First slide

Electrochemical cell

Question

$\large {E^0}_{S{n^{2 + }}/Sn}\, = \, - 1.14V,$ , $\large {E^0}_{P{b^{2 + }}/Pb}\, = \, - 0.126\,V$ The cell reaction of the cell constructed from these two electrodes is

Easy

A

$P{b^{2 + }} + \,S{n^{2 + }}\, \to \,Pb\, + \,Sn$
B

$S{n^{2 + }} + \,Pb\, \to \,\,Sn\, + P{b^{2 + }}$
C

$Pb\, + \,Sn\, \to \,P{b^{2 + }}\, + \,S{n^{2 + }}$
D

$Sn\, + \,P{b^{2 + }}\, \to \,S{n^{2 + }} + \,Pb$

Solution

Low SRP electrode acts as anode undergoes Oxidation.
High SRP electrode acts as cathode undergoes Reduction.

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