10.1 Oxidation States & Balancing Redox Reactions by Half-Reaction Method

10.1 Oxidation States & Balancing Redox Reactions by Half-Reaction Method

Quick Summary: Oxidation-reduction (redox) reactions involve the transfer of electrons between chemical species. Oxidation is the loss of electrons (an increase in oxidation state), while reduction is the gain of electrons (a decrease in oxidation state). Because mass and electrical charge must be conserved simultaneously, balancing complex redox reactions requires the systematic ion-electron (half-reaction) method in acidic or basic aqueous solutions. Disproportionation reactions and redox titrimetry highlight the practical and analytical utility of electron-transfer processes.

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1. Fundamentals of Oxidation-Reduction

Chemical reactions involving the exchange of valence electrons are classified as oxidation-reduction (redox) reactions. Two standard mnemonics track electron flow:

• OIL RIG: Oxidation Is Loss of electrons; Reduction Is Gain of electrons.
• LEO GER: Lose Electrons $\to$ Oxidation; Gain Electrons $\to$ Reduction.

Oxidation and reduction occur concurrently: free electrons do not accumulate in solution. The substance donating electrons undergoes oxidation and serves as the reducing agent (reductant) because it causes the reduction of another substance. Conversely, the substance accepting electrons undergoes reduction and acts as the oxidizing agent (oxidant).

Distinguishing Redox from Non-Redox Reactions

An inspection of oxidation states differentiates redox from non-redox processes:

• Precipitation (Non-Redox): $\text{Ag}^+(aq) + \text{Cl}^-(aq) \to \text{AgCl}(s)$ (Oxidation numbers remain $\text{Ag} = +1, \text{Cl} = -1$).
• Neutralization (Non-Redox): $\text{H}^+(aq) + \text{OH}^-(aq) \to \text{H}_2\text{O}(l)$ (Oxidation numbers remain $\text{H} = +1, \text{O} = -2$).
• Redox Process: $\text{Zn}(s) + 2\text{H}^+(aq) \to \text{Zn}^{2+}(aq) + \text{H}_2(g)$ (Zinc increases from $0 \to +2$; hydrogen decreases from $+1 \to 0$).

Rules for Assigning Oxidation States

1. Pure uncombined elements have an oxidation number of $0$ (e.g., $\text{O}_2, \text{P}_4, \text{Fe} = 0$).
2. Monatomic ions possess oxidation states equal to their ionic charge ($\text{Fe}^{3+} = +3, \text{Cl}^- = -1$).
3. Fluorine is always $-1$ in compounds. Group 1 metals are $+1$, and Group 2 metals are $+2$.
4. Hydrogen is $+1$ with nonmetals ($\text{H}_2\text{O}$) and $-1$ in metal hydrides ($\text{NaH}$).
5. Oxygen is $-2$ in most compounds, but is $-1$ in peroxides ($\text{H}_2\text{O}_2$) and $+2$ in $\text{OF}_2$.
6. The algebraic sum of oxidation numbers equals zero for neutral molecules and equals the net charge for polyatomic ions (e.g., in $\text{Cr}_2\text{O}_7^{2-}$, each $\text{Cr} = +6$).

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2. The Systematic Half-Reaction Balancing Algorithm

The ion-electron method separates the overall transformation into individual oxidation and reduction half-reactions, balancing mass and charge independently before recombining them.

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3. Worked Problems: Acidic and Basic Media

Problem 1: Balancing in Acidic Solution

Balance the reaction between permanganate and iron(II) in acidic solution: $\text{MnO}_4^-(aq) + \text{Fe}^{2+}(aq) \to \text{Mn}^{2+}(aq) + \text{Fe}^{3+}(aq)$

1. Separate half-reactions:
   • Reduction: $\text{MnO}_4^- \to \text{Mn}^{2+}$ (Mn: $+7 \to +2$)
   • Oxidation: $\text{Fe}^{2+} \to \text{Fe}^{3+}$ (Fe: $+2 \to +3$)
2. Balance non-H/O atoms: Both $\text{Mn}$ and $\text{Fe}$ are balanced ($1:1$).
3. Balance oxygen: Add $4\text{ H}_2\text{O}$ to products: $\text{MnO}_4^- \to \text{Mn}^{2+} + 4\text{ H}_2\text{O}$.
4. Balance hydrogen: Add $8\text{ H}^+$ to reactants: $\text{MnO}_4^- + 8\text{ H}^+ \to \text{Mn}^{2+} + 4\text{ H}_2\text{O}$.
5. Balance charge: Left side charge is $+7$; right side is $+2$. Add $5e^-$ to left: $\text{MnO}_4^- + 8\text{ H}^+ + 5e^- \to \text{Mn}^{2+} + 4\text{ H}_2\text{O}$ Oxidation half-reaction: add $1e^-$ to right: $\text{Fe}^{2+} \to \text{Fe}^{3+} + e^-$.
6. Equalize electrons: Multiply the oxidation half-reaction by $5$: $5\text{Fe}^{2+} \to 5\text{Fe}^{3+} + 5e^-$
7. Add and cancel: $\text{MnO}_4^-(aq) + 5\text{Fe}^{2+}(aq) + 8\text{H}^+(aq) \to \text{Mn}^{2+}(aq) + 5\text{Fe}^{3+}(aq) + 4\text{H}_2\text{O}(l)$ Net charge on both sides equals $+17$, confirming complete balance.

