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.

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:

Oxidation: $\text{C}_2\text{O}_4^{2-} \rightarrow \text{CO}_2$
Reduction: $\text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+}$

Oxidation: $\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2$
Reduction: $\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+}$

Oxidation: $\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2$ (O atoms are already balanced)
Reduction: $\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$

Oxidation: $\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2$ (H atoms are already balanced)
Reduction: $14\text{H}^+ + \text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$

Oxidation: $\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2\text{e}^-$ (Charge: $-2$