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Byju's Answer

Standard VIII

Chemistry

Ways to Define Concentration

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 $100$ mL. A silver electrode was dipped in the solution and the cell is set up as
$P t_{(s)}, H_{2 (g)} | H^{+} (1 M) | | A g_{(a q)}^{+} | A g_{(s)}$
The emf of the cell was $0.62$ V. If $E_{C e l l}^{o} = 0.80$ V. What is the percentage of $A g$ in the alloy?[At $25^{o}$ C, $2.303$ RT/F $= 0.06$ ]

25

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2.50

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10

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1

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Solution

The correct option is D $1$
$\begin{matrix} cell reaction is : H_{2} + 2 {A g}^{+} ⟶ 2 H^{+} + 2 A g . \end{matrix}$
$E_{c \in 11} = E_{c e 11}^{0} - \frac{2 \cdot 303 R T}{2 F} log \frac{1}{[A g +]^{2}} (Nernst's Equation)$

$\begin{matrix} \Rightarrow 0.62 & = 0.80 + 0.06 log [A g +] ∴ log [A g^{+}] & = - \frac{0.18}{0.06} = - 3 . \Rightarrow [A g^{+}] = 10^{- 3} m . \end{matrix}$

0r, $[{A g}^{+}] = (10^{- 3} \times 108) g m / L = 0.108 g m / L$ .

$∴$ Amount of $A g$ in $100 m L$ solution $= \frac{0.108}{10} g m = 0.0108$ .

Thus, $1 .$ of $A g$ in alloy $= \frac{0.0108}{1.08} \times 100 = 1 %$
so, option (D) is correct.

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