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

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.