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The oxdation potentiasl of following hal...

The oxdation potentiasl of following half-cell reactions are given `Zn rarr Zn^(2+) + 2e^- , E^o = - 0. 76 V, Fe rarr Fe^(2+) + 2e^(-) , E^@ = 0. 44 V` what will be the emf of cell, whose cell -reaction is ` Fe^(2+) (aq) + Zn(s) -> Zn^(2+) (aq) + Fe`.

A

` -1.20 V`

B

` -0. 32 V`

C

` +1.32 V`

D

` + 1.20 V`

Text Solution

Verified by Experts

The correct Answer is:
C
.
`EMF=E_("cathode" ) - E_("anode ") = 0.44-(0.76) = +0. 32 V`
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