Question

In a hydrogen oxygen fuel cell, electricity is produced. In this process $H_2\left(\text{g}\right)$ is oxidised at anode and $O_2\left(\text{g}\right)$ reduced at cathode. Given : Cathode: $O_2\left(g\right)+2H_{2}O\left(l\right)+4e^{-}\to 4O{H}^{-}\left(aq\right)$ Anode : $H_2\left(g\right)+2O{H}^{-}\left(aq\right)\to 2H_{2}O\left(l\right)+2e^{-}$ $4.48\text{L}$ of $H_2$ at $1\text{atm}$ and $273\text{K}$ is oxidised in $9650\text{sec}$.
If current produced in fuel cell, use for the deposition of $C{u}^{2+}$ in $1\text{L}$, $2\text{M}CuS{O}_4\left(\text{aq}\right)$ solution for $241.25\text{sec}$ using $Pt$ electrode. The $pH$ of solution after electrolysis is:

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

If $CuS{O}_4\left(aq\right)$ solution is electrolysed using Pt electrodes then the reaction that takes place is:
At Anode: $2O{H}^{-}\to H_{2}O+\frac{1}{2}{O}_2+2e^{-}$
At Cathode: $C{u}^{+2}+2e^{-}\to Cu$
And $H_{2}S{O}_4$ will be left in the electrolyte.
$\therefore$
The overall reaction will be:
$C{u}^{+2}+2O{H}^{-}\to H_{2}O+Cu+\frac{1}{2}{O}_2$
OR
$CuS{O}_4+H_{2}O\to Cu+\frac{1}{2}{O}_2+H_{2}S{O}_4$
Now, from the overall cell reaction we can say that moles of Cu formed are equal to the moles of $H_{2}S{O}_4$ formed.
So, Let’s first calculate the moles of Cu formed.
Using Faraday’s equation:
$mole{s}_{Cu}=\frac{1\times i\times t}{n\times 96500}$
Putting values in the above equation:
$No.ofmolesofCu$ =$\frac{1\times 4\times 241.25}{2\times 96500}$
$No.ofmolesofCu=5\times {10}^{-3}$
$\therefore$
Moles of $H_{2}S{O}_4$ which is equal to moles of Cu = $5\times {10}^{-3}$
Hence, moles of $H^{+}$ will be twice the moles of $H_{2}S{O}_4$ = ${10}^{-2}$
Since, Volume = $1L$
$\therefore$
Molarity of $H^{+}$ = $\frac{moles}{volume}={10}^{-2}$
So, pH = -log$\left[H^{+}\right]$
= -log$\left[{10}^{-2}\right]$
= $2$