Justify by giving an example that oxidation and reduction go side by side. Balance the following equation by oxidation number or ion electron method. FeCl3 + H2S FeCl2 + HCl + S

Model Answer

Oxidation and Reduction Go Side by Side

Oxidation and reduction are complementary processes that always occur simultaneously in a redox reaction. This is known as the redox couple. Oxidation involves the loss of electrons (increase in oxidation number), while reduction involves the gain of electrons (decrease in oxidation number). Since electrons cannot exist alone, the electrons lost by one species must be gained by another.

Example: In the reaction between zinc and copper sulfate: $$\mathrm{Zn}+\mathrm{CuSO}{}_4\to \mathrm{ZnSO}{}_4+\mathrm{Cu}$$

• Oxidation: Zinc ($\mathrm{Zn}$) loses electrons and is oxidized to $\mathrm{Zn}{}^{2+}$ (oxidation number increases from 0 to +2).
• Reduction: Copper ($\mathrm{Cu}{}^{2+}$) gains electrons and is reduced to copper metal (oxidation number decreases from +2 to 0).

Thus, oxidation and reduction occur simultaneously in a redox reaction.

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Balancing the Given Equation by Oxidation Number Method

Given equation: $$\mathrm{FeCl}{}_3+\mathrm{H}{}_2\mathrm{S}\to \mathrm{FeCl}{}_2+\mathrm{HCl}+\mathrm{S}$$

Step 1: Assign oxidation numbers

• In $\mathrm{FeCl}{}_3$, Fe has +3, Cl has -1.
• In $\mathrm{H}{}_2\mathrm{S}$, H has +1, S has -2.
• In $\mathrm{FeCl}{}_2$, Fe has +2, Cl has -1.
• In $\mathrm{HCl}$, H has +1, Cl has -1.
• In $\mathrm{S}$, sulfur is in its elemental form (oxidation number = 0).

Step 2: Identify changes in oxidation numbers

• Fe: +3 → +2 (reduction, gains 1 electron)
• S: -2 → 0 (oxidation, loses 2 electrons)

Step 3: Balance electrons lost and gained

• To balance electrons, multiply Fe by 2 (to gain 2 electrons) and S by 1 (loses 2 electrons).
• New half-reactions:
  • Reduction: $\mathrm{Fe}{}^{3+}+\mathrm{e}{}^{-}\to \mathrm{Fe}{}^{2+}$
  • Oxidation: $\mathrm{S}{}^{2-}\to \mathrm{S}+2\mathrm{e}{}^{-}$

Step 4: Combine and balance atoms

• Multiply the reduction half by 2 to balance electrons: $$2\mathrm{Fe}{}^{3+}+2\mathrm{e}{}^{-}\to 2\mathrm{Fe}{}^{2+}$$
• Add to oxidation half: $$2\mathrm{Fe}{}^{3+}+\mathrm{S}{}^{2-}\to 2\mathrm{Fe}{}^{2+}+\mathrm{S}$$
• Now, balance Cl and H:
  • $\mathrm{FeCl}{}_3$ provides 6 Cl⁻, but only 2 Cl⁻ are used in $\mathrm{FeCl}{}_2$. The remaining 4 Cl⁻ form $\mathrm{HCl}$.
  • $\mathrm{H}{}_2\mathrm{S}$ provides 2 H⁺, which combine with 4 Cl⁻ to form 4 $\mathrm{HCl}$.

Final balanced equation: $$\boxed{2\mathrm{FeCl}{}_3+\mathrm{H}{}_2\mathrm{S}\to 2\mathrm{FeCl}{}_2+2\mathrm{HCl}+\mathrm{S}}$$

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Balancing by Ion-Electron Method

Step 1: Write half-reactions

• Reduction half: $\mathrm{Fe}{}^{3+}+\mathrm{e}{}^{-}\to \mathrm{Fe}{}^{2+}$
• Oxidation half: $\mathrm{S}{}^{2-}\to \mathrm{S}+2\mathrm{e}{}^{-}$

Step 2: Balance electrons

• Multiply reduction half by 2: $$2\mathrm{Fe}{}^{3+}+2\mathrm{e}{}^{-}\to 2\mathrm{Fe}{}^{2+}$$
• Add to oxidation half: $$2\mathrm{Fe}{}^{3+}+\mathrm{S}{}^{2-}\to 2\mathrm{Fe}{}^{2+}+\mathrm{S}$$

Step 3: Balance atoms in full equation

• Add $\mathrm{Cl}{}^{-}$ and $\mathrm{H}{}^{+}$ to balance: $$2\mathrm{FeCl}{}_3+\mathrm{H}{}_2\mathrm{S}\to 2\mathrm{FeCl}{}_2+2\mathrm{HCl}+\mathrm{S}$$

Final balanced equation (same as above): $$\boxed{2\mathrm{FeCl}{}_3+\mathrm{H}{}_2\mathrm{S}\to 2\mathrm{FeCl}{}_2+2\mathrm{HCl}+\mathrm{S}}$$