Question

Balance each of the following oxidation-reduction reactions by using the oxidation states method. a. ${\mathrm{Cl}}_2(g)+\mathrm{Al}(s)\to {\mathrm{Al}}^{3+}(aq)+{\mathrm{Cl}}^{-}(aq)$ b. ${\mathbf{O}}_2(g)+{\mathbf{H}}_2\mathbf{O}(l)+\mathbf{P}\mathbf{b}(s)\to \mathbf{P}\mathbf{b}(\mathbf{O}\mathbf{H}{)}_2(s)$ c. ${\mathrm{H}}^{+}(aq)+{\mathrm{MnO}}_4^{-}(aq)+{\mathrm{Fe}}^{2+}(aq)\to$ ${\mathrm{Mn}}^{2+}(aq)+{\mathrm{Fe}}^{3+}(aq)+{\mathrm{H}}_2\mathrm{O}(i)$

Short answer

The balanced equations for the given redox reactions are: a. $3Al(s)+6C{l}_2(g)\to 3A{l}^{3+}(aq)+12C{l}^{-}(aq)$ b. $2Pb(s)+O_2(g)+2H_{2}O(l)\to 2Pb(OH{)}_2(s)$ c. $8H^{+}(aq)+Mn{O}_4^{-}(aq)+5F{e}^{2+}(aq)\to M{n}^{2+}(aq)+5F{e}^{3+}(aq)+4H_{2}O(l)$

Step-by-step solution

(a) Identify oxidation states

: We start by identifying the oxidation states of all the species involved: $C{l}_2$: $C{l}_1$ and $C{l}_2$ have oxidation states of 0 each. $Al(s)$: The oxidation state of metallic Al is 0. $A{l}^{3+}(aq)$: The oxidation state of Al is +3. $C{l}^{-}(aq)$: The oxidation state of Cl is -1.

(a) Determine change in oxidation states

: Now, we'll determine the change in oxidation states between reactants and products: $Al$: Changes from an oxidation state of 0 to +3. $C{l}_2$: Both $C{l}_1$ and $C{l}_2$ change from an oxidation state of 0 to -1.

(a) Balance half-reactions

: Balance the half-reactions by balancing the atoms and adding electrons where appropriate: Oxidation half-reaction: $$Al\to A{l}^{3+}+3e^{-}$$ Reduction half-reaction: $$2C{l}_2+6e^{-}\to 4C{l}^{-}$$

(a) Combine half-reactions

: Combine the balanced half-reactions. To do this, we need to multiply each half-reaction by the appropriate factor so they have the same number of electrons: 3(oxidation half-reaction) + 2(reduction half-reaction): $$3Al+6C{l}_2\to 3A{l}^{3+}+12C{l}^{-}$$ The balanced equation for reaction (a) is: $$3Al(s)+6C{l}_2(g)\to 3A{l}^{3+}(aq)+12C{l}^{-}(aq)$$

(b) Identify oxidation states

: First, we identify the oxidation states of the species involved in the reaction: $O_2$: The oxidation state of O in $O_2$ is 0. $H_{2}O$: The oxidation state of H is +1 and that of O is -2. $Pb(s)$: The oxidation state of metallic Pb is 0. $Pb(OH{)}_2$: The oxidation state of Pb is +2, O is -2 and H is +1.

(b) Determine change in oxidation states

: Next, we determine the change in oxidation states between reactants and products: $O_2$: Changes from an oxidation state of 0 to -2. $Pb(s)$: Changes from an oxidation state of 0 to +2.

