Dr.Bobb222 please help balance the following oxidation-reduction reactions, which occur in acidic solution, using the half-reaction method. (Use the lowest possible coefficients. Include states-of-matter under the given conditions in your answer.)
(a) NO3−(aq) + Pb(s) → NO(g) + Pb2+(aq)(b) O2(g) + Mn(s) → H2O(l) + Mn2+(aq)
(c) ClO2(g) + Mn2+(aq) → ClO2−(aq) + MnO4−(aq)
(d) F2(g) + IO3−(aq) → F −(aq) + IO4−(aq)
Any solution process that you explain to any one of the redox equations I'm sure will help me to solve the rest. The balancing of the oxidation and reduction equations individually is really grilling me. I appreciate your help, thank you.
Well, I'm glad you appreciate my help! Let's dive into balancing these oxidation-reduction reactions.
(a) NO3−(aq) + Pb(s) → NO(g) + Pb2+(aq)
First, let's identify the oxidation and reduction half-reactions:
Oxidation Half-Reaction: NO3−(aq) → NO(g)
Reduction Half-Reaction: Pb(s) → Pb2+(aq)
Now let's balance them individually:
Oxidation Half-Reaction:
Since nitrogen is being reduced from +5 to 0, we need to add 5 electrons to the left side to balance the charge:
5NO3−(aq) → 5NO(g) + 5e−
Reduction Half-Reaction:
Pb(s) → Pb2+(aq)
Since lead is being oxidized from 0 to +2, we lose 2 electrons:
Pb(s) → Pb2+(aq) + 2e−
Now let's balance the number of electrons in both half-reactions by multiplying them accordingly:
5NO3−(aq) + Pb(s) → 5NO(g) + Pb2+(aq) + 2e−
Finally, multiply each half-reaction so that the electrons cancel out:
5NO3−(aq) + 8Pb(s) → 5NO(g) + 4Pb2+(aq)
(b) O2(g) + Mn(s) → H2O(l) + Mn2+(aq)
Oxidation Half-Reaction: Mn(s) → Mn2+(aq)
Reduction Half-Reaction: O2(g) → H2O(l)
Oxidation Half-Reaction:
Mn(s) → Mn2+(aq)
Since manganese is being oxidized from 0 to +2, we lose 2 electrons:
Mn(s) → Mn2+(aq) + 2e−
Reduction Half-Reaction:
O2(g) → 2H2O(l)
To balance the oxygen, add 2 water molecules on the right side:
O2(g) + 4H+(aq) + 4e− → 2H2O(l)
To balance the electrons, multiply the oxidation half-reaction by 4 and the reduction half-reaction by 2:
4Mn(s) → 4Mn2+(aq) + 8e−
2O2(g) + 8H+(aq) + 8e− → 4H2O(l)
Finally, multiplying each half-reaction so that the electrons cancel out, we get:
4Mn(s) + 2O2(g) + 8H+(aq) + 8e− → 4Mn2+(aq) + 8H2O(l)
And that's how you balance these reactions using the half-reaction method! Let me know if you need help with the rest.
Sure, I can help you balance these oxidation-reduction reactions step-by-step. Let's begin.
(a) To balance the equation NO3−(aq) + Pb(s) → NO(g) + Pb2+(aq) in acidic solution using the half-reaction method:
1. Assign oxidation numbers to each element:
NO3−: The oxidation number of O is -2. Since there are three O atoms, the total oxidation number contributed by O is -6. Since the overall charge of NO3− is -1, the oxidation number of N must be +5.
Pb: The oxidation number of Pb is 0.
NO: The oxidation number of O is -2. Since there is one O atom, the total oxidation number contributed by O is -2. Since the overall charge of NO is 0, the oxidation number of N must be +2.
Pb2+: The oxidation number of Pb is +2.
