Balance the redox reaction between Fe2+ and MnO4^- in acidic solution.

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ScanSolve AI Tutor · Accepted Answer

Step 1: Separate the reaction into two half-reactions. Oxidation half-reaction: Fe²⁺ → Fe³⁺ Reduction half-reaction: MnO₄^− → Mn²⁺ Step 2: Balance atoms other than oxygen and hydrogen. For the oxidation half-reaction, Fe atoms are already balanced. Fe²⁺ → Fe³⁺ For the reduction half-reaction, Mn atoms are already balanced. MnO₄^− → Mn²⁺ Step 3: Balance oxygen atoms by adding H₂O. Oxidation: No oxygen atoms. Fe²⁺ → Fe³⁺ Reduction: Add 4H₂O to the right side to balance the 4 oxygen atoms on the left. MnO₄^− → Mn²⁺ + 4H₂O Step 4: Balance hydrogen atoms by adding H^+ (since the reaction is in acidic medium). Oxidation: No hydrogen atoms. Fe²⁺ → Fe³⁺ Reduction: Add 8H^+ to the left side to balance the 8 hydrogen atoms in 4H₂O on the right. 8H^+ + MnO₄^− → Mn²⁺ + 4H₂O Step 5: Balance the charge by adding electrons (e^−). Oxidation: Left side charge: +2 Right side charge: +3 Add 1e^− to the right side to balance the charge. Fe²⁺ → Fe³⁺ + e^− Reduction: Left side charge: (+8) + (−1) = +7 Right side charge: +2 Add 5e^− to the left side to balance the charge. 5e^− + 8H^+ + MnO₄^− → Mn²⁺ + 4H₂O Step 6: Multiply the half-reactions by appropriate integers to equalize the number of electrons. The oxidation half-reaction has 1e^−. The reduction half-reaction has 5e^−. Multiply the oxidation half-reaction by 5: 5(Fe²⁺ → Fe³⁺ + e^−) 5Fe²⁺ → 5Fe³⁺ + 5e^− The reduction half-reaction remains unchanged: 5e^− + 8H^+ + MnO₄^− → Mn²⁺ + 4H₂O Step 7: Add the two balanced half-reactions and cancel common species. 5Fe²⁺ + 5e^− + 8H^+ + MnO₄^− → 5Fe³⁺ + 5e^− + Mn²⁺ + 4H₂O Cancel 5e^− from both sides: 5Fe²⁺ + 8H^+ + MnO₄^− → 5Fe³⁺ + Mn²⁺ + 4H₂O Step 8: Verify that atoms and charges are balanced. Atoms: Mn: 1 on left, 1 on right O: 4 on left, 4 on right Fe: 5 on left, 5 on right H: 8 on left, 8 on right Charge: Left side: 5(+2) + 8(+1) + (−1) = +10 + 8 − 1 = +17; Right side: 5(+3) + (+2) + 0 = +15 + 2 = +17 Both atoms and charges are balanced. The balanced equation is: MnO₄^− + 5Fe²⁺ + 8H^+ → Mn²⁺ + 5Fe³⁺ + 4H₂O

Balance the redox reaction between Fe2+ and MnO4^- in acidic solution.

Step-by-step solution

Separate the reaction into two half-reactions.

Balance atoms other than oxygen and hydrogen.

Balance oxygen atoms by adding $\text{H}_2\text{O}$ .

Balance hydrogen atoms by adding $\text{H}^+$ (since the reaction is in acidic medium).

Balance the charge by adding electrons ( $\text{e}^-$ ).

Multiply the half-reactions by appropriate integers to equalize the number of electrons.

Add the two balanced half-reactions and cancel common species.

Verify that atoms and charges are balanced.

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Balance the redox reaction between Fe2+ and MnO4^- in acidic solution.

Step-by-step solution

Separate the reaction into two half-reactions.

Balance atoms other than oxygen and hydrogen.

Balance oxygen atoms by adding H2O\text{H}_2\text{O}H2​O.

Balance hydrogen atoms by adding H+\text{H}^+H+ (since the reaction is in acidic medium).

Balance the charge by adding electrons (e−\text{e}^-e−).

Multiply the half-reactions by appropriate integers to equalize the number of electrons.

Add the two balanced half-reactions and cancel common species.

Verify that atoms and charges are balanced.

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More Chemistry Questions

Balance oxygen atoms by adding $\text{H}_2\text{O}$ .

Balance hydrogen atoms by adding $\text{H}^+$ (since the reaction is in acidic medium).

Balance the charge by adding electrons ( $\text{e}^-$ ).