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Iron powder is added to 1.0 M solution of $C d C l_{2}$ at 298 K. The reaction occurring is $C d^{2 +} (a q) + F e (s) \to C d (s) + F e^{2 +} (a q) .$ If the standard potential of a cell producing this reaction is 0.037 V, the concentrations of $C d^{2 +}$ and $F e^{2 +}$ ions in the above reaction at equilibrium respectively will be (Hint:- $10^{1.3} = 20$ )

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0.05 M, 0.95 M

0.95 M, 0.05 M

0.40 M, 0.60M

0.60 M, 0.40 M

answer is A.

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Detailed Solution

We have $C d^{2 +} \underset{1.0 M - x}{(a q)} + F e (s) ⇌ C d (s) + \underset{x}{F e^{2 +} (a q)}$

Hence $E = E^{\circ} - \frac{R T}{2 F} \ln \frac{[F e^{2 +}]}{[C d^{2 +}]}$

At equilibrium E = 0. Hence $E H^{o} = \frac{R T}{2 F} \ln \frac{[F e^{2 +}]}{[C d^{2 +}]}$ i.e. $\frac{x}{1.0 M - x} = a n t i \log (\frac{E^{\circ}}{R T / 2 F})$

or $\frac{x}{1.0 M - x} = a n t i \log (\frac{0.037 V}{0.0295 V}) = 10^{1.254} ≃ 18$ or $x = \frac{18}{19} M = 0.95 M$

Thus, $[F e^{2 +}] = 0.95 M$ and $C d^{2 +} (a q) = 0.05 M$

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