The standard Gibbs free energy change of the reaction shown below is -2.7 {kJ} . {mol}^{-1}. {Sn}(s)+P b^{2+}=S n^{2+}+P b(s) Given that E^\circ(P b^{2+} \/ P b)=-0.126 {V}, Find the value of magnitude of E^\circ({Sn}^{2+} \/ {Sn})

JEE Advanced 2021 Full Test 5 Paper 1

SECTION – C (Numerical Answer Type)
This section contains 06 questions. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value (in decimal notation, truncated/rounded‐off to the second decimal place; e.g. XXXXX.XX).

Question : 32 of 54

Marks: +1, -0

The standard Gibbs free energy change of the reaction shown below is −2.7kJ.mol−1.
Sn(s)+Pb2+=Sn2++Pb(s)
Given that E°(Pb2+∕Pb)=−0.126V, Find the value of magnitude of E°(Sn2+∕Sn)

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