Here are the answers to your questions: 1a) Step 1: Determine the oxidation states of Cr and C in the reactants and products. In $\text{Cr}_2\text{O}_7^{2-}$: $2(\text{Cr}) + 7(-2) = -2 \Rightarrow 2(\text{Cr}) = +12 \Rightarrow \text{Cr} = +6$. In $\text{Cr}^{3+}$: $\text{Cr} = +3$. Chromium's oxidation state decreases from $+6$ to $+3$, so $\text{Cr}_2\text{O}_7^{2-}$ is reduced. In $\text{C}_2\text{O}_4^{2-}$: $2(\text{C}) + 4(-2) = -2 \Rightarrow 2(\text{C}) = +6 \Rightarrow \text{C} = +3$. In $\text{CO}_2$: $\text{C} + 2(-2) = 0 \Rightarrow \text{C} = +4$. Carbon's oxidation state increases from $+3$ to $+4$, so $\text{C}_2\text{O}_4^{2-}$ is oxidized. Step 2: Identify the oxidizing and reducing agents. The species that is reduced is the oxidizing agent. The species that is oxidized is the reducing agent. Oxidizing agent: $\boxed{\text{Cr}_2\text{O}_7^{2-}}$ Reducing agent: $\boxed{\text{C}_2\text{O}_4^{2-}}$ 1b) The given reaction is: $$\text{Cr}_2\text{O}_7^{2-} + \text{C}_2\text{O}_4^{2-} + \text{H}^+ \rightarrow \text{Cr}^{3+} + \text{CO}_2 + \text{H}_2\text{O}$$ Step 1: Write the unbalanced half-reactions. Oxidation: $\text{C}_2\text{O}_4^{2-} \rightarrow \text{CO}_2$ Reduction: $\text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+}$ Step 2: Balance atoms other than O and H. Oxidation: $\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2$ Reduction: $\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+}$ Step 3: Balance O atoms by adding $\text{H}_2\text{O}$. Oxidation: $\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2$ (O atoms are already balanced) Reduction: $\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$ Step 4: Balance H atoms by adding $\text{H}^+$ (since it's an acidic medium). Oxidation: $\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2$ (H atoms are already balanced) Reduction: $14\text{H}^+ + \text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$ Step 5: Balance charge by adding electrons ($\text{e}^-$). Oxidation: $\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2\text{e}^-$ (Charge: $-2$