Question

calculate the potential of the cell: ni2+ +fe > ni+fe2+ the half reactions involved are: ni2+...

calculate the potential of the cell: ni2+ +fe > ni+fe2+
the half reactions involved are: ni2+ +2e- > ni E°=-0.26v
and fe2+ +2e- > fe E°=-0.45v
is the reaction spontaneous or nonspontaneous?

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Answer #1

The reaction is:      Ni 2+ + Fe -------> Ni +   Fe 2+

Half rections:

Ni2+ + 2e- --------> Ni       Gain electrons.... Reduction.....Catode

Fe ------------> Fe2+ + 2e-   Loss of electrons....Oxidation....Anode

In addition you know that:

Ni 2+ + 2e- -----> Ni      E°= -0.26 V

Fe 2+ + 2e- ------> Fe   E° = -0.45 V

If   E°>0 The reaction is spontaneous

If   E°< 0 The reaction is nonspontaneous

In this reaction:

E° = -0.26 V - (-0.45 V) = -0.26 V + 0.45 V = 0.19 V

E° is > 0 (positive) ans the reaction is spontaneous.

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