Question

One Faraday of current was passed through the electrolytic cells placed in series containing a solution of $A{g}^{+},N{i}^{2+}$ and $C{r}^{3+}$ respectively. The ratio of amounts of $Ag,Ni$ and $Cr$ deposited will be:

$[At.wt.$ of $Ag=108,Ni=59,Cr=52]$.

• A. $108:29.5:17.4$
• B. $17.4:29.5:108$
• C. $1:2:3$
• D. $108:59:52$

Answer 1

The correct option is A $108:29.5:17.4$

For deposition of $Ag$, reaction is $A{g}^{+}+e^{-}\to Ag$
Thus, $1$ F deposites Ag$=1$ mol $=108$ g

For deposition of $Ni$, the reaction is $N{i}^{2+}+2e^2\to Ni$
Thus, $2$ F depositis $Ni$$=1$ mol$=59$ g
$\therefore$ $1$ F deposits $Ni=0.5$ mol$=29.5$ g

For deposition of $Cr$, the reaction is $C{r}^{3+}+3e^{-}\to Cr$
Thus, $3$F deposits $Cr$ $=1$ mol
$\therefore$ $1$ F deposits $Cr=0.33$ mol$=17.4$ g.

The ratio of amounts of Ag, Ni and Cr deposited will be$:108:29.5:17.4$

Hence, option A is correct.