A galvanic cell is constructed by coupling of an iron and a nickel electrode. If at 298K E∘Ni2+(aq)/Ni(s)=−0.24V E∘Fe2+(aq)/Fe(s)=−0.44V
then calculate the EMF of the cell
A
+0.68V
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B
−0.68V
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C
+0.2V
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D
−0.2V
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Solution
The correct option is C+0.2V Given that : E∘Ni2+(aq)/Ni(s)=−0.24V E∘Fe2+(aq)/Fe(s)=−0.44V
The standard reduction potentials of both are given. Comparing both, E∘Ni2+(aq)/Ni(s)>E∘Fe2+(aq)/Fe(s)
So, nickel will reduce at cathode and iron will oxidise at anode
E∘cell=SRP of substance reduced−SRP of substance oxidisedE∘cell=−0.24−(−0.44)=0.2V