Question

A solution of ${\mathrm{CuSO}}_4$ is electrolysed for 10 minutes with a current of 1.5 amperes. what is the mass of copper deposited at the cathode?

Answer 1

Step 1: Given data

Time(t)$=10\min =10\times 60=600\sec$

Current(A)$=1.5\mathrm{A}$

Step 2: Framing the reaction

From the above question, we can form the reaction at cathode as,

$\underset{\mathrm{Copper}\mathrm{ions}}{{\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)}+\underset{\mathrm{Electrons}}{2{\mathrm{e}}^{-}}\to \underset{\mathrm{Copper}}{\mathrm{Cu}\left(\mathrm{s}\right)}$

Thus 2 electrons are transferred here.

Step 3: Finding the mass of copper

The mass of copper we know is $63.5\mathrm{g}{\mathrm{mol}}^{-1}$

W know that,

$$\begin{gathered} \mathrm{Charge}=\mathrm{time}\times \mathrm{current} \\ \Rightarrow \mathrm{Charge}=600\times 1.5 \\ \Rightarrow \mathrm{Charge}=900\mathrm{C} \end{gathered}$$

Now, Mass of copper deposited

$$\begin{gathered} =\frac{\mathrm{Molar}\mathrm{mass}\times \mathrm{Charge}}{\mathrm{electrons}\mathrm{transferred}\times \mathrm{Faradays}\mathrm{constant}} \\ =\frac{{\displaystyle 63.5\times 900}}{{\displaystyle 2\times 96500}} \\ =\frac{{\displaystyle 57150}}{{\displaystyle 193000}} \\ =0.296\mathrm{g} \end{gathered}$$

Therefore, the mass of copper deposited at the cathode is $0.296\mathrm{g}$