JEE MAIN - Chemistry (2023 - 15th April Morning Shift - No. 21)
The total change in the oxidation state of manganese involved in the reaction of $\mathrm{KMnO}_{4}$ and potassium iodide in the acidic medium is ____________.
Answer
5
Explanation
The reaction of $\mathrm{KMnO}_4$ and potassium iodide ($\mathrm{KI}$) in acidic medium can be represented as follows:
$$\mathrm{MnO_4^-} + \mathrm{I^-} + \mathrm{H^+} \rightarrow \mathrm{Mn^{2+}} + \mathrm{I_2} + \mathrm{H_2O}$$
In this reaction, $\mathrm{KMnO}_4$ acts as an oxidizing agent, while $\mathrm{KI}$ acts as a reducing agent. The oxidation state of manganese changes from +7 in $\mathrm{KMnO_4}$ to +2 in $\mathrm{Mn^{2+}}$, which is a reduction of 5 units.
Therefore, the total change in the oxidation state of manganese involved in the reaction is $\boxed{5}$.
$$\mathrm{MnO_4^-} + \mathrm{I^-} + \mathrm{H^+} \rightarrow \mathrm{Mn^{2+}} + \mathrm{I_2} + \mathrm{H_2O}$$
In this reaction, $\mathrm{KMnO}_4$ acts as an oxidizing agent, while $\mathrm{KI}$ acts as a reducing agent. The oxidation state of manganese changes from +7 in $\mathrm{KMnO_4}$ to +2 in $\mathrm{Mn^{2+}}$, which is a reduction of 5 units.
Therefore, the total change in the oxidation state of manganese involved in the reaction is $\boxed{5}$.
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