1.

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 ……….. .

Answer»

`-0.35 V`
`+0.35 V`
`+1.17 V`
`-1.17V`

Solution :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`.


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