Cu(s) + Sn²⁺ (0.001M) → Cu²⁺ (0.01M) + Sn(s) The Gibbs free energy change for the above reaction at 298 K is x × 10⁻¹ kJ mol⁻¹. The value of x is _ _ _ _ _. [nearest integer] [Given: E_ C u²⁺ /Cu ^Θ = 0.34 V; E_ S n²⁺ /Sn ^Θ = - 0.14 V; F = 96500 C mol⁻¹]

Redox Reactions and Electrochemistry: JEE Main 2022 Chemistry question

JEE Main 2022 · 26 Jun, Evening shift · Q10

JEE Main 202226 Jun, Evening shiftChemistryMediumRedox Reactions and Electrochemistry138× since 2002

Cu(s) + Sn²⁺ (0.001M)

\to

Cu²⁺ (0.01M) + Sn(s) The Gibbs free energy change for the above reaction at 298 K is x

\times

10^

-

1 kJ mol^

-

1. The value of x is __________. [nearest integer] [Given :

E_{C{u^{2 + }}/Cu}^\Theta = 0.34\,V

;

E_{S{n^{2 + }}/Sn}^\Theta = - 0.14\,V

; F = 96500 C mol^

-

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2022 Redox Reactions and Electrochemistry Question with Answer (26 Jun Evening shift, Q10) | JEE Arc