Question

If the cell potential for a voltaic cell is 0.250 V, and the reduction potential for the oxidation reaction is ⎯0.150 V, what is the reduction potential for the reaction occurring at the cathode?

Answer 1

If the cell potential for a voltaic cell is 0.250 V, and the reduction potential for the oxidation reaction is ⎯0.150 V, what is the reduction potential for the reaction occurring at the cathode?

Voltaic cell is made up of 2 half-cells: Oxidation half-cell (Anode/ -ve electrode) and the Reduction half-cell (Cathode/+ve electrode)

These two half-cells differ in electrode potential value. And this difference in the potential of 2 half-cells is known as EMF of the cell or cell potential.

If potential of half-cells are given in terms of reduction potential then cell potential is given as,

Cell potential (E_cell) = Reduction potential of Cathode – Reduction potential of anode.

Ecell = E_cathode – E_anode ……………… (1)

We are given with,

E_cell = 0.250 V,

Reduction potential for the oxidation reaction (i.e. Reduction potential of anode at which oxidation reaction takes place) E_anode = -0.150 V.

Reduction potential for the reduction reaction (i.e. Reduction potential of cathode at which reduction reaction takes place) E_cathode =?

Let us put known values in eq.(1) and solve it form unknown.

0.250 = E_cathode – (-0.150)

0.250 = E_cathode + 0.150

E_cathode = 0.250 - 0.150

E_cathode = +0.100 V

Hence the reduction potential for the reaction occurring at cathode is +0.100 V.

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