Question

The correct information(s) about the n-factor of given species in the reaction: $KMnO_4 + H_2SO_4 + HCl \rightarrow K_2SO_4 + MnSO_4 + Cl_2 + H_2O$ is/are -

• A. $KMnO_4 \rightarrow 5$
• B. $H_2SO_4 \rightarrow 10/3$
• C. $HCl \rightarrow 1$
• D. $K_2SO_4 \rightarrow 2$

Answer 1

Answer: (A), (C)

A, C

Answer 2

To determine the n-factor (also known as the equivalence factor) for each species in a redox reaction, we need to identify the change in the oxidation state of the elements involved in the electron transfer. The n-factor is defined as the number of electrons gained or lost per mole of the substance.

The given unbalanced reaction is: $KMnO_4 + H_2SO_4 + HCl \rightarrow K_2SO_4 + MnSO_4 + Cl_2 + H_2O$

First, let's determine the oxidation states of the relevant elements:

1. Manganese (Mn):

   • In $KMnO_4$: K is +1, O is -2. Let Mn be x. $1 + x + 4(-2) = 0 \Rightarrow x = +7$.
   • In $MnSO_4$: $SO_4$ is a sulfate ion with a -2 charge, so S is +6 and O is -2. Let Mn be y. $y + (-2) = 0 \Rightarrow y = +2$.
   • Change in oxidation state for Mn: From +7 to +2. This is a gain of 5 electrons.
   • n-factor for $KMnO_4$ = 5. (Since 1 mole of $KMnO_4$ contains 1 mole of Mn, and Mn gains 5 electrons).

2. Chlorine (Cl):

   • In $HCl$: H is +1. Let Cl be z. $1 + z = 0 \Rightarrow z = -1$.
   • In $Cl_2$: Elemental chlorine has an oxidation state of 0.
   • Change in oxidation state for Cl: From -1 to 0. This is a loss of 1 electron per Cl atom.
   • n-factor for $HCl$ = 1. (Since 1 mole of $HCl$ contains 1 mole of Cl, and Cl loses 1 electron).

3. Sulfur (S):

   • In $H_2SO_4$: H is +1, O is -2. Let S be w. $2(+1) + w + 4(-2) = 0 \Rightarrow w = +6$.
   • In $K_2SO_4$ and $MnSO_4$: S remains in the +6 oxidation state.
   • Since there is no change in the oxidation state of Sulfur, $H_2SO_4$ acts as an acid (providing $H^+$ ions) and not as a redox species.
   • n-factor for $H_2SO_4$ in this redox reaction = 0.

4. Potassium Sulfate ($K_2SO_4$):

   • This is a product of the reaction. K is +1, S is +6, O is -2. None of these elements change their oxidation states during the reaction. $K_2SO_4$ is a salt formed by spectator ions ($K^+$ and $SO_4^{2-}$).
   • n-factor for $K_2SO_4$ in this redox reaction = 0.

Now, let's evaluate the given options:

(A) $KMnO_4 \rightarrow 5$: This matches our calculation. Correct.

(B) $H_2SO_4 \rightarrow 10/3$: Our calculation shows n-factor = 0. So, this is incorrect.

(C) $HCl \rightarrow 1$: This matches our calculation. Correct.

(D) $K_2SO_4 \rightarrow 2$: Our calculation shows n-factor = 0. So, this is incorrect.

Thus, the correct information(s) are (A) and (C).

The balanced chemical equation for completeness is: $2KMnO_4 + 10HCl + 3H_2SO_4 \rightarrow 2MnSO_4 + 5Cl_2 + K_2SO_4 + 8H_2O$