If the total electricity required to deposit 1 mole of a metal $M$ is equal to that of 10.7 amperes of…
If the total electricity required to deposit 1 mole of a metal $M$ is equal to that of 10.7 amperes of current for 10 hours. The equivalent weight of the metal is (atomic weight $=M \mathrm{u}$)
$M$
$\frac{M}{2}$
$\frac{M}{3}$
$\frac{M}{4}$
Solution
We know that, $q=i t$, where $q$ is the charge in coulomb, $i$ is the current in amperes and $t$ is the time in seconds.
$q=i t$
$\begin{aligned}
10 \mathrm{~h} & =q=10.7 \mathrm{~A} \times 10 \mathrm{~h} \times 60 \mathrm{~min} \mathrm{~h}^{-1} \times 60 \mathrm{~s} \mathrm{~min}^{-1} \\
& =3,85,200 \mathrm{C}
\end{aligned}$
Since, $96500 \mathrm{C}=1$ mole of electrons
$\therefore 3,85200 \mathrm{C}=4$ moles of electrons
Since, 4 moles of electrons are needed to deposit
1 mole of metal, therefore the charge on metal ion is +4. That is, its valency is 4.
$\text {Equivalent weight }=\frac{\text { Atomic weight }}{\text { Valency }}=\frac{M}{4}$