The standard reduction potentials for the Ag+|Ag(s) and Zn2+| Zn(s) half-cell reactions are +0.799 V and -0.762 V, respectively. Calculate the potential for the following electrochemical cell: Zn(s)|Zn2+(0.125 M)||Ag+(0.240 M)|Ag(s)

The potential of the given cell is 1.551 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 = +1.561 V

n = number of electrons exchanged = 2

Putting values in above equation, we get:

Hence, the potential of the given cell is 1.551 V

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