Question

A 55.5kg slab of crude copper from a smelter has a copper content of 98.3%. Estimate the time required to purify it electrochemically if it is used as the anode in a cell that has acidic copper(III) sulfate as its electrolyte and a current $\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{A}$is passed through cell.

Short answer

Time required to purify copper is $8.3\times {10}^{4}sec$.

Step-by-step solution

Step (1):- Mass of copper:-

We can find copper mass by using following method:-

$$\begin{gathered} Mass of copper=55.4kg\times \frac{1000g}{1kg}\times \frac{98.3}{100} \\ = 5.46X{10}^{4}gm \end{gathered}$$

Step (2):- Number of e- involve in it:-

Anode reaction:- $Cu\left(s\right)\to C{u}^{2+}\left(s\right)+2e^{-}$

Molecular mass of copper is$\mathbf{63}\mathbf{.}\mathbf{5}\mathit{g}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}$, so the number of moles of electron that passes through cell can be obtained by following equation:-

Mass of Cu=$n\times \frac{1mol.Cu}{2Mol.e^{-}}\times$molecular mass of Cu

$$\begin{gathered} 54.6\times {10}^{3}gCu=n\times \frac{1mol.Cu}{2mol.e^{-}}\times 63.5g/mol \\ n=\frac{54.6\times {10}^{3}gCu\times 2mol.e^{-}}{63.5g/mol\times 1molCu} \\ =1.72\times {10}^{3}Moles \end{gathered}$$

$\mathit{n}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{72}\mathit{X}{\mathbf{10}}^{\mathbf{3}}\mathit{m}\mathit{o}\mathit{l}\mathbf{.}{\mathit{e}}^{\mathbf{-}}$

Step (3):- Charge Calculation

Since,

$$\begin{gathered} n=\frac{Q}{96,485C/mol} \\ Q=n\times 96,485C/mol \end{gathered}$$

$=1.72\times {10}^3\times 96,485C/mol$

$\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{6}\mathit{X}{\mathbf{10}}^{\mathbf{8}}\mathit{C}$

Step (4):- Time Calculation

Time can be calculated using following method:-

$\text{Q = It}$

$i=2\times {10}^{3}A$

$$\begin{gathered} t=\frac{Q}{i}=\frac{1.66\times {10}^{8}C}{2\times {10}^{3}A} \\ t=8.3\times {10}^{4}C/A \end{gathered}$$

=$\mathbf{8}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathit{s}\mathit{e}\mathit{c}$

Hence, time required to purify ‘Cu’ is $8.3\times {10}^4\sec$.