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JEE Main 20252 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2025 · 2 April · Morning Shift · Chemistry · Electrochemistry

Consider the following electrochemical cell at standard condition....

Consider the following electrochemical cell at standard condition.Au(s)∣QH2,Q∣NH4X(0.01M)∣∣Ag+(1M)∣Ag(s)&Ecell =+0.4V\begin{aligned}& \mathrm{Au}(\mathrm{s})\left|\mathrm{QH}_2, \mathrm{Q}\right| \mathrm{NH}_4 \mathrm{X}(0.01 \mathrm{M})| | \mathrm{Ag}^{+}(1 \mathrm{M}) \mid \mathrm{Ag}(\mathrm{s}) \& \mathrm{E}_{\text{cell }}=+0.4 \mathrm{ V}\end{aligned})The couple (\mathrm{QH}_2 / \mathrm{Q}) represents quinhydrone electrode,thethehalf cell reaction is given belowThe (\mathrm{pK}_{\mathrm{b}}) value of the ammonium halide salt (\left(\mathrm{NH}_4 \mathrm{X}\right)) used here is _________. (nearest integer)$

Correct answer

0

Step-by-step solution

<br/>QH2+2Ag+→2Ag+Q+2H+<br>E=E∘−0.062log⁡[H+]2<br>E=E∘−0.06×log⁡[H+]<br>pH=−log⁡(H+)=E−Eo0.06=0.4−0.10.06<br>=0.30.06=5<br/>\begin{aligned}<br />& \mathrm{QH}_2+2 \mathrm{Ag}^{+} \rightarrow 2 \mathrm{Ag}+\mathrm{Q}+2 \mathrm{H}^{+} \\<br>& \mathrm{E}=\mathrm{E}^{\circ}-\frac{0.06}{2} \log \left[\mathrm{H}^{+}\right]^2 \\<br>& \mathrm{E}=\mathrm{E}^{\circ}-0.06 \times \log \left[\mathrm{H}^{+}\right] \\<br>& \mathrm{pH}=-\log \left(\mathrm{H}^{+}\right)=\frac{\mathrm{E}-\mathrm{E}^{\mathrm{o}}}{0.06}=\frac{0.4-0.1}{0.06} \\<br>& =\frac{0.3}{0.06}=5<br />\end{aligned})
pH+NH4X=7−12pKb−12log⁡C5=7−12×pKb−12log⁡(10−2)pKb=6\begin{aligned} & \mathrm{pH}+\mathrm{NH}_4 \mathrm{X}=7-\frac{1}{2} \mathrm{pK}_{\mathrm{b}}-\frac{1}{2} \log \mathrm{C} \\ & 5=7-\frac{1}{2} \times \mathrm{pK}_{\mathrm{b}}-\frac{1}{2} \log\left(10^{-2}\right) \\ & \mathrm{pK}_{\mathrm{b}}=6\end{aligned})
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