Chapter 19, Problem 1KSP

How much copper metal can be produced by electrolysis of a solution containing ${\text{Cu}}^{\text{2+}}$ ions by a current of 1.85 A applied for exactly four hours?

$\begin{array}{l}\text{(a)}\\ \text{0}\text{.276 mol}\\ \text{(b)}\\ 0.138\text{ mol}\\ \text{(c)}\\ \text{0}\text{.552 mol}\\ \text{(d)}\\ \text{5}\text{.14}\times {\text{10}}^{\text{9}}\text{ mol}\\ \text{(e)}\\ \text{2}\text{.00 mol}\end{array}$

Interpretation Introduction

Interpretation:

The number of moles of copper metal deposited during the electrolysis of copper ion solution is to be determined.

Concept introduction:

The relation between the current, time taken and the charge is as follows:

$\text{Current(A)×}\text{Time(s)}=\text{Charge(C)}$

Here, $\text{A}$ is ampere, the S. I. unit of current, $\text{s}$

is second, the S. I. unit of time, and C is coulomb, the S. I. unit of charge.

The relationship between the number of moles of electrons and the number of moles of metal is as follows:

$\text{Number}\text{of}\text{moles}\text{of}\text{metal}=\text{Number}\text{of}\text{moles}\text{of}{\text{e}}^{-}\times \frac{1\text{mol}\text{metal}}{\text{n}\text{mol}{\text{e}}^{-}}$

Answer to Problem 1KSP

Solution: Option (b).

Explanation of Solution

The reduction half-reaction for copper metal is as follows:

${\text{Cu}}^{\text{2+}}\text{+}{\text{2e}}^{-}\to \text{Cu}$

Therefore, two moles of electrons are needed to reduce one mole of copper metal.

The number of moles of copper metal deposited after electrolysis when 1.85 A of current is passed for four hours, is determined as follows:

$\text{Charge (coulomb) = Current (ampere) }\times \text{ time (second)}$

$\left(1.85\text{ A }\times 4.00\text{h}\times \frac{60\min }{1\text{h}}\times \frac{60\text{s}}{1\text{min}}\right)=26640\text{C}$

The number of moles of electrons required to reduce copper metal is calculated as follows:

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{e}}^{-}=\frac{\text{Charge(C)}}{\text{96,500}\text{C/mol}{\text{e}}^{-}}\\ \text{Number}\text{of}\text{moles}\text{of}{\text{e}}^{-}=\left(\frac{26640\text{C}}{\text{96,500}\text{C/mol}{\text{e}}^{-}}\right)\\ \text{Number}\text{of}\text{moles}\text{of}{\text{e}}^{-}=0.2760\text{mol}{\text{e}}^{-}\end{array}$

Now, the number of moles of copper metal reduced is calculated as follows:

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{metal}=\left(0.2760\text{mol}{\text{e}}^{-}\times \frac{1\text{mol}\text{ Cu}}{\text{2}\text{mol}{\text{e}}^{-}}\right)\\ \text{Number}\text{of}\text{moles}\text{of}\text{metal}=0.138\text{mol}\text{Cu}\end{array}$

Hence, option (b) is correct.

Reason for incorrect options:

Option (a) is incorrect because on solving with the help of the above equation, the answers do not match with option (a).

Option (c) is incorrect because on solving with the help of the above equation, the answers do not match with option (c).

Option (d) is incorrect because on solving with the help of the above equation, the answers do not match with option (d).

Option (e) is incorrect because on solving with the help of the above equation, the answers do not match with option (e).

Hence, options (a), (c), (d), and (e) are incorrect.