-0.14V\left.\right)=0.14V$
$E_{\text{cell }}=E_{\text{cell }}^\circ -\frac{0 . 0591}{n}logQ$
$E_{\text{cell }}=0.14-\frac{0 . 0591}{n}log\frac{0 . 01 \times 1}{1 \times \left(\right. 0 . 01 \left(\left.\right)^{2}}$
$E_{cell }=0.08V$