Calculate current in ampere required to deposit $4.8 \mathrm{~g} \mathrm{Cu}$ from it's salt solution in 30…

Calculate current in ampere required to deposit $4.8 \mathrm{~g} \mathrm{Cu}$ from it's salt solution in 30 minutes. $\left[\right.$ Molar mass of $\left.\mathrm{Cu}=63.5 \mathrm{~g} \mathrm{~mol}^{-1}\right]$
  1. 8.1 ampere
  2. 6.4 ampere
  3. 10.5 ampere
  4. 12.3 ampere

Solution

$\begin{aligned} & \mathrm{Cu}_{(\mathrm{s})}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}_{(\mathrm{s})} \\ & \text { Mole ratio }=\frac{1 \mathrm{~mol}}{2 \mathrm{~mole}^{-}} \\ & \mathrm{W}=\frac{\mathrm{I}(\mathrm{A}) \times \mathrm{t}(\mathrm{s})}{96500\left(\mathrm{C} / \mathrm{mol} \mathrm{e}^{-}\right)} \times \text {mole ratio } \times \text { molar mass } \\ & 4.8 \mathrm{~g}=\frac{\mathrm{I}(\mathrm{A}) \times 30 \times 60}{96500\left(\mathrm{C} / \mathrm{mol} \mathrm{e}^{-}\right)} \times \frac{1 \mathrm{~mol}}{2 \mathrm{~mol} \mathrm{e}^{-}} \times 63.5 \mathrm{~g} \mathrm{~mol}^{-1} \\ & \mathrm{I}(\mathrm{A})=\frac{4.8 \times 96500 \times 2}{63.5 \times 30 \times 60}=8.1 \mathrm{~A}\end{aligned}$

Asked in: MHT CET 2023 (09 May Shift 1)

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