Agalvanic cell is based on the following half-reactions at 279 k: ag+ + e- → ag eo = 0.803 v h2o2 (aq) + 2 h+ + 2 e- → 2 h2o eo = 1.78 v what will the potential of this cell be when [ag+] = 0.559 m, [h+] = 0.00393 m, and [h2o2] = 0.863 m?

The potential of the given cell is 0.856 V

Explanation:

The substance having highest positive potential will always get reduced.

Half reactions for the given cell follows:

Oxidation half reaction:     ( × 2)

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 = +0.977 V

R = Gas constant = 8.314 J/K mol

T = temperature = 279 K

F = Faraday's constant = 96500 C

n = number of electrons exchanged = 2

Putting values in above equation, we get:

Hence, the potential of the given cell is 0.856 V

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