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).
- after doing the question i got 1..but i don't know if im right :S
1. 3
2. 5
3. 1
4. 4
5. 2
No you aren't right.
http://www.chemteam.info/Redox/WS-BalancingRedoxBase.html
I got 4 as my answer, is that correct?
To balance the redox reaction in basic aqueous solution, you need to follow these steps:
Step 1: Assign oxidation numbers to each element in the equation.
Mn(OH)2(s) + MnO4-(aq) = MnO42-(aq)
The oxidation state of Mn in MnO4- is +7, and in MnO42- it is +6. The oxidation state of O in MnO4- and MnO42- is -2. The oxidation state of H in OH- is +1. Therefore, the oxidation state of Mn in Mn(OH)2 is +2 and the oxidation state of H in OH- is -1.
Step 2: Separate the equation into two half-reactions. One for the reduction and one for the oxidation.
Reduction: MnO4-(aq) + 4H2O(l) + 3e- = MnO42-(aq) + 8OH-(aq)
Oxidation: Mn(OH)2(s) = MnO42-(aq) + 4H2O(l) + 2e-
Step 3: Balance the elements except hydrogen (H) and oxygen (O) in each half-reaction.
Reduction: MnO4-(aq) + 4H2O(l) + 3e- = MnO42-(aq) + 8OH-(aq)
Oxidation: 2Mn(OH)2(s) = MnO42-(aq) + 4H2O(l) + 2e-
Step 4: Balance the oxygen atoms by adding water (H2O) molecules to the opposite side of the respective half-reaction.
Reduction: MnO4-(aq) + 4H2O(l) + 3e- = MnO42-(aq) + 8OH-(aq)
Oxidation: 2Mn(OH)2(s) = MnO42-(aq) + 8H2O(l) + 2e-
Step 5: Balance the hydrogen atoms by adding hydrogen ions (H+) to the opposite side of the respective half-reaction.
Reduction: MnO4-(aq) + 4H2O(l) + 3e- = MnO42-(aq) + 8OH-(aq)
Oxidation: 2Mn(OH)2(s) + 8H+ = MnO42-(aq) + 8H2O(l) + 2e-
Step 6: Balance the charges by adding electrons (e-) to the side that needs them.
Reduction: MnO4-(aq) + 4H2O(l) + 3e- = MnO42-(aq) + 8OH-(aq)
Oxidation: 2Mn(OH)2(s) + 8H+ + 10e- = 5MnO42-(aq) + 8H2O(l) + 2e-
Step 7: Multiply each half-reaction by the smallest integer to equalize the number of electrons transferred.
Reduction: 2[MnO4-(aq) + 4H2O(l) + 3e-] = 2[MnO42-(aq) + 8OH-(aq)]
Oxidation: 5[2Mn(OH)2(s) + 8H+ + 10e-] = 5[5MnO42-(aq) + 8H2O(l) + 2e-]
Final balanced equation:
10MnO4-(aq) + 40H2O(l) + 30e- = 10MnO42-(aq) + 80OH-(aq)
50Mn(OH)2(s) + 200H+ + 100e- = 25MnO42-(aq) + 40H2O(l) + 10e-
Comparing the two half-reactions, we can see that the coefficient for OH-(aq) is 80 in the reduction half-reaction.
Therefore, the correct answer is 80.
To balance a redox reaction in basic aqueous solution, you need to follow these steps:
1. Assign oxidation numbers to all elements in the reaction.
2. Identify the element that is being reduced and the element that is being oxidized.
3. Balance the atoms and charges in the half-reactions separately.
4. Balance the oxygen atoms by adding water (H2O) molecules to the side that needs more oxygen atoms.
5. Balance the hydrogen atoms by adding hydrogen ions (H+) to the side that needs more hydrogen atoms.
6. Balance the charges by adding electrons (e-) to the side that needs more negative charge.
7. Make the number of electrons in the oxidation half-reaction equal to the number of electrons in the reduction half-reaction by multiplying one or both of the half-reactions by a whole number.
8. Combine the half-reactions so that the electrons cancel out.
9. If the reaction is in basic solution, add enough hydroxide ions (OH-) to the side with excess hydrogen ions (H+) to neutralize them and form water molecules (H2O).
10. Check that all the atoms and charges are balanced.
Let's use this procedure to balance the given redox reaction:
Mn(OH)2(s) + MnO4-(aq) = MnO42-(aq)
First, assign oxidation numbers:
Mn(OH)2(s) → +2 -2 +1(-2)2 = 0
MnO4-(aq) → +7 -2(4) = -1
MnO42-(aq) → +7 -2(4)2 = -2
From this, we can see that manganese is being reduced from an oxidation state of +7 to +2. So, the reduction half-reaction is:
MnO4-(aq) + 5e- → MnO42-(aq)
Now, let's balance the half-reactions:
Reduction half-reaction:
MnO4-(aq) + 5e- → MnO42-(aq)
Now, let's balance the oxygen atoms by adding water (H2O) molecules:
MnO4-(aq) + 5e- → MnO42-(aq) + 4H2O
Next, balance the hydrogen atoms by adding hydrogen ions (H+):
MnO4-(aq) + 5e- + 8H+ → MnO42-(aq) + 4H2O
Now, balance the charges by adding electrons (e-):
MnO4-(aq) + 5e- + 8H+ → MnO42-(aq) + 4H2O + 5e-
The oxidation half-reaction is:
Mn(OH)2(s) → Mn + 2OH-
Now, let's balance the half-reaction by adding water (H2O) molecules:
Mn(OH)2(s) → Mn + 2OH-
Now, balance the charges by adding electrons (e-):
Mn(OH)2(s) + 2e- → Mn + 2OH-
To make the number of electrons in the oxidation half-reaction equal to the reduction half-reaction, multiply the oxidation half-reaction by 5:
5(Mn(OH)2(s) + 2e-) → 5(Mn + 2OH-)
Now, let's combine the half-reactions:
5(Mn(OH)2(s) + 2e-) + MnO4-(aq) + 5e- + 8H+ → 5(Mn + 2OH-) + MnO42-(aq) + 4H2O + 5e-
To cancel out the electrons, we have:
5Mn(OH)2(s) + MnO4-(aq) + 8H+ → 5Mn + 10OH- + MnO42-(aq) + 4H2O
Now, since the reaction is in basic solution, we need to add hydroxide ions (OH-) to neutralize the hydrogen ions (H+). In this case, we need to add 10 OH- ions:
5Mn(OH)2(s) + MnO4-(aq) + 8H+ + 10OH- → 5Mn + 10OH- + MnO42-(aq) + 4H2O
Finally, cancel out the excess OH- ions:
5Mn(OH)2(s) + MnO4-(aq) + 8H+ + 2OH- → 5Mn + 6OH- + MnO42-(aq) + 4H2O
From this, we can see that the coefficient for OH-(aq) is 6.
Therefore, the correct answer is 6.