Question

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

• 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

$=lA\times (595.l\times 3600)s$

$=2142360C$

$\because 96500Cmol$

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

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

Now, in $H_2-O_2$ fuel cell following reaction occurs,

At anode $2H_2(g)+4O^{-}H(aq)⟶4H_{2}O(l)+4e^{-}$

At cathode $O_2(g)+2H_{2}O(l)+4e^{-}⟶4O{H}^{-}(aq)$

Overall reaction $2H_2(g)+O_2(g)⟶2H_{2}O(l)$

Thus, from above reaction it is clear that

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

$\because 22mol$ of $e^{-}\equiv llmol$ of $H_{2}O$

$=(11\times 18)g$ of $H_{2}O$

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

$=198g$ or $mL$ of $H_{2}O$

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

We know that,

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

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