Balance the following redox reactions: a) Zn (s) + Ag+(aq) → Zn2+(aq) + Ag (s) b) Cu2+(aq) + Cl-(aq) → Cu (s)…

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Balance the following redox reactions: a) Zn (s) + Ag+(aq) → Zn2+(aq) + Ag (s) b) Cu2+(aq) + Cl-(aq) → Cu (s) + Cl2(g) c) Pb2+(aq) + Br-(aq) → Pb (s) + Br2(aq) d) HCl (aq) + MnO2(s)→Cl2(g) + Mn2+(aq) This reaction takes place in an acidic medium.

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Here are the balanced redox reactions: a) Zn (s) + Ag⁺(aq) → Zn²⁺(aq) + Ag (s) Step 1: Separate into half-reactions. Oxidation: Zn (s) → Zn²⁺(aq) Reduction: Ag⁺(aq) → Ag (s) Step 2: Balance atoms and charge for each half-reaction. Oxidation: Zn (s) → Zn²⁺(aq) + 2e⁻ Reduction: Ag⁺(aq) + e⁻ → Ag (s) Step 3: Equalize electrons by multiplying the reduction half-reaction by 2. 2Ag⁺(aq) + 2e⁻ → 2Ag (s) Step 4: Add the half-reactions and cancel electrons. Zn (s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag (s) b) Cu²⁺(aq) + Cl⁻(aq) → Cu (s) + Cl₂(g) Step 1: Separate into half-reactions. Oxidation: Cl⁻(aq) → Cl₂(g) Reduction: Cu²⁺(aq) → Cu (s) Step 2: Balance atoms and charge for each half-reaction. Oxidation: 2Cl⁻(aq) → Cl₂(g) + 2e⁻ Reduction: Cu²⁺(aq) + 2e⁻ → Cu (s) Step 3: Electrons are already equal. Add the half-reactions and cancel electrons. Cu²⁺(aq) + 2Cl⁻(aq) → Cu (s) + Cl₂(g) c) Pb²⁺(aq) + Br⁻(aq) → Pb (s) + Br₂(aq) Step 1: Separate into half-reactions. Oxidation: Br⁻(aq) → Br₂(aq) Reduction: Pb²⁺(aq) → Pb (s) Step 2: Balance atoms and charge for each half-reaction. Oxidation: 2Br⁻(aq) → Br₂(aq) + 2e⁻ Reduction: Pb²⁺(aq) + 2e⁻ → Pb (s) Step 3: Electrons are already equal. Add the half-reactions and cancel electrons. Pb²⁺(aq) + 2Br⁻(aq) → Pb (s) + Br₂(aq) d) HCl (aq) + MnO₂(s) → Cl₂(g) + Mn²⁺(aq) (acidic medium) Step 1: Separate into half-reactions. Oxidation: Cl⁻(aq) → Cl₂(g) Reduction: MnO₂(s) → Mn²⁺(aq) Step 2: Balance atoms other than O and H. Oxidation: 2Cl⁻(aq) → Cl₂(g) Reduction: MnO₂(s) → Mn²⁺(aq) Step 3: Balance O atoms by adding H₂O. Reduction: MnO₂(s) → Mn²⁺(aq) + 2H₂O (l) Step 4: Balance H atoms by adding H⁺ (acidic medium). Reduction: MnO₂(s) + 4H⁺(aq) → Mn²⁺(aq) + 2H₂O (l) Step 5: Balance charge by adding electrons. Oxidation: 2Cl⁻(aq) → Cl₂(g) + 2e⁻ Reduction: MnO₂(s) + 4H⁺(aq) + 2e⁻ → Mn²⁺(aq) + 2H₂O (l) Step 6: Electrons are already equal. Add the half-reactions and cancel electrons. 2Cl⁻(aq) + MnO₂(s) + 4H⁺(aq) → Cl₂(g) + Mn²⁺(aq) + 2H₂O (l) Step 7: Convert to the full molecular equation using HCl as the source of H⁺ and Cl⁻, and forming MnCl₂ from Mn²⁺ and spectator Cl⁻ ions. We need 4H⁺ and 4Cl⁻ (2 for oxidation, 2 for MnCl₂) from 4HCl. 4HCl (aq) + MnO₂(s) → Cl₂(g) + MnCl₂(aq) + 2H₂O (l)