2. Determine the half-reactions:
Oxidation half-reaction: Pb(s) → Pb2+(aq)
Reduction half-reaction: NO3−(aq) → NO(g)
3. Balance the elements other than H and O in each half-reaction:
Oxidation half-reaction: Pb(s) → Pb2+(aq)
Reduction half-reaction: 2NO3−(aq) → 2NO(g)
4. Balance the O atoms by adding H2O molecules:
Oxidation half-reaction: Pb(s) → Pb2+(aq)
Reduction half-reaction: 2NO3−(aq) + 4H+(aq) → 2NO(g) + 2H2O(l)
5. Balance the H atoms by adding H+ ions:
Oxidation half-reaction: Pb(s) + 4H+(aq) → Pb2+(aq)
Reduction half-reaction: 2NO3−(aq) + 4H+(aq) → 2NO(g) + 2H2O(l)
6. Balance the charges by adding electrons (e^-):
Oxidation half-reaction: Pb(s) + 4H+(aq) → Pb2+(aq) + 2e^-
Reduction half-reaction: 2NO3−(aq) + 4H+(aq) + 6e^- → 2NO(g) + 2H2O(l)
7. Multiply the half-reactions by appropriate factors to ensure the numbers of electrons lost in the oxidation half-reaction equal the number of electrons gained in the reduction half-reaction (in this case, multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2):
3(Pb(s) + 4H+(aq) → Pb2+(aq) + 2e^-)
2(2NO3−(aq) + 4H+(aq) + 6e^- → 2NO(g) + 2H2O(l))
8. Add the balanced half-reactions together and cancel out common terms:
3Pb(s) + 12H+(aq) + 6NO3−(aq) → 3Pb2+(aq) + 6e^- + 4NO(g) + 4H2O(l)
To summarize, the balanced equation for reaction (a) is:
3Pb(s) + 12H+(aq) + 6NO3−(aq) → 3Pb2+(aq) + 6e^- + 4NO(g) + 4H2O(l)
To balance these oxidation-reduction reactions, also known as redox reactions, using the half-reaction method, follow these steps:
Step 1: Separate the reaction into half-reactions.
Step 2: Balance the atoms other than hydrogen and oxygen in each half-reaction.
Step 3: Balance the charges in each half-reaction by adding electrons (e-).
Step 4: Equalize the number of electrons in both half-reactions by multiplying one or both of the half-reactions.
Step 5: Combine the half-reactions in a way that cancels out the electrons.
Step 6: Verify the balanced equation by checking that mass and charge are conserved.
Let's go through the process for the first reaction:
(a) NO3−(aq) + Pb(s) → NO(g) + Pb2+(aq)
Step 1: Separate the reaction into half-reactions:
Oxidation half-reaction: Pb(s) → Pb2+(aq)
Reduction half-reaction: NO3−(aq) → NO(g)
Step 2: Balance the atoms other than hydrogen and oxygen in each half-reaction:
Oxidation half-reaction: Pb(s) → Pb2+(aq) (already balanced)
Reduction half-reaction: 3NO3−(aq) → NO(g) (add coefficient 3 in front of NO3−)
Step 3: Balance the charges in each half-reaction by adding electrons (e-):
Oxidation half-reaction: Pb(s) → Pb2+(aq) + 2e-
Reduction half-reaction: 3NO3−(aq) → NO(g) + 6e-
Step 4: Equalize the number of electrons in both half-reactions:
Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to equalize the electrons:
3Pb(s) → 3Pb2+(aq) + 6e-
6NO3−(aq) → 2NO(g) + 12e-
Step 5: Combine the half-reactions in a way that cancels out the electrons:
Add the two half-reactions together:
3Pb(s) + 6NO3−(aq) → 3Pb2+(aq) + 2NO(g) + 12e-
3Pb(s) + 6NO3−(aq) → 3Pb2+(aq) + 2NO(g) + 12e-
Cancel out the electrons:
3Pb(s) + 6NO3−(aq) → 3Pb2+(aq) + 2NO(g)
Step 6: Verify the balanced equation:
Count the atoms on each side of the equation to ensure they are balanced:
Left side: 3 Pb, 6 N, 18 O
Right side: 3 Pb, 6 N, 18 O
The equation is now balanced for reaction (a):
3Pb(s) + 6NO3−(aq) → 3Pb2+(aq) + 2NO(g)
Follow these steps to balance the remaining reactions (b), (c), and (d) in a similar manner.
3Pb + 2NO3- + 8H+ → 3Pb2+ + 2NO + 4H2O
2Mn + O2 + 4H+ → 2Mn2+ + 2H2O
Mn2+ + 5ClO2 + 4H2O → MnO4- + 5ClO2- + 8H+
IO3- + F2 + H2O → IO4- + 2F- + 2H+