Complete the following reaction:

\(2H_2O + 5SO_2 + 2MNO_4^- \longrightarrow _{------} + _{------} + _{------}\)

\(5SO_4^{2-}, \, \ 4H^+, \, \ 2Mn^{2+}\)

The correct answer is option 1. \(5SO_4^{2-}, \, \ 4H^+, \, \ 2Mn^{2+}\).

Let us analyze the reaction and identify the products step by step.

The given reaction is:

\( 2H_2O + 5SO_2 + 2MnO_4^- \longrightarrow \text{Products} \)

Step 1: Identify the Oxidation and Reduction Processes

Sulfur Dioxide (SO\(_2\)):

Oxidation: Sulfur in SO\(_2\) has an oxidation state of +4.

Products: Upon oxidation, SO\(_2\) is likely converted to sulfate ion \((SO_4^{2-})\), where sulfur has an oxidation state of +6.

Permanganate Ion (MnO\(_4^-\)):

Reduction: The manganese in MnO\(_4^-\) has an oxidation state of +7.

Products: MnO\(_4^-\) is typically reduced to Mn\(^{2+}\) in acidic solutions.

Step 2: Write Half-Reactions

Oxidation Half-Reaction (SO\(_2\) to SO\(_4^{2-}\)):

\(5SO_2 + 5H_2O \longrightarrow 5SO_4^{2-} + 10H^+ + 10e^- \)

Reduction Half-Reaction (MnO\(_4^-\) to Mn\(^{2+}\)):

\(2MnO_4^- + 16H^+ + 10e^- \longrightarrow 2Mn^{2+} + 8H_2O \)

Step 3: Combine the Half-Reactions

Balance the number of electrons to combine the half-reactions:

Oxidation: \(10e^-\) released

Reduction: \(10e^-\) consumed

\(5SO_2 + 2MnO_4^- + 2H_2O \longrightarrow 5SO_4^{2-} + 4H^+ + 2Mn^{2+}\)

Step 4: Verify the Products and Charges

Sulfur: \(5 \times SO_2\) oxidized to \(5 \times SO_4^{2-}\).

Manganese: \(2 \times MnO_4^-\) reduced to \(2 \times Mn^{2+}\).

Water and Hydrogen Ions: \(H_2O\) is used in the reaction and produces \(H^+\).

Conclusion:

The overall reaction will be as:

\(2H_2O + 5SO_2 + 2MnO_4^- \longrightarrow  5SO_4^{2-} + 4H^+ + 2Mn^{2+}\)