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Byju's Answer

Standard XII

Chemistry

Redox Reactions and Couple

One Faraday o...

Question

One Faraday of current was passed through the electrolytic cell placed in series containing solution of $A g^{+}, N i^{+ +}$ and $C r^{+ + +}$ respectively. The amounts of $A g$ (at. wt. = 108), $N i$ (at. wt. = 59) and $C r$ (at. wt. = 52) deposited will be:

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Solution

Reactions involved to deposit $A g, N i$ and $C r$ on the electrodes :
${A g}^{+} + e^{-} ⟶ A g$
${N i}^{+ +} + 2 e^{-} ⟶ N i$
${C r}^{+ + +} + 3 e^{-} ⟶ C r$
$1 F, 2 F$ and $3 F$ of charge will be required to deposit $108$ gm of $A g, 58.7$ gm of $N i$ and $52$ gm of $C r$ respectively.
Therefore if one Faraday $(1 F)$ was passed, The amount of $A g$ deposited will be $108$ gm.
The amount of $N i$ deposited will be $\frac{58.7}{2} g m$ i.e, $29.35$ gm
The amount of $C r$ deposited will be $\frac{52}{3}$ gm i.e. $17.33$ gm.

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\begin{matrix} A g (i n g m) & N i (i n g m) & C r (i n g m) a) & 108 & 59 & 52 b) & 108 & 29.5 & 17.3 c) & 36 & 19.6 & 17.3 d) & 108 & 108 & 108 \end{matrix}

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One Faraday of current was passed through the electrolytic cell placed in series containing solution of Ag+,Ni++ and Cr+++ respectively. The amounts of Ag(at. wt. = 108), Ni(at. wt. = 59) and Cr(at. wt. = 52) deposited will be:

One Faraday of current was passed through the electrolytic cell placed in series containing solution of $A g^{+}, N i^{+ +}$ and $C r^{+ + +}$ respectively. The amounts of $A g$ (at. wt. = 108), $N i$ (at. wt. = 59) and $C r$ (at. wt. = 52) deposited will be: