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. $$\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2$$ Step 2: Balance oxygen atoms by adding $\text{H}_2\text{O}$ (already balanced). Step 3: Balance hydrogen atoms by adding $\text{H}^{+}$ (not needed). Step 4: Balance charge by adding electrons. $$\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2\text{e}^-$$ The half-equation for oxidation is: $$\boxed{\text{C}_2\text{O}_4^{2-}(\text{aq}) \rightarrow 2\text{CO}_2(\text{g}) + 2\text{e}^-}$$ B: Reduction: Step 1: Balance atoms other than O and H. $$\text{MnO}_4^{-} \rightarrow \text{Mn}^{2+}$$ Step 2: Balance oxygen atoms by adding $\text{H}_2\text{O}$. $$\text{MnO}_4^{-} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$$ Step 3: Balance hydrogen atoms by adding $\text{H}^{+}$. $$\text{MnO}_4^{-} + 8\text{H}^{+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$$ Step 4: Balance charge by adding electrons. $$\text{MnO}_4^{-} + 8\text{H}^{+} + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$$ The half-equation for reduction is: $$\boxed{\text{MnO}_4^{-}(\text{aq}) + 8\text{H}^{+}(\text{aq}) + 5\text{e}^- \rightarrow \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})}$$ ii) Write down the overall balanced equation: Step 1: Multiply the half-equations to balance the electrons. Oxidation: $(\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2\text{e}^-) \times 5$$ $$5\text{C}_2\text{O}_4^{2-} \rightarrow 10\text{CO}_2 + 10\text{e}^-$$ Reduction: $(\text{MnO}_4^{-} + 8\text{H}^{+} + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}) \times 2$$ $$2\text{MnO}_4^{-} + 16\text{H}^{+} + 10\text{e}^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}$$ Step 2: Add the two balanced half-equations and cancel electrons. $$5\text{C}_2\text{O}_4^{2-}(\text{aq}) + 2\text{MnO}_4^{-}(\text{aq}) + 16\text{H}^{+}(\text{aq}) \rightarrow 10\text{CO}_2(\text{g}) + 2\text{Mn}^{2+}(\text{aq}) + 8\text{H}_2\text{O}(\text{l})$$ The overall balanced equation is: $$\boxed{5\text{C}_2\text{O}_4^{2-}(\text{aq}) + 2\text{MnO}_4^{-}(\text{aq}) + 16\text{H}^{+}(\text{aq}) \rightarrow 10\text{CO}_2(\text{g}) + 2\text{Mn}^{2+}(\text{aq}) + 8\text{H}_2\text{O}(\text{l})}$$ iii) Calculate the concentration of $\text{KMnO}_4$ solution. Given: Volume of $\text{KMnO}_4$ solution ($V_{\text{KMnO}_4}$) = $31.0 \text{ cm}^3 = 0.0310 \text{ dm}^3$ Volume of oxalic acid ($V_{\text{H}_2\text{C}_2\text{O}_4}$) = $50 \text{ cm}^3 = 0.050 \text{ dm}^3$ Concentration of oxalic acid ($C_{\text{H}_2\text{C}_2\text{O}_4}$) = $0.25 \text{ M}$ Step 1: Calculate the moles of oxalic acid ($\text{C}_2\text{O}_4^{2-}$). $$n_{\text{C}_2\text{O}_4^{2-}} = C_{\text{H}_2\text{C}_2\text{O}_4} \times V_{\text{H}_2\text{C}_2\text{O}_4}$$ $$n_{\text{C}_2\text{O}_4^{2-}} = 0.25 \text{ mol/dm}^3 \times 0.050 \text{ dm}^3 = 0.0125 \text{ mol}$$ Step 2: Use the mole ratio from the balanced equation to find the moles of $\text{MnO}_4^{-}$. From the balanced equation: $5\text{C}_2\text{O}_4^{2-}$ reacts with $2\text{MnO}_4^{-}$. $$n_{\text{MnO}_4^{-}} = n_{\text{C}_2\text{O}_4^{2-}} \times \frac{2}{5}$$ $$n_{\text{MnO}_4^{-}} = 0.0125 \text{ mol} \times \frac{2}{5} = 0.005 \text{ mol}$$ Step 3: Calculate the concentration of $\text{KMnO}_4$ solution (which is the concentration of $\text{MnO}_4^{-}$). $$C_{\text{KMnO}_4} = \frac{n_{\text{MnO}_4^{-}}}{V_{\text{KMnO}_4}}$$ $$C_{\text{KMnO}_4} = \frac{0.005 \text{ mol}}{0.0310 \text{ dm}^3}$$ $$C_{\text{KMnO}_4} \approx 0.16129 \text{ M}$$ Rounding to three significant figures: $$C_{\text{KMnO}_4} \approx 0.161 \text{ M}$$ The concentration of $\text{KMnO}_4$ solution is: $$\boxed{\text{0.161 M}}$$ That's 2 down. 3 left today — send the next one.