a) A redox reaction is a chemical reaction in which there is a change in the oxidation states of atoms…

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a) A redox reaction is a chemical reaction in which there is a change in the oxidation states of atoms involved. This involves both reduction (gain of electrons) and oxidation (loss of electrons

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a) A redox reaction is a chemical reaction in which there is a change in the oxidation states of atoms involved. This involves both reduction (gain of electrons) and oxidation (loss of electrons) occurring simultaneously. b) i) A: Oxidation: Step 1: Balance atoms other than O and H. C₂O₄²⁻ → 2CO₂ Step 2: Balance oxygen atoms by adding H₂O (already balanced). Step 3: Balance hydrogen atoms by adding H⁺ (not needed). Step 4: Balance charge by adding electrons. C₂O₄²⁻ → 2CO₂ + 2e^− The half-equation for oxidation is: C₂O₄²⁻(aq) → 2CO₂(g) + 2e^− B: Reduction: Step 1: Balance atoms other than O and H. MnO₄⁻ → Mn²⁺ Step 2: Balance oxygen atoms by adding H₂O. MnO₄⁻ → Mn²⁺ + 4H₂O Step 3: Balance hydrogen atoms by adding H⁺. MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O Step 4: Balance charge by adding electrons. MnO₄⁻ + 8H⁺ + 5e^− → Mn²⁺ + 4H₂O The half-equation for reduction is: MnO₄⁻(aq) + 8H⁺(aq) + 5e^− → Mn²⁺(aq) + 4H₂O(l) ii) Write down the overall balanced equation: Step 1: Multiply the half-equations to balance the electrons. Oxidation: (C₂O₄²⁻ → 2CO₂ + 2e^−) × 5 5C₂O₄²⁻ → 10CO₂ + 10e^− Reduction: (MnO₄⁻ + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O) × 2 2MnO₄⁻ + 16H⁺ + 10e^- → 2Mn²⁺ + 8H₂O Step 2: Add the two balanced half−equations and cancel electrons. 5C₂O₄²⁻(aq) + 2MnO₄⁻(aq) + 16H⁺(aq) → 10CO₂(g) + 2Mn²⁺(aq) + 8H₂O(l) The overall balanced equation is: 5C₂O₄²⁻(aq) + 2MnO₄⁻(aq) + 16H⁺(aq) → 10CO₂(g) + 2Mn²⁺(aq) + 8H₂O(l) iii) Calculate the concentration of KMnO₄ solution. Given: Volume of KMnO₄ solution (VKMnO₄) = 31.0 cm³ = 0.0310 dm³; Volume of oxalic acid (VH₂C₂O₄) = 50 cm³ = 0.050 dm³; Concentration of oxalic acid (CH₂C₂O₄) = 0.25 M Step 1: Calculate the moles of oxalic acid (C₂O₄²⁻). nC₂O₄²⁻ = CH₂C₂O₄ × VH₂C₂O₄; nC₂O₄²⁻ = 0.25 mol/dm³ × 0.050 dm³ = 0.0125 mol Step 2: Use the mole ratio from the balanced equation to find the moles of MnO₄⁻. From the balanced equation: 5C₂O₄²⁻ reacts with 2MnO₄⁻. nMnO₄⁻ = nC₂O₄²⁻ × 2/5; nMnO₄⁻ = 0.0125 mol × 2/5 = 0.005 mol Step 3: Calculate the concentration of KMnO₄ solution (which is the concentration of MnO₄⁻). CKMnO₄ = nMnO₄⁻VKMnO₄; CKMnO₄ = 0.005 mol0.0310 dm³ CKMnO₄ ≈ 0.16129 M Rounding to three significant figures: CKMnO₄ ≈ 0.161 M The concentration of KMnO₄ solution is: 0.161 M

A) A redox reaction is a chemical reaction in which there is a change in the oxidation states of atoms involved. This involves both reduction (gain of electrons) and oxidation (loss of electrons

Step-by-step solution

Part b(i)· 8 steps

Balance atoms other than O and H.

Balance oxygen atoms by adding $\text{H}_2\text{O}$ (already balanced).

Balance hydrogen atoms by adding $\text{H}^{+}$ (not needed).

Balance charge by adding electrons.

Balance atoms other than O and H.

Balance oxygen atoms by adding $\text{H}_2\text{O}$ .

Balance hydrogen atoms by adding $\text{H}^{+}$ .

Balance charge by adding electrons.

Part b(ii)· 2 steps

Multiply the half-equations to balance the electrons.

Add the two balanced half-equations and cancel electrons.

Part b(iii)· 3 steps

Calculate the moles of oxalic acid ( $\text{C}_2\text{O}_4^{2-}$ ).

Use the mole ratio from the balanced equation to find the moles of $\text{MnO}_4^{-}$ .

Calculate the concentration of $\text{KMnO}_4$ solution (which is the concentration of $\text{MnO}_4^{-}$ ).

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A) A redox reaction is a chemical reaction in which there is a change in the oxidation states of atoms involved. This involves both reduction (gain of electrons) and oxidation (loss of electrons

Step-by-step solution

Part b(i)· 8 steps

Balance atoms other than O and H.

Balance oxygen atoms by adding H2O\text{H}_2\text{O}H2​O (already balanced).

Balance hydrogen atoms by adding H+\text{H}^{+}H+ (not needed).

Balance charge by adding electrons.

Balance atoms other than O and H.

Balance oxygen atoms by adding H2O\text{H}_2\text{O}H2​O.

Balance hydrogen atoms by adding H+\text{H}^{+}H+.

Balance charge by adding electrons.

Part b(ii)· 2 steps

Multiply the half-equations to balance the electrons.

Add the two balanced half-equations and cancel electrons.

Part b(iii)· 3 steps

Calculate the moles of oxalic acid (C2O42−\text{C}_2\text{O}_4^{2-}C2​O42−​).

Use the mole ratio from the balanced equation to find the moles of MnO4−\text{MnO}_4^{-}MnO4−​.

Calculate the concentration of KMnO4\text{KMnO}_4KMnO4​ solution (which is the concentration of MnO4−\text{MnO}_4^{-}MnO4−​).

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Balance oxygen atoms by adding $\text{H}_2\text{O}$ (already balanced).

Balance hydrogen atoms by adding $\text{H}^{+}$ (not needed).

Balance oxygen atoms by adding $\text{H}_2\text{O}$ .

Balance hydrogen atoms by adding $\text{H}^{+}$ .

Calculate the moles of oxalic acid ( $\text{C}_2\text{O}_4^{2-}$ ).

Use the mole ratio from the balanced equation to find the moles of $\text{MnO}_4^{-}$ .

Calculate the concentration of $\text{KMnO}_4$ solution (which is the concentration of $\text{MnO}_4^{-}$ ).