What is the cell potential (emf, in v) of this cell at 25°c? cu cu2+ (0.0257 m) | br2 [br" (0.392 m)

The voltage of the cell is 0.80 V.

Explanation:

The given cell is:

Half reactions for the given cell follows:

Oxidation half reaction:

Reduction half reaction:

Net 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 = +0.73 V

n = number of electrons exchanged = 2

Putting values in above equation, we get:

Hence, the voltage of the cell is 0.80 V.

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