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Byju's Answer

Standard XII

Chemistry

Electrolysis and Electrolytes

An alloy of ...

Question

An alloy of $P b - A g$ weighing $1.08 g$ was dissolved in dilute $H N O_{3}$ and the volume made to $100 m L$ . A silver electrode was dipped in the solution and the emf of the cell set-up
Pt(s), $H_{2} (g) | H^{+} (1 M) | | A g^{+} (a q .) | A g (s)$ was $0.62 V$ . If $E_{c e l l}^{\circ}$ is $0.80 V$ , what is the percentage of $A g$ in the alloy?
$(A t 25^{\circ} C, R T / F = 0.06)$ .

25

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2.50

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10

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1

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50

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Solution

The correct option is E $50$
Overall cell reaction is:
$H_{2} (g) + 2 A g^{+} ⇌ 2 A g (s) + 2 H^{+} (a q .)$
$E = E^{\circ} - \frac{0.06 \times 2.303}{2} log \frac{[H^{+}]^{2}}{[A g^{+}]^{2} p H_{2}}$
$0.62 = 0.80 + \frac{2 \times 0.06 \times 2.303}{2} log [A g^{+}]$
$[A g^{+}] = 0.05 M$
Number of moles of $A g^{+}$ in $100 m L$
$= \frac{M V}{1000} = \frac{0.05 \times 100}{1000} = 0.005$
Mass of silver $= 0.005 \times 108 g$
Percentage of $A g$ in $1.08 g$ of alloy $= \frac{0.005 \times 108 \times 100}{1.08} = 50$ %.

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