Consider the following half cell reaction Cr₂O₇²⁻(aq) + 6e⁻⁺ 14H⁺(aq) → 2Cr³⁺(aq) + 7H₂O(l). The reaction was conducted with the ratio of [ Cr³⁺ ]² [ Cr₂O₇²⁻ ] = 10⁻⁶. The pH value at which the EMF of the half cell will become zero is _ _ _. (nearest integer value) [Given: standard half cell reduction potential E°_ Cr₂O₇²⁻, H⁺/Cr³⁺ = 1.33 V, 2.303RT/F = 0.059 V.]

Redox Reactions and Electrochemistry: JEE Main 2025 Chemistry question

JEE Main 2025 · 8 Apr, Evening shift · Q75

JEE Main 20258 Apr, Evening shiftChemistryHardRedox Reactions and Electrochemistry138× since 2002

Consider the following half cell reaction

\mathrm{Cr_2O_7^{2-}(aq) + 6e^- + 14H^+(aq) \rightarrow 2Cr^{3+}(aq) + 7H_2O(l)}

. The reaction was conducted with the ratio of

\frac{[\mathrm{Cr^{3+}}]^2}{[\mathrm{Cr_2O_7^{2-}}]} = 10^{-6}

. The pH value at which the EMF of the half cell will become zero is ______. (nearest integer value) [Given: standard half cell reduction potential

E^\circ_{\mathrm{Cr_2O_7^{2-}, H^+/Cr^{3+}}} = 1.33

\frac{2.303RT}{F} = 0.059

V.]

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2025 Redox Reactions and Electrochemistry Question with Solution (8 Apr Evening shift, Q75) | JEE Arc