NVAMediumJEE Main 2025 · 2 April, Shift 1Nernst Equation

Chemistry Question from JEE Main 2025 · 2 April, Shift 1

Consider the following electrochemical cell at standard condition.

Au(s)QH2,QNH4X(0.01M)Ag+(1M)Ag(s),Ecell=+0.4V\mathrm{Au(s)} \mid \mathrm{QH_2}, \mathrm{Q} \parallel \mathrm{NH_4X}\,(0.01 \, \text{M}) \parallel \mathrm{Ag^+}\,(1 \, \text{M}) \mid \mathrm{Ag(s)}, \qquad E_{cell} = +0.4 \, \text{V}

The couple QH2/Q\mathrm{QH_2}/\mathrm{Q} represents the quinhydrone electrode; the half cell reaction is given below.

Quinhydrone half-cell reaction: quinone (Q) plus 2 electrons plus 2 protons gives hydroquinone (QH2), with standard potential E°(Q/QH2) = +0.7 V.

The pKbpK_b value of the ammonium halide salt (NH4X)(\mathrm{NH_4X}) used here is _____ (nearest integer).

(Given: EAg+/Ag=+0.8VE^\circ_{\mathrm{Ag^+}/\mathrm{Ag}} = +0.8 \, \text{V} and 2.303RTF=0.06V\frac{2.303RT}{F} = 0.06 \, \text{V})

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