Question

$0.05F$ 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 $303g/mol$) ?

• A. $30.3g$
• B. $15.5g$
• C. $7.6g$
• D. $60.6g$

Answer 1

The correct option is C

$7.6g$

Step 1: Charge$=0.05F$

Step 2: Amount of ${\mathrm{PbSO}}_4$ precipitated $=W$

Step 3: Molar mass of ${\mathrm{PbSO}}_4$ $303g/mol$

According to Faraday’s 1st law of electrolysis:

$1^{st}$method:

$W=E\times Q$

$E=M/2$

$W=\left(303\times 0.05\right)/2=7.6g$

$2^{nd}$ method:

${\mathrm{Pb}}^{2+}+{{\mathrm{SO}}_4}^{2-}\to {\mathrm{PbSO}}_4$

For $2F$ current passed, ${\mathrm{PbSO}}_4$ deposited$=303g/mol$

For $0.05F$current passed, ${\mathrm{PbSO}}_4$ deposited$=W$

$W=\left(303\times 0.05\right)/2=7.6g$

Therefore, the amount of ${\mathrm{PbSO}}_4$ precipitated $=7.6g$