Question

The reduction electrode potential E, of 0.1 M solution
of M* ions (Eºrp = -2.36 V) is:
1)-2.41
(3)-4.82
(4) None
(2) +2.41​

Answer 1

Answer:

Option (1) -2.41 V

Explanation:

Using Nernst Equation

$\boxed{E=E^0+\frac{0.059}{n}\log\frac{a_{ox}}{a}}$

Number of electrons = 1 (for single electrode potential)

[M⁺] = 0.1 M

E° = -2.36V

Thus

$E=E^0+\frac{0.059}{n}\log[M^+]$

$\implies E=-2.36+\frac{0.059}{1}\log0.1$

$\implies E=-2.36-0.059=-2.419$

Thus the reduction electrode potential E of 0.1 M solution is -2.41

Thus option (1) is correct.