Calculate the voltage of the cell: zn 1 zn2+ (10m) 11 cu2+(0.20m) 1 cu

The voltage of the cell is 1.05 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 = +1.1 V

n = number of electrons exchanged = 2

Putting values in above equation, we get:

Hence, the voltage of the cell is 1.05 V.

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