Azinc rod is placed in 0.1 mznso4 solution at 298 k. write the electrode reaction and calculate the potential of the electrode. (ez24/2 = -0.76v.)

The potential of electrode is -0.79 V

Explanation:

When zinc is dipped in zinc sulfate solution, the electrode formed is

Reduction reaction follows:  

To calculate the potential of electrode, we use the equation given by Nernst equation:

where,

= electrode potential of the cell = ?V

= standard electrode potential of the cell = -0.76 V

n = number of electrons exchanged = 2

   (concentration of pure solids are taken as 1)

Putting values in above equation, we get:

Hence, the potential of electrode is -0.79 V

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