To identify the oxidizing agent, we need to determine the change in oxidation states for the elements involved in the redox reaction. An oxidizing agent is the substance that gets reduced (gains electrons) and causes another substance to be oxidized. The given reaction is: $$2MnO_4^{-} + 10Cl^{-} + 16H^{+} \rightarrow 2Mn^{2+} + 5Cl_2 + 8H_2O$$ Step 1: Determine the oxidation states of Manganese (Mn) and Chlorine (Cl) in the reactants and products. For $MnO_4^{-}$: Let the oxidation state of Mn be $x$. Oxygen typically has an oxidation state of -2. The sum of oxidation states in the $MnO_4^{-}$ ion must equal its charge, -1. $$x + 4(-2) = -1$$ $$x - 8 = -1$$ $$x = +7$$ So, Mn in $MnO_4^{-}$ has an oxidation state of +7. For $Cl^{-}$: The oxidation state of a monatomic ion is equal to its charge. So, Cl in $Cl^{-}$ has an oxidation state of -1. For $Mn^{2+}$: The oxidation state of a monatomic ion is equal to its charge. So, Mn in $Mn^{2+}$ has an oxidation state of +2. For $Cl_2$: Chlorine is in its elemental form, so its oxidation state is 0. Step 2: Identify which species is oxidized and which is reduced. Manganese (Mn): The oxidation state of Mn changes from +7 in $MnO_4^{-}$ to +2 in $Mn^{2+}$. This is a decrease in oxidation state, meaning Mn has gained electrons and undergone reduction*. Chlorine (Cl): The oxidation state of Cl changes from -1 in $Cl^{-}$ to 0 in $Cl_2$. This is an increase in oxidation state, meaning Cl has lost electrons and undergone oxidation*. Step 3: Identify the oxidizing agent. The species that gets reduced is the oxidizing agent. Since Manganese in $MnO_4^{-}$ is reduced, $MnO_4^{-}$ acts as the oxidizing agent. Comparing this with the given options: A. $Mn^{2+}$ (product of reduction) B. $Cl^{-}$ (reducing agent) C. $H_2O$ (product, not directly involved in redox change) D. $MnO_4^{-}$ (oxidizing agent) The correct option is D. The substance that serves as the oxidizing agent is $\boxed{\text{D. } MnO_4^{-}}$. That's 2 down. 3 left today — send the next one.