JEE Chemistry 2025 Question with Solution

Given: The cell potential is $E_{cell}=+0.4\, \text{V}$. The standard potential of the quinhydrone electrode is $E^\circ_{QH/Q}=+0.7\, \text{V}$. The silver electrode is $Ag^+(1\, \text{M})|Ag$, so $E_{Ag^+/Ag}=E^\circ_{Ag^+/Ag}=+0.8\, \text{V}$.

Find: The $pK_b$ value corresponding to the ammonium halide salt in the quinhydrone half-cell.

The overall cell reaction is

For the quinhydrone electrode, since quinone and hydroquinone are present together, the potential depends on $[H^+]$.

First calculate the anode potential using

So,

For the quinhydrone electrode,

which gives

Substituting $E_{anode}=0.4\, \text{V}$,

Now use the hydrolysis of ammonium ion:

With concentration $C=0.01\, \text{M}=10^{-2}\, \text{M}$ and $[H^+]=10^{-5}\, \text{M}$,

Assuming small hydrolysis,

Using

Therefore,

So, the correct option is A.

The solution gives the final value as $6$, and this matches option A.

Given: The cathode is the silver electrode and the anode is the quinhydrone electrode.

Find: Convert the cell potential into $\text{pH}$, then use salt hydrolysis to obtain $pK_b$.

Because $[Ag^+]=1\, \text{M}$, the cathode potential remains its standard value:

Then,

For the quinhydrone electrode,

Hence,

Since $[Q]=[QH_2]$,

Therefore,

Now,

For ammonium salt solution of concentration $10^{-2}\, \text{M}$,

Then,

So,

Thus the final answer is $6$.