Consider the following electrochemical cell at standard condition.

$$Au(s)|QH_2,Q|NH_4X(0.01M)||Ag^+(1M)|Ag(s)$$

$$E_{cell} = +0.4V$$

The couple $$QH_2/Q$$ represents quinhydrone electrode, the half cell reaction is given below:

[Given: $$E^o_{Ag^+/Ag} = +0.8V$$ and $$\frac{2.303RT}{F} = 0.06V$$]

The $$pK_b$$ value of the ammonium halide salt ($$NH_4X$$) used here is ______. (nearest integer)

The quinhydrone electrode acts as the anode and the silver electrode acts as the cathode.

Anode reaction

$$QH_2 \rightarrow Q+2H^++2e^-$$

$$E^\circ_{Q/QH_2}=+0.7\text{ V}$$

Cathode reaction

$$2Ag^++2e^- \rightarrow 2Ag$$

$$E^\circ_{Ag^+/Ag}=+0.8\text{ V}$$

The standard cell potential is

$$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$$

$$E^\circ_{cell}=0.8-0.7=0.1\text{ V}$$

The overall cell reaction is

$$QH_2+2Ag^+ \rightarrow Q+2H^++2Ag$$

For the quinhydrone electrode, $$[Q]=[QH_2]$$, therefore they cancel out in the reaction quotient. Also, $$[Ag^+]=1\text{ M}$$.

The Nernst equation becomes

$$E_{cell}=E^\circ_{cell}-\frac{0.06}{2}\log[H^+]^2$$

Substituting the given values,

$$0.4=0.1-0.03\log[H^+]^2$$

$$0.4=0.1-0.06\log[H^+]$$

Since

$$pH=-\log[H^+]$$

we get

$$0.4=0.1+0.06(pH)$$

$$0.3=0.06(pH)$$

$$pH=5$$

The solution in the anodic compartment contains $$NH_4X$$, a salt of a weak base and a strong acid.

For such a salt,

$$pH=7-\frac{1}{2}pK_b-\frac{1}{2}\log C$$

Given,

$$pH=5$$

$$C=0.01\text{ M}=10^{-2}\text{ M}$$

Substituting,

$$5=7-\frac{1}{2}pK_b-\frac{1}{2}\log(10^{-2})$$

$$5=7-\frac{1}{2}pK_b+1$$

$$5=8-\frac{1}{2}pK_b$$

$$\frac{1}{2}pK_b=3$$

$$pK_b=6$$

Therefore,

$$\boxed{pK_b=6}$$