Figure out the oxidation numbers of all the atoms in the below equation. The figure out what is being oxidized and what is being reduced and write out the two half-reactions and explain what happens in each. Mn(s) + Pb(NO3)2(aq) ---> Mn(N3O)2(aq) + Pb(s)
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To determine the oxidation numbers of the atoms in the equation, we first need to identify the elements and compounds involved.
The equation is: Mn(s) + Pb(NO3)2(aq) ---> Mn(N3O)2(aq) + Pb(s)
Let's start by assigning oxidation numbers to each atom:
1. Mn(s): Since Mn is a metal, its oxidation number is typically 0 in its elemental form.
2. Pb(NO3)2(aq): The NO3 ion is known to have a -1 charge. Considering there are two NO3 ions in this compound, each oxygen atom has an oxidation number of -2. Therefore, the sum of the oxidation numbers in Pb(NO3)2 is -2*3 + 1*2 = -6 + 2 = -4. Since the compound has a neutral charge, the oxidation number of Pb must be +4 to balance it out.
3. Mn(N3O)2(aq): The N3O ion is not common and doesn't have a specific oxidation number. However, we know that the overall charge of the compound is 0, so the sum of the oxidation numbers must add up to 0. Therefore, the oxidation number of Mn in Mn(N3O)2 must be +2.
4. Pb(s): In its elemental form, Pb has an oxidation number of 0.
Now that we have determined the oxidation numbers of the atoms, let's analyze the oxidation and reduction processes.
Oxidation is the process of losing electrons, while reduction is the process of gaining electrons. Looking at the equation:
Mn(s) + Pb(NO3)2(aq) ---> Mn(N3O)2(aq) + Pb(s)
We can observe that Mn is going from an oxidation number of 0 to +2, meaning it is gaining electrons and undergoing reduction (it is being reduced from Mn(0) to Mn(+2)).
On the other hand, Pb is going from an oxidation number of +4 to 0, meaning it is losing electrons and undergoing oxidation (it is being oxidized from Pb(+4) to Pb(0)).
Now, let's write the two half-reactions based on the changes in oxidation numbers:
Reduction half-reaction: Mn(0) ---> Mn(+2) + 2e^-
Oxidation half-reaction: Pb(+4) + 4e^- ---> Pb(0)
The reduction half-reaction shows that Mn is gaining two electrons (hence the 2e^-) to go from an oxidation number of 0 to +2. The oxidation half-reaction shows that Pb is losing four electrons (hence the 4e^-) to go from an oxidation number of +4 to 0.
In summary:
- The Mn atom is being reduced from an oxidation number of 0 to +2.
- The Pb atom is being oxidized from an oxidation number of +4 to 0.
Hope this helps!