Problem 2: Balancing in Basic Solution

Balance the oxidation of sulfite by permanganate in basic solution: $\text{MnO}_4^-(aq) + \text{SO}_3^{2-}(aq) \to \text{MnO}_2(s) + \text{SO}_4^{2-}(aq)$

1. Balance half-reactions as if in acid:
   • Reduction: $\text{MnO}_4^- + 4\text{ H}^+ + 3e^- \to \text{MnO}_2 + 2\text{ H}_2\text{O}$
   • Oxidation: $\text{SO}_3^{2-} + \text{H}_2\text{O} \to \text{SO}_4^{2-} + 2\text{ H}^+ + 2e^-$
2. Equalize electrons ($6e^-$ transferred): Multiply reduction by $2$ and oxidation by $3$: $2\text{MnO}_4^- + 8\text{ H}^+ + 6e^- \to 2\text{MnO}_2 + 4\text{ H}_2\text{O}$ $3\text{SO}_3^{2-} + 3\text{ H}_2\text{O} \to 3\text{SO}_4^{2-} + 6\text{ H}^+ + 6e^-$
3. Combine and simplify in acid: $2\text{MnO}_4^- + 3\text{SO}_3^{2-} + 2\text{H}^+ \to 2\text{MnO}_2 + 3\text{SO}_4^{2-} + \text{H}_2\text{O}$
4. Convert to basic medium: Add $2\text{ OH}^-$ to both sides. On the left, $2\text{H}^+ + 2\text{OH}^- = 2\text{H}_2\text{O}$: $2\text{MnO}_4^- + 3\text{SO}_3^{2-} + 2\text{H}_2\text{O} \to 2\text{MnO}_2 + 3\text{SO}_4^{2-} + \text{H}_2\text{O} + 2\text{OH}^-$ Subtract $1\text{ H}_2\text{O}$ from both sides to obtain the final equation: $2\text{MnO}_4^-(aq) + 3\text{SO}_3^{2-}(aq) + \text{H}_2\text{O}(l) \to 2\text{MnO}_2(s) + 3\text{SO}_4^{2-}(aq) + 2\text{OH}^-(aq)$ Net charge on both sides is $-8$, and all atoms balance.

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4. Disproportionation Reactions & Redox Titrations

Disproportionation Reactions

A disproportionation reaction occurs when a single chemical species is simultaneously oxidized and reduced, forming two distinct products:

• Hydrogen Peroxide Decomposition: In $\text{H}_2\text{O}_2$, oxygen has an oxidation state of $-1$. Upon decomposition, it disproportionates into water ($\text{O} = -2$) and oxygen gas ($\text{O} = 0$): $2\text{H}_2\text{O}_2(aq) \to 2\text{H}_2\text{O}(l) + \text{O}_2(g)$
• Halogen disproportionation in base: Elemental chlorine dissolves in cold base to yield chloride and hypochlorite: $\text{Cl}_2(g) + 2\text{OH}^-(aq) \to \text{Cl}^-(aq) + \text{ClO}^-(aq) + \text{H}_2\text{O}(l)$ Chlorine ($0$) is reduced to $-1$ in $\text{Cl}^-$ and oxidized to $+1$ in $\text{ClO}^-$.

Quantitative Redox Titrations

Redox titrimetry quantifies unknown analyte concentrations via precise electron stoichiometry:

• Permanganate ($\text{MnO}_4^-$): In acidic solution, potassium permanganate acts as a self-indicating titrant. The deeply purple $\text{MnO}_4^-$ ion reduces to nearly colorless $\text{Mn}^{2+}$. A persistent faint pink color signals the stoichiometric equivalence point.
• Dichromate ($\text{Cr}_2\text{O}_7^{2-}$): Orange potassium dichromate reduces to green $\text{Cr}^{3+}$. Because the orange-to-green transition is gradual, an auxiliary redox indicator (such as diphenylamine sulfonate) is required to detect the endpoint.