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Standard VIII

Physics

Heating Effect of Electric Current

1.05 g of a l...

Question

$1.05 g$ of a lead ore containing impurity of $A g$ was dissolved in quantity of $H N O_{3}$ and the volume was made $350 m L$ . $A A g$ electrode was dipped in the solution and $E_{c e l l}$ of
$P t (H_{2}) | H^{+} (1 M) | | A g^{+} | A g$
was $0.503 V$ at $298 K$ . Calculate % of $A g$ in the ore.
$E_{A g^{+} / A g}^{\circ} = 0.80 V$ .

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Solution

$A n o d e \frac{1}{2} H_{2} (g) \to H^{+} (1 M) + e^{-}$
$C a t h o d e A g^{+} (x) + e^{-} \to A g (s) - ------------------------------- -$
$\frac{1}{2} H_{2} (g) + A g^{+} (x) ⇌ A g (s) + H^{+} (1 M) - ----------------------------------------- - (n = 1)$
$E^{\circ} = 0.80 - 0 = 0.80 v o l t$
$Q = \frac{[H^{+}]}{[A g^{+}]} = \frac{1}{x}$
$E = E^{\circ} - \frac{0.0591}{n} {log}_{10} Q$
$0.503 = 0.80 - \frac{0.0591}{1} {log}_{10} (\frac{1}{x})$
$x = 9.43 \times 10^{- 6} M$
Number of moles of $A g^{+}$ in $350 m L$
$= \frac{M V}{1000} = \frac{9.43 \times 10^{- 6} \times 350}{1000}$
$= 3.3 \times 10^{- 6}$
Mass of $A g = 3.3 \times 10^{- 6} \times 108 = 3.56 \times 10^{- 4} g$
% $A g$ in the ore $= \frac{3.56 \times 10^{- 4}}{1.05} \times 100 = 0.0339$ %.

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1.05 g of a lead ore containing impurity of Ag was dissolved in quantity of HNO3 and the volume was made 350 mL. A Ag electrode was dipped in the solution and Ecell ofPt(H2)|H+(1 M)||Ag+|Agwas 0.503 V at 298 K. Calculate % of Ag in the ore.E∘Ag+/Ag=0.80 V.