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Question

A galvanic cell is constructed by coupling of an iron and a nickel electrode. If at 298 K
ENi2+(aq)/Ni(s)=0.24 V
EFe2+(aq)/Fe(s)=0.44 V
then calculate the EMF of the cell

A
+0.68 V
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B
0.68 V
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C
+0.2 V
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D
0.2 V
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Solution

The correct option is C +0.2 V
Given that :
ENi2+(aq)/Ni(s)=0.24 V
EFe2+(aq)/Fe(s)=0.44 V
The standard reduction potentials of both are given. Comparing both,
ENi2+(aq)/Ni(s)>EFe2+(aq)/Fe(s)
So, nickel will reduce at cathode and iron will oxidise at anode

Ecell=SRP of substance reducedSRP of substance oxidisedEcell=0.24(0.44)=0.2 V

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