Am I headed in the right direction.
Sn + MnO4- --> Sn2+ + Mn2+
Sn --> Sn2+ + 2e-
5e + 8H + MnO4- --> Mn2+ + 4H2O
If Im right what is the whole number that Im multiplying by
Both half reactions are right. Electrons gained by one half reaction must get them from somewhere which is just another way of saying electrons lost must equal electrons gained. The only number for both 2 and 5 is 10, so multiply one half cell by 5 and the other by 2, then add them, then cancel anything common to both sides.
To determine the whole number you are multiplying by, you need to balance the redox equation. Let's start by balancing the reduction half-reaction:
Sn → Sn2+ + 2e-
The number of electrons (e-) is already balanced. Next, balance the oxidation half-reaction:
5e- + 8H+ + MnO4- → Mn2+ + 4H2O
To balance the number of oxygen (O) atoms, you need to add 4 water (H2O) molecules to the reactant side:
5e- + 8H+ + MnO4- → Mn2+ + 4H2O
Now, the number of oxygen atoms is balanced. To balance the number of hydrogen (H) atoms, you need to add 8 hydrogen ions (H+) to the product side:
5e- + 8H+ + MnO4- → Mn2+ + 4H2O
Finally, balance the charges by adding 5 electrons (e-) to the product side:
Sn → Sn2+ + 2e-
5e- + 8H+ + MnO4- → Mn2+ + 4H2O + 5e-
Now, you can multiply both half-reactions by appropriate numbers to make the number of electrons equal on both sides. In this case, you need to multiply the first half-reaction by 5 and the second half-reaction by 2:
5(Sn → Sn2+ + 2e-)
2(5e- + 8H+ + MnO4- → Mn2+ + 4H2O + 5e-)
After multiplying, you will get the following balanced equation:
5Sn + 2MnO4- + 16H+ → 5Sn2+ + 2Mn2+ + 8H2O
Now, you can see that the whole number you are multiplying by is 5 for Sn.
To balance the given redox reaction:
Sn + MnO4- → Sn2+ + Mn2+
1. Balance the equation for the oxidation half-reaction:
Sn → Sn2+ + 2e-
2. Balance the equation for the reduction half-reaction:
5e + 8H+ + MnO4- → Mn2+ + 4H2O
Now, multiply these half-reactions by the necessary whole numbers to make the number of electrons equal on both sides:
Sn → Sn2+ + 2e- (multiply by 5)
5Sn → 5Sn2+ + 10e-
5e + 8H+ + MnO4- → Mn2+ + 4H2O (multiply by 2)
10e + 16H+ + 2MnO4- → 2Mn2+ + 8H2O
Now, the number you multiply the first half-reaction by is 5, and the number you multiply the second half-reaction by is 2.