In the redox reaction, which is the reducing agent? #MnO2(s) + 4H(aq) + 2Cl^(-)(aq) -> Mn^(2+)(aq) + 2H_2O + Cl_2(g)#?

1 Answer
Apr 12, 2017

By checking the changes that occur in their oxidation numbers, we see that #Cl^-# is the reducing agent, and #MnO_2# is the oxidizing agent.


The simplest way to determine the reducing agent (or the oxidizing agent for that mattter) is to determine the oxidation number of each atom in the equation.

By the way, I think the equation you have given should have the second substance written as #H^+# rather than just #H#. Free hydrogen atoms do not exist in aqueous solution, but hydrogen ions do.

So, we will focus on the Mn atoms and the Cl atoms as these will have changes in their oxidation numbers.

In #MnO_2#, since we assign -2 as the value of the two oxygen atoms, and since the total of the oxidation numbers must be zero (the real charge on the formula), the Mn atom has a value of +4.

In the #Cl^-# ion, the oxidation number is the real charge, -1.

Looking at the products, the #Mn^(2+)# ion now has oxidation number +2. This decrease in the oxidation number tells us that the Mn atom has been reduced. Therefore, the compound in which it is found #(MnO_2)# is the oxidizing agent!

At the same time, the chlorine atom increases to an oxidation number of zero (in its element form as #Cl_2#). The increase in oxidation number means the #Cl^-# ion has been oxidized, and so, must be the reducing agent.