Question

The reduction electrode potential, E of 0.1 M solution of ${M^ + }$ ions ($E_{RP}^o$= -2.36 V) is :
a.) - 4.82 V
b.) - 2.41 V
c.) + 2.41 V
d.) None of these

Answer 1

Hint:. The electrode potential is the potential developed across electrodes due to change in charge. It can be found by formula -
E = ${E^0} - \dfrac{{0.0591}}{n}\log \dfrac{1}{{{M^ + }}}$

Complete step by step answer:
First, let us write what is given to us and what we need to find out.
Given :
Molarity of solution = 0.1 M
$E_{RP}^o$= -2.36 V
To find :
Reduction electrode potential
We have the formula to find reduction potential as -
E = ${E^0} - \dfrac{{0.0591}}{n}\log \dfrac{1}{{{M^ + }}}$
We have been given that the metal is ${M^ + }$. This means n = 1
So, filling the values; we have
E = $ - 2.36 - \dfrac{{0.0591}}{1}\log \dfrac{1}{{0.1}}$
Simplifying the equation, we get
E = $ - 2.36 - 0.0591\log 10$
Further, the log 10 = 1
So, E = $ - 2.36 - 0.0591$
E = -2.41 V
So, the correct answer is “Option B”.

Note: We know that all the electrochemical cells are based on the redox reaction. This means that each cell consists of two half cell reactions- the reduction half cell reaction and the oxidation half cell reaction. The oxidation i.e. the gain of electrons take place at anode while the reduction i.e. the loss of electrons take place at cathode. Both the anode and cathode are in separate compartments but are connected by a channel that allows the movement of electrons. Some electric potential arises between the two electrodes due to potential difference between them.