This business/management problem is solved step by step below, with detailed explanations to help you understand the method and arrive at the correct answer.
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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. C_2O_4^2- → 2CO_2 Step 2: Balance oxygen atoms by adding H_2O (already balanced). Step 3: Balance hydrogen atoms by adding H^+ (not needed). Step 4: Balance charge by adding electrons. C_2O_4^2- → 2CO_2 + 2e^- The half-equation for oxidation is: C_2O_4^2-(aq) → 2CO_2(g) + 2e^- B: Reduction: Step 1: Balance atoms other than O and H. MnO_4^- → Mn^2+ Step 2: Balance oxygen atoms by adding H_2O. MnO_4^- → Mn^2+ + 4H_2O Step 3: Balance hydrogen atoms by adding H^+. MnO_4^- + 8H^+ → Mn^2+ + 4H_2O Step 4: Balance charge by adding electrons. MnO_4^- + 8H^+ + 5e^- → Mn^2+ + 4H_2O The half-equation for reduction is: MnO_4^-(aq) + 8H^+(aq) + 5e^- → Mn^2+(aq) + 4H_2O(l) ii) Write down the overall balanced equation: Step 1: Multiply the half-equations to balance the electrons. Oxidation: (C_2O_4^2- → 2CO_2 + 2e^-) × 5 5C_2O_4^2- → 10CO_2 + 10e^- Reduction: (MnO_4^- + 8H^+ + 5e^- → Mn^2+ + 4H_2O) × 2 2MnO_4^- + 16H^+ + 10e^- → 2Mn^2+ + 8H_2O Step 2: Add the two balanced half-equations and cancel electrons. 5C_2O_4^2-(aq) + 2MnO_4^-(aq) + 16H^+(aq) → 10CO_2(g) + 2Mn^2+(aq) + 8H_2O(l) The overall balanced equation is: 5C_2O_4^2-(aq) + 2MnO_4^-(aq) + 16H^+(aq) → 10CO_2(g) + 2Mn^2+(aq) + 8H_2O(l) iii) Calculate the concentration of KMnO_4 solution. Given: Volume of KMnO_4 solution (V_KMnO_4) = 31.0 cm^3 = 0.0310 dm^3 Volume of oxalic acid (V_H_2C_2O_4) = 50 cm^3 = 0.050 dm^3 Concentration of oxalic acid (C_H_2C_2O_4) = 0.25 M Step 1: Calculate the moles of oxalic acid (C_2O_4^2-). n_C_2O_4^2- = C_H_2C_2O_4 × V_H_2C_2O_4 n_C_2O_4^2- = 0.25 mol/dm^3 × 0.050 dm^3 = 0.0125 mol Step 2: Use the mole ratio from the balanced equation to find the moles of MnO_4^-. From the balanced equation: 5C_2O_4^2- reacts with 2MnO_4^-. n_MnO_4^- = n_C_2O_4^2- × (2)/(5) n_MnO_4^- = 0.0125 mol × (2)/(5) = 0.005 mol Step 3: Calculate the concentration of KMnO_4 solution (which is the concentration of MnO_4^-). C_KMnO_4 = n_MnO_4^-V_KMnO_4 C_KMnO_4 = 0.005 mol0.0310 dm^3 C_KMnO_4 ≈ 0.16129 M Rounding to three significant figures: C_KMnO_4 ≈ 0.161 M The concentration of KMnO_4 solution is: 0.161 M That's 2 down. 3 left today — send the next one.