1) Balance this redox reaction using the oxidation-number method. Zn + O2 + HCl ® ZnCl2 + H2O
2) Balance this redox reaction using the half-reaction method. (basic solution) Zn + MnO4- ® Zn(OH)2 + MnO2
Instead of us doing the work for you and you just copying it down, I suggest you try to tell us exactly what you don't understand about the problem and we can go from there. I also suggest you post at the top of the page since this post is WAY down the list.
1) To balance the redox reaction using the oxidation-number method, we need to assign oxidation numbers to each element in the equation.
Starting with the reactants:
Zn: This is a neutral element, so its oxidation number is 0.
O2: Oxygen typically has an oxidation number of -2. Since there are two oxygen atoms, the total oxidation number is -4.
HCl: Hydrogen and chlorine have oxidation numbers of +1 and -1, respectively.
Moving on to the products:
ZnCl2: Chlorine has an oxidation number of -1, so two chlorine atoms will have a total oxidation number of -2. This means that zinc must have an oxidation number of +2 to balance the charge.
H2O: Oxygen typically has an oxidation number of -2, and there are two of them in water. So the total oxidation number is -4. This means that hydrogen must have an oxidation number of +1 to balance the charge.
Now we can see the changes in oxidation numbers:
Zn: Changes from 0 to +2, so it is being oxidized (losing electrons).
O2: Does not change its oxidation number.
HCl: Changes from 0 to -1, so it is being reduced (gaining electrons).
ZnCl2: Does not change its oxidation number.
H2O: Does not change its oxidation number.
Since zinc is being oxidized and hydrogen chloride is being reduced, we need to balance this by adding electrons to the appropriate side of the equation.
Now we can write the half-reactions:
Oxidation half-reaction: Zn → Zn2+ + 2e-
Reduction half-reaction: 2H+ + 2e- → H2
To balance the number of electrons transferred, we need to multiply the oxidation half-reaction by 2:
2Zn → 2Zn2+ + 4e-
Now we can add the two half-reactions together:
2Zn + O2 + 4H+ → 2Zn2+ + 2H2O
Lastly, we need to balance the number of H+ ions by adding HCl to the reactant side:
2Zn + O2 + 4HCl → 2ZnCl2 + 2H2O
Therefore, the balanced redox reaction using the oxidation-number method is:
2Zn + O2 + 4HCl → 2ZnCl2 + 2H2O
2) To balance the redox reaction in basic solution using the half-reaction method, the first step is to split the reaction into two half-reactions: one for the oxidation and one for the reduction.
Oxidation half-reaction: Zn → Zn2+
Reduction half-reaction: MnO4- → MnO2
Now we need to balance each half-reaction individually.
Oxidation half-reaction:
Zn → Zn2+
The number of Zn atoms on both sides is balanced, but the charge is not. To balance the charge, we need to add 2 electrons to the left side:
Zn → Zn2+ + 2e-
Reduction half-reaction:
MnO4- → MnO2
The number of Mn atoms on both sides is balanced, but the oxygens are not. To balance the oxygens, we need to add 2 hydroxide ions (OH-) to the right side:
MnO4- + 2OH- → MnO2 + H2O
Next, we need to balance the charge by adding electrons. In this case, we need to add 4 electrons to the left side:
MnO4- + 2OH- + 4e- → MnO2 + H2O
Now we can combine the two half-reactions and eliminate the electrons by multiplying the equations accordingly:
2Zn + 4OH- → 2ZnO2- + 2H2O
MnO4- + 2OH- + 4e- → MnO2 + H2O
By multiplying the first equation by 2 and the second equation by 4, we get:
4Zn + 8OH- → 4ZnO2- + 4H2O
4MnO4- + 8OH- + 16e- → 4MnO2 + 4H2O
Now we can combine the two equations:
4Zn + 4MnO4- + 8OH- + 16e- → 4ZnO2- + 4MnO2 + 4H2O
Finally, we can simplify the equation by canceling out common terms:
2Zn + 2MnO4- + 4OH- + 8e- → 2ZnO2- + 2MnO2 + 2H2O
Therefore, the balanced redox reaction in basic solution using the half-reaction method is:
2Zn + 2MnO4- + 4OH- → 2ZnO2- + 2MnO2 + 2H2O