The following redox reaction occurs in basic aqueous solution. When balanced with the smallest whole number coefficients, what is the coefficient for OH-(aq)? Mn(OH)2(s) + MnO4-(aq) = MnO42-(aq).
1. ) 2
2. ) 3
3. ) 4
4. ) 6
5. ) 8
the answer i got was 6...but im no 100% sure :S
yes i think this is correct.
I think its 4
To balance a redox reaction, you need to follow these steps:
1. Write down the unbalanced equation:
Mn(OH)2(s) + MnO4-(aq) → MnO42-(aq)
2. Assign oxidation numbers to each element:
In Mn(OH)2, Mn has an oxidation number of +2.
In MnO4-, Mn has an oxidation number of +7.
3. Identify the atoms undergoing oxidation and reduction:
In this case, Mn is being reduced from +7 to +4, so it is the reducing agent.
Mn(OH)2 is being oxidized from +2 to +4, so it is the oxidizing agent.
4. Split the equation into two half-reactions:
Oxidation half-reaction: Mn(OH)2 → MnO42-
Reduction half-reaction: MnO4- → MnO4-
5. Balance the atoms (excluding oxygen and hydrogen) in each half-reaction:
Oxidation half-reaction: Mn(OH)2 → MnO4- + 2H2O
Reduction half-reaction: MnO4- → MnO42-
6. Balance the oxygen atoms by adding water molecules:
Oxidation half-reaction: Mn(OH)2 → MnO4- + 2H2O
Reduction half-reaction: MnO4- + 2H2O → MnO42-
7. Balance the hydrogen atoms by adding H+ ions:
Oxidation half-reaction: Mn(OH)2 → MnO4- + 2H2O
Reduction half-reaction: MnO4- + 2H2O → MnO42- + 8H+
8. Balance the charges by adding electrons (e-):
Oxidation half-reaction: Mn(OH)2 → MnO4- + 2H2O + 6e-
Reduction half-reaction: MnO4- + 2H2O + 6e- → MnO42+ + 8H+
9. Multiply the half-reactions by appropriate coefficients so that the number of electrons gained in the reduction half-reaction equals the number of electrons lost in the oxidation half-reaction:
Oxidation half-reaction: 3Mn(OH)2 → 3MnO4- + 6H2O + 18e-
Reduction half-reaction: 2MnO4- + 4H2O + 12e- → 2MnO42+ + 16H+
10. Multiply the half-reactions by appropriate coefficients so that the number of electrons cancels out when the half-reactions are combined:
Oxidation half-reaction: 3Mn(OH)2 → 3MnO4- + 6H2O + 18e-
Reduction half-reaction: 4MnO4- + 8H2O + 24e- → 4MnO42+ + 32H+
11. Combine the half-reactions and cancel out any species that appear on both sides of the equation:
3Mn(OH)2 + 4MnO4- + 8H2O → 3MnO4- + 4MnO42+ + 32H2O
12. Simplify the equation and remove any unnecessary water molecules:
3Mn(OH)2 → 4MnO42+ + 24H2O
From the balanced equation, you can see that the coefficient for OH-(aq) is 24.
Therefore, the correct answer is not listed among your options.