What is the charge required for the reduction of two moles of $\mathrm{Cu}^{2+}$ to $\mathrm{Cu}$ ?

What is the charge required for the reduction of two moles of $\mathrm{Cu}^{2+}$ to $\mathrm{Cu}$ ?
  1. $2.89 \times 10^5 \mathrm{C}$
  2. $1.93 \times 10^5 \mathrm{C}$
  3. $9.65 \times 10^5 \mathrm{C}$
  4. $3.86\times 10^5 \mathrm{C}$

Solution

$\begin{aligned} & \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu} \\ & 1 \mathrm{~mole} \quad 2 \mathrm{~mole}^{-} \\ & 2 \mathrm{~mole} \quad 4 \mathrm{~mole}^{-} \\ & \text {Charge of } 1 \mathrm{~mole} \text { electrons }=96500 \mathrm{C} \\ & \text { Charge required for the reduction of two moles of } \mathrm{Cu}^{2+} \\ & =4 \times 96500=3.86 \times 10^5 \mathrm{C}\end{aligned}$

Asked in: MHT CET 2021 (24 Sep Shift 2)

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