(b) Balance half-reactions

: Balance the half-reactions by balancing the atoms and adding electrons where appropriate: Oxidation half-reaction: $$Pb\to P{b}^{2+}+2e^{-}$$ Reduction half-reaction: $$O_2+2H_{2}O+4e^{-}\to 4O{H}^{-}$$

(b) Combine half-reactions

: Combine the balanced half-reactions: 2(oxidation half-reaction) + 1(reduction half-reaction): $$2Pb+O_2+2H_{2}O\to 2P{b}^{2+}+4O{H}^{-}$$ The balanced equation for reaction (b) is: $$2Pb(s)+O_2(g)+2H_{2}O(l)\to 2Pb(OH{)}_2(s)$$

(c) Identify oxidation states

: Identify the oxidation states of species involved in the reaction: $H^{+}$: The oxidation state of H is +1. $Mn{O}_4^{-}$: The oxidation state of Mn is +7, O is -2. $F{e}^{2+}$: The oxidation state of Fe is +2. $M{n}^{2+}$: The oxidation state of Mn is +2. $F{e}^{3+}$: The oxidation state of Fe is +3. $H_{2}O$: The oxidation state of H is +1 and that of O is -2.

(c) Determine change in oxidation states

: Determine the change in oxidation states between reactants and products: $Mn{O}_4^{-}$: Mn changes from an oxidation state of +7 to +2. $F{e}^{2+}$: Changes from an oxidation state of +2 to +3.

(c) Balance half-reactions

: Balance the half-reactions by balancing the atoms and adding electrons where appropriate: Oxidation half-reaction: $$5F{e}^{2+}\to 5F{e}^{3+}+5e^{-}$$ Reduction half-reaction: $$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$$

(c) Combine half-reactions

: Combine the balanced half-reactions: 1(oxidation half-reaction) + 1(reduction half-reaction): $$Mn{O}_4^{-}+8H^{+}+5F{e}^{2+}\to M{n}^{2+}+5F{e}^{3+}+4H_{2}O$$ The balanced equation for reaction (c) is: $$8H^{+}(aq)+Mn{O}_4^{-}(aq)+5F{e}^{2+}(aq)\to M{n}^{2+}(aq)+5F{e}^{3+}(aq)+4H_{2}O(l)$$

Key concepts

Oxidation States Method

The oxidation states method is a powerful tool used to balance oxidation-reduction reactions. It involves assigning oxidation numbers to each element in the reaction and analyzing changes in these numbers to identify which elements are oxidized and which are reduced.

• **Identify Changes:** Determine the oxidation state of each atom in the reactants and products. This helps in tracking electron transfer.
• **Recognize Oxidation and Reduction:** Oxidation involves an increase in oxidation state, while reduction involves a decrease.
• **Calculate Electron Transfer:** For instance, if aluminum (Al) changes from 0 in the reactant to +3 in Al³⁺ (product), it implies a loss of three electrons.

Balancing oxidation and reduction events separately makes the overall balancing much clearer and systematic.

Balancing Chemical Equations

Balancing chemical equations is essential in ensuring that the number of atoms for each element is the same on both sides of the equation.

• **Conserve Mass:** The first rule is the law of conservation of mass. A balanced equation must have equal numbers of each type of atom on both sides.
• **Use Coefficients:** Adjust coefficients before the chemical formulas to balance the number of atoms. For example, balancing reaction (a) requires that both sides have three Al atoms and six Cl₂ molecules.
• **Iterative Process:** This often involves trial and error, balancing one element at a time.

Balancing equations is crucial for accurately representing chemical reactions.

Half-Reaction Method

The half-reaction method is another systematic way to balance redox reactions. It splits the overall reaction into two half-reactions: oxidation and reduction.

• **Split and Balance:** Begin by separating the reaction into oxidation and reduction parts. For instance, the oxidation of Fe²⁺ to Fe³⁺ and the reduction of MnO₄^- to Mn²⁺.
• **Electron Balance:** Add electrons to balance the electron loss and gain in each half-reaction. This ensures both mass and charge are balanced.
• **Combine:** Finally, combine the half-reactions by equalizing the electron transfer. This often involves multiplying each half-reaction by an appropriate factor.

By focusing on each half-reaction separately, balancing complex redox reactions becomes more manageable.