This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Alright nkemdivo — let's do this. b) i) Write down the half equation for A: Oxidation: C_2O_4^2-(aq) → 2CO_2(g) + 2e^- B: Reduction: MnO_4^-(aq) + 8H^+(aq) + 5e^- → Mn^2+(aq) + 4H_2O(l) ii) Write down the overall balanced equation Step 1: Multiply the oxidation half-equation by 5 and the reduction half-equation by 2 to balance the electrons. 5C_2O_4^2-(aq) → 10CO_2(g) + 10e^- 2MnO_4^-(aq) + 16H^+(aq) + 10e^- → 2Mn^2+(aq) + 8H_2O(l) Step 2: Add the two half-equations and cancel out the electrons. 5C_2O_4^2-(aq) + 2MnO_4^-(aq) + 16H^+(aq) → 10CO_2(g) + 2Mn^2+(aq) + 8H_2O(l) iii) Calculate the concentration of the KMnO_4 solution. Step 1: Calculate the moles of oxalic acid (H_2C_2O_4). Volume of oxalic acid = 50 cm^3 = 0.050 dm^3. Concentration of oxalic acid = 0.25 M. Moles of C_2O_4^2- = Concentration × Volume = 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, 2 moles of MnO_4^- react with 5 moles of C_2O_4^2-. Moles of MnO_4^- = 0.0125 mol C_2O_4^2- × 2 mol MnO_4^-5 mol C_2O_4^2- = 0.005 mol Step 3: Calculate the concentration of the KMnO_4 solution. Volume of KMnO_4 solution = 31.0 cm^3 = 0.0310 dm^3. Concentration of KMnO_4 = Moles of MnO_4^-Volume of KMnO_4 = 0.005 mol0.0310 dm^3 ≈ 0.16129 M Rounding to three significant figures: 0.161 M c) Bond energy can provide information about the energy changes that accompany a chemical reaction. i) What do you understand by the term bond energy? Bond energy is the amount of energy required to break one mole of a specific type of covalent bond in the gaseous state. It is also the energy released when one mole of that same bond is formed. Send me the next one 📸