Question

$0.05\mathrm{F}$ charge is passed through a lead storage battery. In the anodic reaction, what is the amount of ${\mathrm{PbSO}}_4$ precipitated (Molar mass of ${\mathrm{PbSO}}_4$ is $303\mathrm{g}/\mathrm{mol}$)?

Answer 1

Step 1: Given data:

Amount of charge passed through a lead storage battery is $0.05\mathrm{F}$

Step 2: Find the reaction that takes place at the anode:

$\mathrm{Pb}+{\mathrm{SO}}_4^{-2}\to {\mathrm{PbSO}}_4+2{\mathrm{e}}^{-}\left[\because \mathrm{At}\mathrm{anode}\mathrm{oxidation}\mathrm{takes}\mathrm{place}\right]$

Step 3: Find the number of moles of ${\mathbf{PbSO}}_{\mathbf{4}}$when $\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{}\mathbf{F}$ charge is passed:

For $1{\mathrm{e}}^{-}$ the charge will be $1\mathrm{F}$

But it is given that the charge is $0.05\mathrm{F}$

$\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{PbSO}}_4\mathrm{formed}=\frac{0.05}{2}\mathrm{mol}$

Step 4: Find the mass of ${\mathbf{PbSO}}_{\mathbf{4}}$ formed:

$$\begin{gathered} \mathrm{Mass}\mathrm{of}{\mathrm{PbSO}}_4\mathrm{formed}=\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{PbSO}}_4\times \mathrm{Molecular}\mathrm{weight}\mathrm{of}{\mathrm{PbSO}}_4 \\ \Rightarrow \mathrm{Mass}\mathrm{of}{\mathrm{PbSO}}_4\mathrm{formed}=\frac{0.05}{2}\mathrm{mol}\times 303\mathrm{g}{\mathrm{mol}}^{-1} \\ \Rightarrow \mathrm{Mass}\mathrm{of}{\mathrm{PbSO}}_4\mathrm{formed}=7.57\mathrm{g} \end{gathered}$$

Hence, the amount of ${\mathbf{PbSO}}_{\mathbf{4}}$ precipitated is ${\mathbf{W}}_{{\mathbf{PbSO}}_{\mathbf{4}}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{7}\mathbf{.}\mathbf{57}\mathbf{}\mathbf{g}$