How much charge in terms of Faraday is required for the following :…

How much charge in terms of Faraday is required for the following :

(i) 1 mol of MnO, to Mn2+
(ii) 1 mol of H,0 to 02

Ans.

To calculate the charge in terms of Faraday required for the given reactions, we need to determine the number of moles of electrons involved in each reaction and then use Faraday’s constant, $F$, which is the charge of one mole of electrons.

The charge $Q$ in terms of Faraday is given by:

$Q=n\times F$

where:

• $Q$ is the charge in coulombs,

• $n$ is the number of moles of electrons,

• $F$ is Faraday’s constant (1 mol−1×�1mol−1×e), where $e$ is the elementary charge (1.602×10−19 C1.602×10−19C).

Let’s calculate for each reaction:

(i) MnO4−→Mn2+MnO4−​→Mn2+

The balanced equation for this reaction is:

MnO4−+8�++5�−→Mn2++4�2�MnO4−​+8H++5e−→Mn2++4H2​O

Therefore, 5 moles of electrons are involved in the reduction of 1 mole of MnO4−MnO4−​ to Mn2+Mn2+.

(ii) H2O→O2H2​O→O2​

The balanced equation for this reaction is:

2H2O→O2+4�++4�−2H2​O→O2​+4H++4e−

Therefore, 4 moles of electrons are involved in the oxidation of 1 mole of H2OH2​O to O2O2​.

Now, we use Faraday’s constant:

$F=96485\text{C/mol}$

(i) For the reduction of MnO4−MnO4−​ to Mn2+Mn2+:

$Q=5\times 96485\text{C/mol}=482425\text{C}$

(ii) For the oxidation of H2OH2​O to O2O2​:

$Q=4\times 96485\text{C/mol}=385940\text{C}$

Therefore, the charge required in terms of Faraday for the given reactions are: (i) $482425\text{C}$ for the reduction of MnO4−MnO4−​ to Mn2+Mn2+, and (ii) $385940\text{C}$ for the oxidation of H2OH2​O to O2O2​.