Question

A current of $9.65$ A is drawn from a Daniel cell for exactly one hour. Given molar masses of $Cu$ and $Zn$ a $63.5$ g $mo{l}^{-1}$ and $65.4$ g $mo{l}^{-1}$ respectively, the decreases in the mass of anode and increase in the mass of cathode will be respectively:

• A. $11.43$g, $11.77$g
• B. $11.77$g, $11.43$g
• C. $22.86$g, $23.54$g
• D. $23.45$g, $22.86$g

Answer 1

The correct option is B $11.77$g, $11.43$g

The reaction occurring in Daniell cell is $Zn+{Cu}^{2+}\to {Zn}^{2+}+Cu$ Zine is lost at anode and $Cu$ is deposited at cathode amount of current with drawn is $\left(\left(9.65A\right)(60\times 60S)/96500C{mol}^{-1}\right)$, which is $0.36$ mol, Since two electrons are involved in the reduction and oxidation reaction, $0.18$ ml of $Zn$ is lost at anode and $0.18$ mol of $Cu$ is deposited at cathode. The masses are $11.77g$ and $11.43g$ for $Zn$ and $Cu$ respectively.