Give an example to show oxidation and reduction go simultaneously. Balance the following equation by oxidation number or ion electron method HNO3 + H2S SO2 + NO2 + H2O [2+3]

Model Answer

Example of Simultaneous Oxidation and Reduction (Redox Reaction)

A classic example of oxidation and reduction occurring simultaneously is the reaction between zinc and hydrochloric acid: $$\mathrm{Zn}+2\mathrm{HCl}\to \mathrm{ZnCl}{}_2+\mathrm{H}{}_2$$

• Oxidation: Zinc ($\mathrm{Zn}$) loses electrons and is oxidized to $\mathrm{Zn}{}^{2+}$ (oxidation state increases from 0 to +2).
• Reduction: Hydrogen ($\mathrm{H}{}^{+}$) gains electrons and is reduced to $\mathrm{H}{}_2$ (oxidation state decreases from +1 to 0).

This shows that oxidation and reduction always occur together in a redox reaction.

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Balancing the Given Equation by Oxidation Number Method

The given equation is: $$\mathrm{HNO}{}_3+\mathrm{H}{}_2\mathrm{S}\to \mathrm{SO}{}_2+\mathrm{NO}{}_2+\mathrm{H}{}_2\mathrm{O}$$

Step 1: Assign oxidation numbers

• In $\mathrm{HNO}{}_3$: H = +1, O = −2, N = +5 (since $+1+x+3(-2)=0\Rightarrow x=+5$)
• In $\mathrm{H}{}_2\mathrm{S}$: H = +1, S = −2 (since $2(+1)+x=0\Rightarrow x=-2$)
• In $\mathrm{SO}{}_2$: S = +4 (since $x+2(-2)=0\Rightarrow x=+4$)
• In $\mathrm{NO}{}_2$: N = +4 (since $x+2(-2)=0\Rightarrow x=+4$)
• In $\mathrm{H}{}_2\mathrm{O}$: H = +1, O = −2

Step 2: Identify oxidation and reduction half-reactions

• Sulfur (S) is oxidized from −2 to +4 (loses 6 electrons per S atom).
• Nitrogen (N) is reduced from +5 to +4 (gains 1 electron per N atom).

Step 3: Balance electrons

• To balance electrons, multiply the oxidation half by 1 and the reduction half by 6 (LCM of 6 and 1):
  • Oxidation: $\mathrm{S}{}^{-2}\to \mathrm{S}{}^{+4}+6\mathrm{e}{}^{-}$
  • Reduction: $6\mathrm{N}{}^{+5}+6\mathrm{e}{}^{-}\to 6\mathrm{N}{}^{+4}$

Step 4: Write the balanced ionic equation $$\mathrm{H}{}_2\mathrm{S}+6\mathrm{HNO}{}_3\to \mathrm{SO}{}_2+6\mathrm{NO}{}_2+6\mathrm{H}{}_2\mathrm{O}$$

Step 5: Verify the balance

• Atoms: H = 2 + 6 = 8 (right side: 6 × 2 = 12 → Adjust coefficients) Correcting by balancing H atoms: $$\mathrm{H}{}_2\mathrm{S}+6\mathrm{HNO}{}_3\to \mathrm{SO}{}_2+6\mathrm{NO}{}_2+4\mathrm{H}{}_2\mathrm{O}$$ Now, H = 2 + 6 = 8 (right side: 4 × 2 = 8) ✔️
• Charges: Both sides are neutral.

Final balanced equation: $$\boxed{\mathrm{H}{}_2\mathrm{S}+6\mathrm{HNO}{}_3\to \mathrm{SO}{}_2+6\mathrm{NO}{}_2+4\mathrm{H}{}_2\mathrm{O}}$$

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Balancing by Ion-Electron Method

Step 1: Write half-reactions in acidic medium

• Oxidation half (Sulfur): $$\mathrm{H}{}_2\mathrm{S}\to \mathrm{SO}{}_2$$ Balance S and O: $$\mathrm{H}{}_2\mathrm{S}+2\mathrm{H}{}_2\mathrm{O}\to \mathrm{SO}{}_2+6\mathrm{H}{}^{+}+6\mathrm{e}{}^{-}$$ (Balance H by adding $\mathrm{H}{}_2\mathrm{O}$ and $\mathrm{H}{}^{+}$)

• Reduction half (Nitrogen): $$\mathrm{HNO}{}_3\to \mathrm{NO}{}_2$$ Balance N and O: $$\mathrm{HNO}{}_3+\mathrm{H}{}^{+}+\mathrm{e}{}^{-}\to \mathrm{NO}{}_2+\mathrm{H}{}_2\mathrm{O}$$ Multiply by 6 to balance electrons: $$6\mathrm{HNO}{}_3+6\mathrm{H}{}^{+}+6\mathrm{e}{}^{-}\to 6\mathrm{NO}{}_2+6\mathrm{H}{}_2\mathrm{O}$$

Step 2: Combine half-reactions $$\mathrm{H}{}_2\mathrm{S}+2\mathrm{H}{}_2\mathrm{O}+6\mathrm{HNO}{}_3\to \mathrm{SO}{}_2+6\mathrm{NO}{}_2+6\mathrm{H}{}_2\mathrm{O}+4\mathrm{H}{}^{+}$$ Simplify by canceling $\mathrm{H}{}_2\mathrm{O}$ and $\mathrm{H}{}^{+}$: $$\mathrm{H}{}_2\mathrm{S}+6\mathrm{HNO}{}_3\to \mathrm{SO}{}_2+6\mathrm{NO}{}_2+4\mathrm{H}{}_2\mathrm{O}$$

Final balanced equation (same as above): $$\boxed{\mathrm{H}{}_2\mathrm{S}+6\mathrm{HNO}{}_3\to \mathrm{SO}{}_2+6\mathrm{NO}{}_2+4\mathrm{H}{}_2\mathrm{O}}$$