Question

It is given that for an electrochemical cell, the half-cell potentials are, e anode
= 0.43V and E cathode = -0.321. What is the emf of the cell?​

Answer 1

Answer:

In an electrochemical cell,

The half cell potentials are,

Ecathode=-0.321v

Eanode=0.43v

The EMF of the cell,E°=Ecathode-Eanode

E°=-0.321V-0.43V

E°=-0.751V

So the EMF of the cell=-0.751V

Explanation:

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