First slide

Electrochemical cells

Question

An alloy of Pb-Ag weighing 1.08 g was dissolved in dilute HNO₃ and the volume made to 100 mL. A silver electrode was dipped in the solution and EMF of the cell set up was
${Pt}_{(s)}, H_{2 (g)} |H^{+} (1 M) ∥ {Ag}_{(a q .)}^{+}| {Ag}_{(s)}$
0.62 V. The percentage of Ag in the alloy is:
$[E_{cell}^{\circ} = 0 .80 V, 2 .303 RT / F = 0 .06 at 25^{\circ} C]$

Moderate

Solution

$\begin{array}{l} H_{2} + 2 {Ag}^{+} ⟶ 2 H^{+} + 2 Ag \\ E_{cell} = E_{cell}^{\circ} - \frac{2 .303 RT}{nF} \log \frac{1}{{[{Ag}^{+}]}^{2}} \\ 0 .62 = 0 .80 + 0 .06 \log [{Ag}^{+}] \\ or \log [{Ag}^{+}] = \frac{- 0 .18}{0 .06} = - 3 \\ [{Ag}^{+}] = 1 .0 \times 10^{- 3} M \\ = 1 .0 \times 10^{- 3} \times 108 = 0 .108 {gL}^{- 1} \end{array}$
$∴$ Amount of Ag in 100 mL solution = 0.0108 g
$∴$ $% Ag = \frac{0.0108}{1.08} \times 100 = 1$

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