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JEE Main 2-April-2025 Paper - Shift 1 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 2-April-2025 Paper - Shift 14 Marks
Question
Consider the following electrochemical cell at standard condition.
$\begin{array}{l}Au(s)\left|QH_2, Q\right| NH_4 X(0.01 M)| | Ag^{+}(1 M) \mid Ag(s) \\
E_{\text {cell }}=+0.4 V\end{array}$
The couple $QH _2 / Q$ represents quinhydrone electrode, the half cell reaction is given below
Image
$\left[\right.$ Given : $E _{ Ag ^{+} / Ag }^{ o }=+0.8 V$ and $\left.\frac{2.303 RT }{ F }=0.06 V\right]$
The $pK _{ b }$ value of the ammonium halide salt $\left( NH _4 X \right)$ used here is __________. (nearest integer)
✓

Answer

(6)
Explanation: $QH _2+2 Ag ^{+} \rightarrow 2 Ag + Q +2 H ^{+}$
$E = E ^{\circ}-\frac{0.06}{2} \log \left[ H ^{+}\right]^2$
$E = E ^{\circ}-0.06 \times \log \left[ H ^{+}\right]$
$pH =-\log \left( H ^{+}\right)=\frac{ E - E ^{ o }}{0.06}=\frac{0.4-0.1}{0.06}$
$=\frac{0.3}{0.06}=5$
$\begin{array}{l} pH + NH _4 X =7-\frac{1}{2} pK _{ b }-\frac{1}{2} \log C \\
5=7-\frac{1}{2} \times pK _{ b }-\frac{1}{2} \log \left(10^{-2}\right) \\ pK _{ b }=6 .\end{array}$

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