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Electrochemistry question

2025 · 2 Apr · Shift 1 · Q23

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Electrochemistry
2025 · 2 Apr · Shift 1 · Q23

Electrochemistry question

2025 · 2 Apr · Shift 1 · Q23

JEE MainChemistryElectrochemistryNumerical+4 / −1

Consider the following electrochemical cell at standard condition.

\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}

The couple

\mathrm{QH}_2 / \mathrm{Q}

represents quinhydrone electrode, the half cell reaction is given below: JEE Main 2025 (Online) 2nd April Morning Shift Chemistry - Electrochemistry Question 8 English

JEE Main 2025 (Online) 2nd April Morning Shift Chemistry - Electrochemistry Question 8 English

\left[\text { Given : } \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^0=+0.8 \mathrm{~V} \text { and } \frac{2.303 \mathrm{RT}}{\mathrm{~F}}=0.06 \mathrm{~V}\right]

The

\mathrm{pK}_{\mathrm{b}}

value of the ammonium halide salt

\left(\mathrm{NH}_4 \mathrm{X}\right)

used here is

\underline{\hspace{2cm}}

. (nearest integer)

Numerical answer

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