Question

A current of $9.65A$ is drawn from a Daniell cell for exactly one hour. If molar masses of Cu and Zn are $63.5g{mol}^{-1}$ respectively, the loss in mass at anode and gain in mass at cathode respectively are:

• A. $11.43g$ and $11.77g$
• B. $11.77g$ and $11.43g$
• C. $22.86g$ and $23.54g$
• D. $23.54g$ and $22.86g$

Answer 1

Answer: (B)

$11.77g$ and $11.43g$

The reaction occured in Daniel cell is-

$Zn+{Cu}^{2+}⟶{Zn}^{2+}+Cu$

Zinc is lost at anode and Cu is deposited at the cathode.

Quantity of electricity passed, $\left(q\right)=I\times t_{\left(\text{in s}\right)}$

whereas,

$I=$ Current drawn $=9.65A$

$t=$ time $=1h=60\times 60=3600s$

$\therefore q=9.65\times 3600=34740C$

Amount of electron withdrawn $=\frac{34740}{96500}=0.36\text{moles}$

Since two electrons are involved in the reduction and oxidation reactions.

$\therefore$ Amount of $Zn$ lost at anode or $Cu$ deposited at cathode $=\frac{0.36}{2}=0.18\text{moles}$

Given that:-

Molar mass of $Cu=63.5g{mol}^{-1}$

Molar mass of $Zn=65g{mol}^{-1}$

$\therefore$ Mass of $Zn$ lost at anode $=0.18\times 65.4=11.77g$

Mass of $Cu$ deposited at cathode $=0.18\times 63.5=11.43g$

Hence, the loss in mass at anode and gain in mass at the cathode are $11.77g$ and $11.43g$, respectively.