The standard reduction potential for the half reactions are as
`Zn to Zn^(2+) + 2e^(-) E^@ = +0.76 V`
`Fe to Fe^(2+) + 2e^(-) E^@ = +0.41 V`. So for cell reaction `Fe^(2+) + Zn to Zn^(2+) + Fe` is ……….. .

Options

A. `-0.35 V`
B. `+0.35 V`
C. `+1.17 V`
D. `-1.17V`

Answer

B

Explanation

In the reaction `Fe^(2+) + Zn^@ to Zn^(2+) + Fe^(@)`
`emf = E_("cathode") -E_("anode")`
`= -0.41 - (0.76)`
`= -0.41 + 0.76`
emf = `+ 0.35 V`.