Consider the following electrochemical cell at standard condition. Au(s)|QH₂, Q|NH₄X(0.01 M)||Ag⁺(1 M)|Ag(s) E cell = +0.4 V. The couple QH₂/Q represents quinhydrone electrode, the half cell reaction is given below: Q + 2e⁻⁺ 2H⁺ → QH₂ (Q = p-benzoquinone, QH₂ = hydroquinone), E° Q/QH₂ = +0.7 V. [Given: E° Ag⁺/Ag = +0.8 V and 2.303RT/F = 0.06 V]. The pK b value of the ammonium halide salt (NH₄X) used here is _ _ _. (nearest integer)

Redox Reactions and Electrochemistry: JEE Main 2025 Chemistry question

JEE Main 2025 · 2 Apr, Morning shift · Q73

JEE Main 20252 Apr, Morning shiftChemistryHardRedox Reactions and Electrochemistry138× since 2002

Consider the following electrochemical cell at standard condition.

\mathrm{Au(s)|QH_2, Q|NH_4X(0.01\,M)||Ag^+(1\,M)|Ag(s)}

E_{cell} = +0.4

V. The couple

\mathrm{QH_2/Q}

represents quinhydrone electrode, the half cell reaction is given below :

\mathrm{Q + 2e^- + 2H^+ \rightarrow QH_2}

(Q = p-benzoquinone,

\mathrm{QH_2}

= hydroquinone),

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

V. [Given :

E^\circ_{Ag^+/Ag} = +0.8

V and

\frac{2.303RT}{F} = 0.06

V]. The

\mathrm{p}K_b

value of the ammonium halide salt (

\mathrm{NH_4X}

) used here is ______. (nearest integer)

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2025 Redox Reactions and Electrochemistry Question with Solution (2 Apr Morning shift, Q73) | JEE Arc