Question

Calculate the molarity of a solution containing $5\, g$ of NaOH dissolved in the product of a $H_2 - O_2$ fuel cell operated at $1\, A$ current for 595.1 hours. (Assume $1\, F = 96500 C / mol$ of electrons and molecular weight of NaOH as $49\, g\, mol^{-1}$)

• A. 0.05M
• B. 0.025M
• C. 0.1 M
• D. 0.075 M
• E. none of these

Answer 1

Answer: (E)

Total charge produced by cell

$= l A \times(595 . l \times 3600)\, s$

$=2142360\, C$

$\because 96500\,C mol$

$\because 2142360\, C = \frac{2142360}{96500}$

$= 22.20\, mol$ of $e^{-}$

Now, in $H _{2}- O _{2}$ fuel cell following reaction occurs,

At anode $2 H _{2}(g)+4 O ^{-} H (a q) \longrightarrow 4 H _{2} O (l)+4 e^{-}$

At cathode $O _{2}(g)+2 H _{2} O (l)+4 e^{-} \longrightarrow 4 OH ^{-}(a q)$

Overall reaction $2 H _{2}(g)+ O _{2}(g) \longrightarrow 2 H _{2} O (l)$

Thus, from above reaction it is clear that

2 mol of $e^{-} \equiv 1 mol$ of $H _{2} O$

$\because 22 mol$ of $e^{-} \equiv l l mol$ of $H _{2} O$

$=(11 \times 18)\, g$ of $H _{2} O$

[$\because$ No. of moles $=\frac{\text { Weight}}{\text { Molecular weight}}$]

$=198\, g$ or $mL$ of $H _{2} O$

Now, number of mol of $NaOH =\frac{5}{40}=0.125\, mol$

We know that,

Molarity $=\frac{\text { No. of moles of solute }}{\text { Volume of solution (in L) }}$

$=\frac{0.125}{198} \times 1000=0.63\, M$