Calculate the time required in second to deposit $6.35 \mathrm{~g}$ copper from its salt solution by passing…

Calculate the time required in second to deposit $6.35 \mathrm{~g}$ copper from its salt solution by passing 5 ampere current. [Molar mass of $\mathrm{Cu}=63.5 \mathrm{~g} \mathrm{~mol}^{-1}$ ]
  1. 3600
  2. 3700
  3. 3860
  4. 4000

Solution

$\begin{aligned} & \mathrm{Cu}_{(\mathrm{s})}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}_{(\mathrm{s})} \\ & \text { Mole ratio }=\frac{1 \mathrm{~mol}}{2 \mathrm{~mole}^{-}} \\ & \mathrm{W}=\frac{\mathrm{I}(\mathrm{A}) \times \mathrm{t}(\mathrm{s})}{96500\left(\mathrm{C} / \mathrm{mol} \mathrm{e}^{-}\right)} \times \text {mole ratio } \times \text { molar mass } \\ & 6.35 \mathrm{~g}=\frac{5 \times \mathrm{t}}{96500\left(\mathrm{C} / \mathrm{mol} \mathrm{e}^{-}\right)} \times \frac{1 \mathrm{~mol}}{2 \mathrm{~mol} \mathrm{e}} \times 63.5 \mathrm{~g} \mathrm{~mol}^{-1} \\ & \mathrm{t}=\frac{6.35 \times 96500 \times 2}{5 \times 63.5}=3860 \text { seconds }\end{aligned}$

Asked in: MHT CET 2023 (13 May Shift 1)

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