Question

The same amount of electricity was passed through two separate electrolytic cells containing solutions of nickel nitrate $\left[Ni{\left({NO}_3\right)}_2\right]$ and chromium nitrate $\left[Cr{\left({NO}_3\right)}_3\right]$ respectively. If $0.3g$ of nickel was deposited in the first cell, the amount of chromium deposited is :
$(at.wt.$ of $Ni=59,$ $at.wt.$ of $Cr=52)$

• A. $0.1g$
• B. $0.17g$
• C. $0.3g$
• D. $0.6g$

Answer 1

The correct option is B $0.17g$

$\left[Ni\right(N{O}_3{)}_2]N{i}^{2+}+2e^{-}\to Ni\Rightarrow 0.3g$
$nn+2(-1)=0$
$n=+2$
$\left[Cr\right(N{O}_3{)}_3]C{r}^{3+}+3e^{-}\to Cr$
$x+3(-1)=0$
$x=+3$

Let the electricity charge passed be ${}^{\prime }{c}^{\prime }$
moles of electric charge $=\frac{c}{96500}$
For $Ni\frac{C}{96500}=\frac{Massdiposited}{equivalentmass}=\frac{0.3}{58.7/2}...\left(I\right)$

For $Cr\frac{C}{96500}=\frac{x}{52/3}....\left(II\right)$

Comparing (I) & (II)

$\frac{0.3}{58.7/2}=\frac{x}{52/3}$

$\frac{0.3\times 2}{58.7}=\frac{3x}{52}$

$\frac{0.6}{58.7}\times \frac{52}{3}=x$

$0.17=x$