Question

A lead storage battery containing 5.0L of $H_{2}S{O}_4$ solution is operated for $9.65\times {10}^{5}s$ with a steady current of 100 mA. Assuming volume of the solution remains constant, normality of $H_{2}S{O}_4$ will :

• A. increase by unity
• B. increase by 0.20
• C. decrease by 0.40
• D. remain unchanged

Answer 1

The correct option is C

decrease by 0.40

The reactions at cathode and anode are:

Anode :
$P{b}_{\left(s\right)}+H_{2}S{O}_{4\left(aq\right)}\to PbS{O}_{4\left(S\right)}+2H_{\left(aq\right)}^{+}+2e^{-}$
Cathode :
$Pb{O}_{2\left(s\right)}+H_{2}S{O}_{4\left(aq\right)}+2H^{+}+2e^{-}\to PbS{O}_4\left(s\right)+2H_{2}O$

From Faraday's Law,

No. of moles $=\frac{m}{M}=\frac{Q}{FZ}$
Hence, equiv. of $H_{2}S{O}_4$ consumed
$=\frac{4\times 100\times {10}^{-3}\times 9.65\times {10}^5}{2\times 96500}=2;$
decrease in normality $=\frac{2}{5}$