Complete and balance the following redox equation. The sum of the coefficients when they are all whole numbers is
MnO4- + H+ + Br-® Mn2+ + Br2 + H2O (acidic solution)
6
17
21
29
43
It's actually 43
To balance the given redox equation in an acidic solution, we must follow these steps:
1. Assign oxidation numbers to each element:
MnO4- : Mn has an oxidation number of +7
H+ : H has an oxidation number of +1
Br- : Br has an oxidation number of -1
Mn2+ : Mn has an oxidation number of +2
Br2 : Br has an oxidation number of 0
H2O : H has an oxidation number of +1, and O has an oxidation number of -2
2. Identify the elements undergoing oxidation and reduction:
Mn is reduced from +7 to +2, so it is undergoing reduction (oxidation number decreases).
Br is oxidized from -1 to 0, so it is undergoing oxidation (oxidation number increases).
3. Write separate half-reactions for oxidation and reduction:
Reduction half-reaction: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
Oxidation half-reaction: 2Br- → Br2 + 2e-
4. Balance the atoms other than hydrogen and oxygen:
Reduction: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O (already balanced)
Oxidation: 2Br- → Br2 + 2e- (already balanced)
5. Balance the oxygen atoms by adding water molecules (H2O):
Reduction: (Already balanced)
Oxidation: 2Br- + 2H2O → Br2 + 4OH-
6. Balance the hydrogen atoms by adding hydrogen ions (H+):
Reduction: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O (already balanced)
Oxidation: 2Br- + 2H2O → Br2 + 4OH- (already balanced)
7. Balance the charges by adding electrons:
Reduction: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O (already balanced)
Oxidation: 2Br- + 2H2O → Br2 + 4OH- + 4e-
8. Multiply the half-reactions by appropriate coefficients to make the number of electrons equal in both half-reactions:
Reduction: 2(MnO4-) + 16H+ + 10e- → 2(Mn2+) + 8H2O
Oxidation: 10(Br-) + 10H2O → 5(Br2) + 20OH- + 20e-
9. Add the half-reactions together and cancel out any common terms:
2(MnO4-) + 16H+ + 10(Br-) + 10H2O → 2(Mn2+) + 8H2O + 5(Br2) + 20OH- + 20e-
2(MnO4-) + 16H+ + 10(Br-) + 9H2O → 2(Mn2+) + 5(Br2) + 20OH- + 20e-
10. Simplify the equation by removing unnecessary terms:
Cancel out the water molecules and hydrogen ions:
2(MnO4-) + 6H+ + 10(Br-) → 2(Mn2+) + 5(Br2) + 20OH- + 20e-
11. The sum of the coefficients when they are all whole numbers is calculated by multiplying each coefficient by the smallest whole number to eliminate any fractions or decimals:
2(MnO4-) + 6H+ + 10(Br-) → 2(Mn2+) + 5(Br2) + 20OH- + 20e-
Multiply by 2:
4(MnO4-) + 12H+ + 20(Br-) → 4(Mn2+) + 10(Br2) + 40OH- + 40e-
12. The sum of coefficients is 4 + 12 + 20 + 4 + 10 + 40 + 40 = 130.
Therefore, the sum of the coefficients when they are all whole numbers is 130.
To balance a redox equation, you need to make sure that both the number of atoms and the charges are balanced on both sides of the equation.
Let's start by identifying the atoms and charges involved:
On the left-hand side (reactants):
- MnO4- (Manganese oxide ion)
- H+ (Hydrogen ion)
- Br- (Bromide ion)
On the right-hand side (products):
- Mn2+ (Manganese ion)
- Br2 (Bromine)
- H2O (Water)
Now, let's balance the atoms one by one. Since there are multiple oxygen atoms on the left-hand side (4 in MnO4-) and only one on the right-hand side (in H2O), we need to add water molecules on the right-hand side to balance the oxygen atoms:
MnO4- + H+ + Br- -> Mn2+ + Br2 + H2O
MnO4- + H+ + Br- -> Mn2+ + Br2 + 4H2O
Now, let's focus on balancing the hydrogen atoms. There are 1 hydrogen ion on the left-hand side (H+) and 8 hydrogen atoms on the right-hand side (4 in H2O). To balance the hydrogen, we can add 8 hydrogen ions on the left-hand side:
MnO4- + 8H+ + Br- -> Mn2+ + Br2 + 4H2O
Now, let's balance the charges. On the left-hand side, the total charge is -1 for the MnO4- ion, -8 for the 8H+ ions, and -1 for the Br- ion, which sums up to -10. On the right-hand side, the total charge is +2 for the Mn2+ ion, 0 for the Br2 molecule, and 0 for the H2O molecules, which sums up to +2. We can balance the charges by adding 12 electrons (e-) on the left-hand side:
MnO4- + 8H+ + Br- + 12e- -> Mn2+ + Br2 + 4H2O
Now, the number of atoms and charges are balanced on both sides of the equation.
To find the sum of the coefficients, we add up the coefficients in front of each compound:
1 + 8 + 1 + 12 + 1 + 4 + 8 = 35
Therefore, the sum of the coefficients in the balanced equation, when they are all whole numbers, is 35.
Mn in MnO4^- goes from +7 on the left to +2 on the right.
Br goes from -2 (for 2 Br^-) on the left to 0 on the right (for two Br atoms).
Now you balance.