Question

When Manganeses (II) reacts with peroxodisulphate, the respective products obtained are:

• A. \(MnO_4^{2-}\) only
• B. \(MnO_4^{-}\) and \(SO_4^{2-}\)
• C. \(MnO_2\) and \(S_2O_4^{2-}\)
• D. \(MnO_4^{2-}\) and \(SO_4^{2-}\)

Hint:

Peroxodisulphate is a strong oxidant. It turns Mn(II) into permanganate and itself becomes sulphate.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Peroxodisulphate, \(S_2O_8^{2-}\), is a very strong oxidising agent. It oxidises \(Mn^{2+}\) to permanganate in acidic solution. It is itself reduced to sulphate.

Step 2: Write the reaction:
\[ 2Mn^{2+} + 5S_2O_8^{2-} + 8H_2O \to 2MnO_4^{-} + 10SO_4^{2-} + 16H^{+} \]

Step 3: Check the electron change:
Manganese goes from +2 to +7, a loss of 5 electrons per Mn. Each \(S_2O_8^{2-}\) gains 2 electrons to give two \(SO_4^{2-}\). For 2 Mn we lose 10 electrons, and 5 peroxodisulphate ions gain 10 electrons. The reaction balances.

Step 4: Check the options:
Option 1 and option 4 give manganate \(MnO_4^{2-}\) (Mn +6), which is not formed here. Option 3 gives \(S_2O_4^{2-}\) (dithionite), which is a reduced species and cannot come from an oxidant. Option 2 matches.

Final Answer:
The products are \(MnO_4^{-}\) and \(SO_4^{2-}\). \[ \boxed{MnO_4^{-} \text{ and } SO_4^{2-}} \]