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Question

Chemistry Question on EMF

If E°(Zn2+,Zn)=0.763 VE°(Zn^{2+},Zn)=-0.763\ V and E°(Fe2+,Fe)=0.44 VE°(Fe^{2+},Fe)=-0.44\ V, then the emf of the cell Zn/Zn2+(a=0.001)Fe2+(a=0.005)FeZn/Zn^{2+}(a=0.001)||Fe^{2+}(a=0.005) Fe is

A

equal to 0.323 V

B

less than 0.323 V

C

greater than 0.323 V

D

equal to 1.103 V

Answer

greater than 0.323 V

Explanation

Solution

The correct option is (C): greater than 0.323 V.