JEE Main 2020 · ChemistryNumerical answerNumerical valueEasyCalculation

JEE Main 2 September 2020, Shift 1, Chemistry Q47

Question 47 of 75 in this shift, Chemistry question 22 of 25.

The Gibbs energy change (in J) for the given reaction at

[\mathrm{Cu^{2+}}]=[\mathrm{Sn^{2+}}]=1

M and 298 K is :

\mathrm{Cu(s)+Sn^{2+}(aq.)\rightarrow Cu^{2+}(aq.)+Sn(s)}

; (

E^0_{\mathrm{Sn^{2+}|Sn}}=-0.16

E^0_{\mathrm{Cu^{2+}|Cu}}=0.34

V, Take

F=96500\ \mathrm{C\ mol^{-1}}

)

Official answer

96500.00

NTA final key (Sep 2020).

Chapter: Electrochemistry
Topic: Nernst Equation
Idea tested: Gibbs Energy and EMF

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.