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Byju's Answer

Standard XII

Chemistry

Nernst Equation

An alloy weig...

Question

An alloy weighing $1.05$ g of $P b - A g$ was dissolved in desired amount of $H N O$ $_{3}$ and the volume was made $350$ mL. An Ag electrode was dipped in solution and $E$ $_{c e l l}$ of the cell $P t H_{2} 1 a t m | H^{+} 1 M | | A g^{+} | A g$ was $0.503 V$ at 298 K. The percentage of lead in alloy is (divide answer by 25) :
Given: $E_{A g^{+} / A g}^{o} = 0.80 V$

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Solution

The cell reactions are
Anode : $H_{2} \to 2 H^{+} + 2 e$
Cathode: $2 A g^{+} + 2 e \to 2 A g$
Thus, $E_{c e l l} = E_{O P_{H_{2}}}^{o} + E_{R P_{A g}}^{o} + \frac{0.059}{2} l o g_{10} \frac{[A g^{+}]^{2} \times P_{H_{2}}}{[H^{+}]^{2}}$
or $0.503 = 0 + 0.80 + \frac{0.059}{2} l o g_{10} [A g^{+}]^{2}$
or $[A g^{+}] = 9.25 \times 10^{- 6} M$
$∴$ Mole of Ag $^{+}$ in 350 mL
$= 9.25 \times 10^{- 6} \times \frac{350}{1000}$
$∴$ Mass of Ag $^{+}$ in 350 mL
$= 9.25 \times 10^{- 6} \times \frac{350}{1000} \times 108$
$= 3.497 \times 10^{- 4} g$
$∴$ % of Ag in 1.05 g alloy
$= \frac{3.497 \times 10^{- 4}}{1.05} \times 100 = 0.033$ %
$∴$ $%$ of lead in alloy $=$ $99.957 %$

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An alloy weighing 1.05 g of Pb−Ag was dissolved in desired amount of HNO3 and the volume was made 350 mL. An Ag electrode was dipped in solution and Ecell of the cell PtH21atm|H+1M||Ag+|Ag was 0.503V at 298 K. The percentage of lead in alloy is (divide answer by 25) :Given: EoAg+/Ag=0.80V