Balance the following redox reactions by ion – electron method : (a) MnO4– (aq) + I– (aq) ? MnO2 (s) + I2(s) (in basic medium) (b) MnO4– (aq) + SO2 (g) ? Mn2+ (aq) + HSO4– (aq) (in acidic solution) (c) H2O2 (aq) + Fe2+ (aq) ? Fe3+ (aq) + H2O (l) (in acidic solution) (d) Cr2O7 2– + SO2(g) ? Cr3+ (aq) + SO42– (aq) (in acidic solution)

Balancing redox reactions using the ion-electron method involves separating the oxidation and reduction half-reactions, balancing them individually, and then combining them to achieve a balanced overall reaction. Let's tackle each of the reactions you've provided step by step.

Balancing MnO4– + I– in Basic Medium

For the reaction:

MnO4– (aq) + I– (aq) ? MnO2 (s) + I2 (s)

Step 1: Identify Oxidation and Reduction

Manganese in MnO4– is reduced from +7 to +4 in MnO2, while iodide (I–) is oxidized to iodine (I2), changing from -1 to 0.

Step 2: Write Half-Reactions

• Reduction: MnO4– + 8e– + 4H2O ? MnO2 + 8OH–
• Oxidation: 2I– ? I2 + 2e–

Step 3: Balance Electrons

To balance the electrons, multiply the oxidation half-reaction by 4:

• 4(2I– ? I2 + 2e–) → 8I– ? 4I2 + 8e–

Step 4: Combine Half-Reactions

Now, combine the half-reactions:

MnO4– + 8I– + 4H2O ? MnO2 + 4I2 + 8OH–

Balancing MnO4– + SO2 in Acidic Medium

For the reaction:

MnO4– (aq) + SO2 (g) ? Mn2+ (aq) + HSO4– (aq)

Step 1: Identify Oxidation and Reduction

Manganese is reduced from +7 in MnO4– to +2 in Mn2+, while sulfur in SO2 is oxidized from +4 to +6 in HSO4–.

Step 2: Write Half-Reactions

• Reduction: MnO4– + 8H+ + 5e– ? Mn2+ + 4H2O
• Oxidation: SO2 + 2H2O ? HSO4– + 2H+ + 2e–

Step 3: Balance Electrons

To balance the electrons, multiply the oxidation half-reaction by 5:

• 5(SO2 + 2H2O ? HSO4– + 2H+ + 2e–) → 5SO2 + 10H2O ? 5HSO4– + 10H+ + 10e–

Step 4: Combine Half-Reactions

Now, combine the half-reactions:

MnO4– + 5SO2 + 6H+ ? Mn2+ + 5HSO4– + 3H2O

Balancing H2O2 + Fe2+ in Acidic Medium

For the reaction:

H2O2 (aq) + Fe2+ (aq) ? Fe3+ (aq) + H2O (l)

Step 1: Identify Oxidation and Reduction

Hydrogen peroxide (H2O2) is oxidized to water (H2O), while Fe2+ is oxidized to Fe3+.

Step 2: Write Half-Reactions

• Reduction: H2O2 + 2H+ + 2e– ? 2H2O
• Oxidation: Fe2+ ? Fe3+ + e–

Step 3: Balance Electrons

To balance the electrons, multiply the oxidation half-reaction by 2:

• 2(Fe2+ ? Fe3+ + e–) → 2Fe2+ ? 2Fe3+ + 2e–

Step 4: Combine Half-Reactions

Now, combine the half-reactions:

H2O2 + 2Fe2+ + 2H+ ? 2H2O + 2Fe3+

Balancing Cr2O7 2– + SO2 in Acidic Medium

For the reaction:

Cr2O7 2– + SO2 (g) ? Cr3+ (aq) + SO4 2– (aq)

Step 1: Identify Oxidation and Reduction

Chromium is reduced from +6 in Cr2O7 2– to +3 in Cr3+, while sulfur is oxidized from +4 in SO2 to +6 in SO4 2–.

Step 2: Write Half-Reactions

• Reduction: Cr2O7 2– + 14H+ + 6e– ? 2Cr3+ + 7H2O
• Oxidation: SO2 + 2H2O ? SO4 2– + 2H+ + 2e–

Step 3: Balance Electrons

To balance the electrons, multiply the oxidation half-reaction by 3:

• 3(SO2 + 2H2O ? SO4 2– + 2H+ + 2e–) → 3SO2 + 6H2O ? 3SO4 2– + 6H+ + 6e–

Step 4: Combine Half-Reactions

Now, combine the half-reactions:

Cr2O7 2– + 3SO2 + 8H+