Question

In a particular electrolysis process, eocathode = -1. 151 v and eoanode = 1. 548 v. what minimum voltage must be applied to drive the nonspontaneous process?

Answer 1

In a particular electrolysis process, Eο cathode = -1.151 v and Eο anode = 1.548 V. The minimum voltage must be applied to drive the non-spontaneous process is 2.699V.

Electrode Potential Or Voltage is the emf of a galvanic cell where emf is called as the Electromotive force.

Electrode Potential of a spontaneous cell is positive. Hence to make a non-spontaneous cell into a spontaneous cell, the voltage or potential applied should be positive to make the cell working.

We know that,

Electrode Potential = Electrode Potential of cathode - Electrode Potential of anode

where, Electrode Potential of cathode = -1.151V

            Electrode Potential of anode = 1.548V

On substituting,

⇒ Electrode Potential = -1.151V - 1.548V

⇒ Electrode Potential = - 2.699V

Hence, the minimum voltage to drive a non-spontaneous cell into a spontaneous cell should be at least 2.699V.

Learn more about Electrode Potential here, https://brainly.com/question/17060277

#SPJ4