Home
Class 12
CHEMISTRY
Consider the following relations for emf...

Consider the following relations for emf of a electrochemical cell :
(a) emf of cell = (Oxidation potential of anode) - (Reduction potential of cathode)
(b) emf of cell = (Oxidation potential of anode) + (Reduction potential of cathod)
(c) emf of cell = (Reduction potential of anode) + (Reduction potential of cathode)
(d) emf of cell = (Oxidation potential of anode) - (Oxidation potential of cathode)
Which of the above relations are correct ?

A

(c ) and (a)

B

(a) and (b)

C

(c ) and (d)

D

(b) and (d)

Text Solution

Verified by Experts

The correct Answer is:
D
Promotional Banner

Similar Questions

Explore conceptually related problems

Consider the following relations for emf of a electrochemical cell (i) emf of cell = (Oxidation potential of anode)-(Reduction potential of cathode) (ii) emf of cell = (Oxidation potential of anode)+(Reduction potential of cathode) (iii) emf of cell = (Reduction potential of anode)+(Reduction potential of cathode) (iv) emf of cell = (Oxidation potential of anode)-(Oxidation potential of cathode) Which of the above realtions are correct?

Define oxidation and reduction potentials.

If the orders of reduction potential is then :

Standard reduction potential is calculated at

For the half-cell : At pH= 2, electrode oxidation potential is :

The potential of the following cell at 25^(@)C is

The reduction potential of hydrogen half cell will be negative if :

Define electrode potential, oxidation potential and reduction potential. Why is it not possible to calculate the absolute value of electrode potential?

The half cell reduction potential of a hydrogen electrode at pH = 5 will be :

The oxidation potential of hydrogen half-cell will be negative if: