Q59 / 180 · COMEDK 2024 (Afternoon Shift) · Chemistry
When 0.1 mole of $\mathrm{MnO}_4{ }^{2-}$ is oxidised, the quantity of electricity required to completely oxidise $\mathrm{MnO}_4{ }^{2-}$ to $\mathrm{MnO}_4^{-}$ is
- A2 $\times$ 96500 C
- B96500 C
- C48250 C
- Correct answerD9650 C
Solution
The oxidation reaction of manganate ion to permanganate ion is given by:
$\mathrm{MnO}_{4}^{2-} \rightarrow \mathrm{MnO}_{4}^{-} + e^{-}$
In this reaction, the oxidation state of manganese changes from $+6$ to $+7$. Thus, $1$ mole of $\mathrm{MnO}_{4}^{2-}$ requires $1$ mole of electrons for complete oxidation to $\mathrm{MnO}_{4}^{-}$.
Given that $0.1$ mole of $\mathrm{MnO}_{4}^{2-}$ is to be oxidized, the number of moles of electrons required is $0.1$ mole.
The quantity of electricity $Q$ is given by $Q = n \times F$, where $n$ is the number of moles of electrons and $F$ is the Faraday constant ($96500 \text{ C mol}^{-1}$).
$Q = 0.1 \text{ mol} \times 96500 \text{ C mol}^{-1} = 9650 \text{ C}$.
Answer: 9650 C
$\mathrm{MnO}_{4}^{2-} \rightarrow \mathrm{MnO}_{4}^{-} + e^{-}$
In this reaction, the oxidation state of manganese changes from $+6$ to $+7$. Thus, $1$ mole of $\mathrm{MnO}_{4}^{2-}$ requires $1$ mole of electrons for complete oxidation to $\mathrm{MnO}_{4}^{-}$.
Given that $0.1$ mole of $\mathrm{MnO}_{4}^{2-}$ is to be oxidized, the number of moles of electrons required is $0.1$ mole.
The quantity of electricity $Q$ is given by $Q = n \times F$, where $n$ is the number of moles of electrons and $F$ is the Faraday constant ($96500 \text{ C mol}^{-1}$).
$Q = 0.1 \text{ mol} \times 96500 \text{ C mol}^{-1} = 9650 \text{ C}$.
Answer: 9650 C
Hence, the correct answer is option: 'D'.