Onsider the galvanic cell described by (N and M are metals): N(s)|N2+(aq)||M+(aq)|M(s) If Eocathode = 0.576 V and Eoanode = 1.129 V, and [N2+(aq)] = 0.307 M and [M+(aq)] = 0.973 M, what is Ecell, using the Nernst equation? V

The cell potential of the given cell is -0.567 V

Explanation:

The given chemical cell follows:

Oxidation half reaction:  

Reduction half reaction:       ( × 2)

Net cell reaction:

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the of the reaction, we use the equation:

Putting values in above equation, we get:

To calculate the EMF of the cell, we use the Nernst equation, which is:

where,

= electrode potential of the cell = ? V

= standard electrode potential of the cell = -0.553 V

n = number of electrons exchanged = 2

Putting values in above equation, we get:

Hence, the cell potential of the given cell is -0.567 V

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