Writing Oxidation and Reduction Half-Equations
What Are Half-Equations, and Why Do We Use Them?
Half-equation in a redox reaction
A half-equation focuses on one part of a redox reaction: either the oxidation or reduction process.
By breaking a redox reaction into these two components, we can better track the movement of electrons.
Let’s take the reaction between zinc and copper(II) sulfate as an example:
Full Reaction:
$$
\text{Zn}(s) + \text{Cu}^{2+}(aq) \to \text{Zn}^{2+}(aq) + \text{Cu}(s)
$$
This reaction involves two simultaneous processes:
- Zinc metal ($\text{Zn}$) is oxidized to form zinc ions ($\text{Zn}^{2+}$).
- Copper ions ($\text{Cu}^{2+}$) are reduced to form copper metal ($\text{Cu}$).
Writing Oxidation and Reduction Half-Equations
Step-by-Step Guide:
- Identify the species being oxidized and reduced:
- Oxidation: The species losing electrons.
- Reduction: The species gaining electrons.
- Write the unbalanced equation for each process:
- For oxidation, place the electrons ($e^-$) on the right-hand side of the equation.
- For reduction, place the electrons ($e^-$) on the left-hand side.
- Balance the atoms:
- Ensure the number of atoms of each element is the same on both sides of the equation.
- Balance the charges:
- Add electrons to one side of the equation to balance the overall charge.
- Combine the half-equations:
- Ensure the number of electrons lost in oxidation equals the number of electrons gained in reduction.
- Add the two half-equations together, canceling out the electrons.
Zinc and Copper(II) Sulfate Reaction
Step 1: Identify the species being oxidized and reduced
- Zinc($\text{Zn}$) is oxidized: it loses electrons to form $\text{Zn}^{2+}$.
- Copper ions($\text{Cu}^{2+}$) are reduced: they gain electrons to form $\text{Cu}$.
Step 2: Write the unbalanced half-equations
- Oxidation: $\text{Zn} \to \text{Zn}^{2+}$
- Reduction: $\text{Cu}^{2+} \to \text{Cu}$
Step 3: Balance the atoms
- Both equations already have balanced atoms.
Step 4: Balance the charges
- Oxidation: $\text{Zn} \to \text{Zn}^{2+} + 2e^-$
- Reduction: $\text{Cu}^{2+} + 2e^- \to \text{Cu}$
Step 5: Combine the half-equations
- Add the two half-equations together:
$$
\text{Zn} + \text{Cu}^{2+} \to \text{Zn}^{2+} + \text{Cu}
$$ - The electrons cancel out, and the overall reaction is balanced.
Balancing Half-Equations in Acidic or Neutral Solutions
When redox reactions occur in aqueous solutions, water ($\text{H}_2\text{O}$), hydrogen ions ($\text{H}^+$), or hydroxide ions ($\text{OH}^-$) may need to be included to balance oxygen and hydrogen atoms in the half-equations.
Steps for Acidic Solutions:
- Balance all atoms except hydrogen and oxygen.
- Balance oxygen atoms by adding $\text{H}_2\text{O}$ molecules.
- Balance hydrogen atoms by adding $\text{H}^+$ ions.
- Balance the charges by adding electrons ($e^-$).
Reduction of Dichromate Ions ($\text{Cr}_2\text{O}_7^{2-}$) in Acidic Solution
Step 1: Write the unbalanced equation
$$
\text{Cr}_2\text{O}_7^{2-} \to \text{Cr}^{3+}
$$
Step 2: Balance chromium atoms
$$
\text{Cr}_2\text{O}_7^{2-} \to 2\text{Cr}^{3+}
$$
Step 3: Balance oxygen atoms by adding $\text{H}_2\text{O}$
$$
\text{Cr}_2\text{O}_7^{2-} \to 2\text{Cr}^{3+} + 7\text{H}_2\text{O}
$$
Step 4: Balance hydrogen atoms by adding $\text{H}^+$
$$
\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \to 2\text{Cr}^{3+} + 7\text{H}_2\text{O}
$$
Step 5: Balance the charges by adding electrons
- Left-hand side: $2-$ (from $\text{Cr}_2\text{O}_7^{2-}$) + $14+$ (from $\text{H}^+$) = $12+$
- Right-hand side: $6+$ (from $2\text{Cr}^{3+}$)
- Add 6 electrons to the left-hand side:
$$
\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \to 2\text{Cr}^{3+} + 7\text{H}_2\text{O}
$$
When balancing half-equations in acidic solutions, always use $\text{H}_2\text{O}$ to balance oxygen and $\text{H}^+$ to balance hydrogen.
Steps for Neutral Solutions:
- Balance all atoms except hydrogen and oxygen.
- Balance oxygen atoms by adding $H_2O$.
- Balance hydrogen atoms by adding $H_2O$ or $OH^-$.
- Balance charges by adding electrons ($e^-$).
Reduction of Manganese Dioxide ($MnO_2$) in Neutral Solution
Step 1: Write the unbalanced equation
$$MnO_2 (s) \rightarrow Mn^{2+} (aq)$$
Step 2: Balance manganese atoms
$$MnO_2 (s) \rightarrow Mn^{2+} (aq)$$
(No changes needed as manganese is already balanced.)
Step 3: Balance oxygen atoms by adding $H_2O$
$$MnO_2 (s) \rightarrow Mn^{2+} (aq) + 2H_2O (l)$$
Step 4: Balance hydrogen atoms by adding $OH^-$
$$MnO_2 (s) + 4H^+ (aq) \rightarrow Mn^{2+} (aq) + 2H_2O (l)$$
Step 5: Balance charges by adding electrons
- Left side: $+4$ (from $4H^+$)
- Right side: $+2$ (from $Mn^{2+}$)
- Add 2 electrons to the right side:
$$MnO_2 (s) + 4H^+ (aq) + 2e^- \rightarrow Mn^{2+} (aq) + 2H_2O (l) $$
Applications of Half-Equations
Electrochemical Cells
Half-equations describe the reactions at the anode (oxidation) and cathode (reduction) in batteries and fuel cells. For example:
- In a zinc-copper voltaic cell:
- Anode: $\text{Zn} \to \text{Zn}^{2+} + 2e^-$
- Cathode: $\text{Cu}^{2+} + 2e^- \to \text{Cu}$
Corrosion
The rusting of iron involves redox reactions:
- Oxidation: $\text{Fe} \to \text{Fe}^{2+} + 2e^-$
- Reduction: $\text{O}_2 + 4\text{H}^+ + 4e^- \to 2\text{H}_2\text{O}$
- Forgetting to balance charges: Ensure the total charge is the same on both sides of the half-equation.
- Ignoring the solution type: Use $\text{H}^+$ for acidic solutions and $\text{OH}^-$ for basic solutions.
- Not canceling electrons: When combining half-equations, the number of electrons must match.
- Write the oxidation and reduction half-equations for the reaction between magnesium and hydrochloric acid.
- Balance the following half-equation in acidic solution: $\text{MnO}_4^- \to \text{Mn}^{2+}$.
- Why is it impossible to have oxidation without reduction in a redox reaction?
