Separate the reaction into half-reactions.

Question

Separate the reaction into half-reactions.

Asked on April 24, 2026Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

Accepted Answer · 2026-04-24T10:00:27.709Z

Right Dàññý, let's go. To balance the redox reaction $\text{MnO}_4^- (aq) + \text{Fe}^{2+} (aq) \rightarrow \text{Mn}^{2+} (aq) + \text{Fe}^{3+} (aq)$ in an acidic medium using the half-reaction method: Step 1: Separate the reaction into half-reactions. Oxidation half-reaction: $\text{Fe}^{2+} (aq) \rightarrow \text{Fe}^{3+} (aq)$ Reduction half-reaction: $\text{MnO}_4^- (aq) \rightarrow \text{Mn}^{2+} (aq)$ Step 2: Balance atoms other than oxygen and hydrogen. The Fe atoms are balanced. The Mn atoms are balanced. Step 3: Balance oxygen atoms by adding $\text{H}_2\text{O}$. Oxidation: $\text{Fe}^{2+} (aq) \rightarrow \text{Fe}^{3+} (aq)$ Reduction: $\text{MnO}_4^- (aq) \rightarrow \text{Mn}^{2+} (aq) + 4\text{H}_2\text{O} (l)$ Step 4: Balance hydrogen atoms by adding $\text{H}^+$. Oxidation: $\text{Fe}^{2+} (aq) \rightarrow \text{Fe}^{3+} (aq)$ Reduction: $\text{MnO}_4^- (aq) + 8\text{H}^+ (aq) \rightarrow \text{Mn}^{2+} (aq) + 4\text{H}_2\text{O} (l)$ Step 5: Balance the charge by adding electrons ($\text{e}^-$). Oxidation: Reactant charge: $+2$ Product charge: $+3$ Add 1 electron to the product side: $$\text{Fe}^{2+} (aq) \rightarrow \text{Fe}^{3+} (aq) + \text{e}^-$$ Reduction: Reactant charge: $(-1) + 8(+1) = +7$ Product charge: $+2$ Add 5 electrons to the reactant side: $$\text{MnO}_4^- (aq) + 8\text{H}^+ (aq) + 5\text{e}^- \rightarrow \text{Mn}^{2+} (aq) + 4\text{H}_2\text{O} (l)$$ Step 6: Multiply the half-reactions to equalize the number of electrons. Multiply the oxidation half-reaction by 5: $$5\text{Fe}^{2+} (aq) \rightarrow 5\text{Fe}^{3+} (aq) + 5\text{e}^-$$ The reduction half-reaction remains the same: $$\text{MnO}_4^- (aq) + 8\text{H}^+ (aq) + 5\text{e}^- \rightarrow \text{Mn}^{2+} (aq) + 4\text{H}_2\text{O} (l)$$ Step 7: Add the balanced half-reactions and cancel common species. $$5\text{Fe}^{2+} (aq) + \text{MnO}_4^- (aq) + 8\text{H}^+ (aq) + 5\text{e}^- \rightarrow 5\text{Fe}^{3+} (aq) + 5\text{e}^- + \text{Mn}^{2+} (aq) + 4\text{H}_2\text{O} (l)$$ Cancel the 5 electrons from both sides: $$5\text{Fe}^{2+} (aq) + \text{MnO}_4^- (aq) + 8\text{H}^+ (aq) \rightarrow 5\text{Fe}^{3+} (aq) + \text{Mn}^{2+} (aq) + 4\text{H}_2\text{O} (l)$$ Step 8: Verify that atoms and charges are balanced. Atoms: Fe (5), Mn (1), O (4), H (8) are balanced on both sides. Charge: Reactants: $5(+2) + (-1) + 8(+1) = +10 - 1 + 8 = +17$. Products: $5(+3) + (+2) + 0 = +15 + 2 = +17$. Charges are balanced. The balanced redox reaction is: $$\boxed{5\text{Fe}^{2+} (aq) + \text{MnO}_4^- (aq) + 8\text{H}^+ (aq) \rightarrow 5\text{Fe}^{3+} (aq) + \text{Mn}^{2+} (aq) + 4\text{H}_2\text{O} (l)}$$ Identifying the oxidizing and reducing agents: • The species that is oxidized (loses electrons) is the reducing agent. Here, $\text{Fe}^{2+}$ is oxidized to $\text{Fe}^{3+}$. • The species that is reduced (gains electrons) is the oxidizing agent. Here, $\text{MnO}_4^-$ is reduced to $\text{Mn}^{2+}$. Oxidizing agent: $\boxed{\text{MnO}_4^-}$ Reducing agent: $\boxed{\text{Fe}^{2+}}$ What's next?