Balance the following redox reactions: a) Zn (s) + Ag+(aq) → Zn2+(aq) + Ag (s) b) Cu2+(aq) + Cl-(aq) → Cu (s) + Cl2(g) c) Pb2+(aq) + Br-(aq) → Pb (s) + Br2(aq) d) HCl (aq) + MnO2(s)→Cl2(g) + Mn2+(aq) This reaction takes place in an acidic medium.

Step-by-step solution

Part a· 4 steps

Separate into half-reactions.

Balance atoms and charge for each half-reaction.

Equalize electrons by multiplying the reduction half-reaction by 2.

Add the half-reactions and cancel electrons.

Part b· 3 steps

Separate into half-reactions.

Balance atoms and charge for each half-reaction.

Electrons are already equal.

Part c· 3 steps

Separate into half-reactions.

Balance atoms and charge for each half-reaction.

Electrons are already equal.

Part d· 7 steps

Separate into half-reactions.

Balance atoms other than O and H.

Balance O atoms by adding $\text{H}_{2}\text{O}$ .

Balance H atoms by adding $\text{H}^{+}$ (acidic medium).

Balance charge by adding electrons.

Electrons are already equal.

Convert to the full molecular equation using $\text{HCl}$ as the source of $\text{H}^{+}$ and $\text{Cl}^{-}$ , and forming $\text{MnCl}_{2}$ from $\text{Mn}^{2+}$ and spectator $\text{Cl}^{-}$ ions.

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More Chemistry Questions

Balance the following redox reactions: a) Zn (s) + Ag+(aq) → Zn2+(aq) + Ag (s) b) Cu2+(aq) + Cl-(aq) → Cu (s) + Cl2(g) c) Pb2+(aq) + Br-(aq) → Pb (s) + Br2(aq) d) HCl (aq) + MnO2(s)→Cl2(g) + Mn2+(aq) This reaction takes place in an acidic medium.

Step-by-step solution

Part a· 4 steps

Separate into half-reactions.

Balance atoms and charge for each half-reaction.

Equalize electrons by multiplying the reduction half-reaction by 2.

Add the half-reactions and cancel electrons.

Part b· 3 steps

Separate into half-reactions.

Balance atoms and charge for each half-reaction.

Electrons are already equal.

Part c· 3 steps

Separate into half-reactions.

Balance atoms and charge for each half-reaction.

Electrons are already equal.

Part d· 7 steps

Separate into half-reactions.

Balance atoms other than O and H.

Balance O atoms by adding H2O\text{H}_{2}\text{O}H2​O.

Balance H atoms by adding H+\text{H}^{+}H+ (acidic medium).

Balance charge by adding electrons.

Electrons are already equal.

Convert to the full molecular equation using HCl\text{HCl}HCl as the source of H+\text{H}^{+}H+ and Cl−\text{Cl}^{-}Cl−, and forming MnCl2\text{MnCl}_{2}MnCl2​ from Mn2+\text{Mn}^{2+}Mn2+ and spectator Cl−\text{Cl}^{-}Cl− ions.

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Balance O atoms by adding $\text{H}_{2}\text{O}$ .

Balance H atoms by adding $\text{H}^{+}$ (acidic medium).

Convert to the full molecular equation using $\text{HCl}$ as the source of $\text{H}^{+}$ and $\text{Cl}^{-}$ , and forming $\text{MnCl}_{2}$ from $\text{Mn}^{2+}$ and spectator $\text{Cl}^{-}$ ions.