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
12 years ago
1 year ago
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.
1 year ago
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.