Accepted Answer · 2026-04-24T10:00:27.709Z

Right Dàññý, let's go. To balance the redox reaction $\text{MnO}_4^- (aq) + \text{Fe}^{2+} (aq) \rightarrow \text{Mn}^{2+} (aq) + \text{Fe}^{3+} (aq)$ in an acidic medium using the half-reaction method: Step 1: Separate the reaction into half-reactions. Oxidation half-reaction: $\text{Fe}^{2+} (aq) \rightarrow \text{Fe}^{3+} (aq)$ Reduction half-reaction: $\text{MnO}_4^- (aq) \rightarrow \text{Mn}^{2+} (aq)$ Step 2: Balance atoms other than oxygen and hydrogen. The Fe atoms are balanced. The Mn atoms are balanced. Step 3: Balance oxygen atoms by adding $\text{H}_2\text{O}$. Oxidation: $\text{Fe}^{2+} (aq) \rightarrow \text{Fe}^{3+} (aq)$ Reduction: $\text{MnO}_4^- (aq) \rightarrow \text{Mn}^{2+} (aq) + 4\text{H}_2\text{O} (l)$ Step 4: Balance hydrogen atoms by adding $\text{H}^+$. Oxidation: $\text{Fe}^{2+} (aq) \rightarrow \text{Fe}^{3+} (aq)$ Reduction: $\text{MnO}_4^- (aq) + 8\text{H}^+ (aq) \rightarrow \text{Mn}^{2+} (aq) + 4\text{H}_2\text{O} (l)$ Step 5: Balance the charge by adding electrons ($\text{e}^-$). Oxidation: Reactant charge: $+2$ Product charge: $+3$ Add 1 electron to the product side: $$\text{Fe}^{2+} (aq) \rightarrow \text{Fe}^{3+} (aq) + \text{e}^-$$ Reduction: Reactant charge: $(-1) + 8(+1) = +7$ Product charge: $+2$ Add 5 electrons to the reactant side: $$\text{MnO}_4^- (aq) + 8\text{H}^+ (aq) + 5\text{e}^- \rightarrow \text{Mn}^{2+} (aq) + 4\text{H}_2\text{O} (l)$$ Step 6: Multiply the half-reactions to equalize the number of electrons. Multiply the oxidation half-reaction by 5: $$5\text{Fe}^{2+} (aq) \rightarrow 5\text{Fe}^{3+} (aq) + 5\text{e}^-$$ The reduction half-reaction remains the same: $$\text{MnO}_4^- (aq) + 8\text{H}^+ (aq) + 5\text{e}^- \rightarrow \text{Mn}^{2+} (aq) + 4\text{H}_2\text{O} (l)$$ Step 7: Add the balanced half-reactions and cancel common species. $$5\text{Fe}^{2+} (aq) + \text{MnO}_4^- (aq) + 8\text{H}^+ (aq) + 5\text{e}^- \rightarrow 5\text{Fe}^{3+} (aq) + 5\text{e}^- + \text{Mn}^{2+} (aq) + 4\text{H}_2\text{O} (l)$$ Cancel the 5 electrons from both sides: $$5\text{Fe}^{2+} (aq) + \text{MnO}_4^- (aq) + 8\text{H}^+ (aq) \rightarrow 5\text{Fe}^{3+} (aq) + \text{Mn}^{2+} (aq) + 4\text{H}_2\text{O} (l)$$ Step 8: Verify that atoms and charges are balanced. Atoms: Fe (5), Mn (1), O (4), H (8) are balanced on both sides. Charge: Reactants: $5(+2) + (-1) + 8(+1) = +10 - 1 + 8 = +17$. Products: $5(+3) + (+2) + 0 = +15 + 2 = +17$. Charges are balanced. The balanced redox reaction is: $$\boxed{5\text{Fe}^{2+} (aq) + \text{MnO}_4^- (aq) + 8\text{H}^+ (aq) \rightarrow 5\text{Fe}^{3+} (aq) + \text{Mn}^{2+} (aq) + 4\text{H}_2\text{O} (l)}$$ Identifying the oxidizing and reducing agents: • The species that is oxidized (loses electrons) is the reducing agent. Here, $\text{Fe}^{2+}$ is oxidized to $\text{Fe}^{3+}$. • The species that is reduced (gains electrons) is the oxidizing agent. Here, $\text{MnO}_4^-$ is reduced to $\text{Mn}^{2+}$. Oxidizing agent: $\boxed{\text{MnO}_4^-}$ Reducing agent: $\boxed{\text{Fe}^{2+}}$ What's next?

Separate the reaction into half-reactions.