I know nothing about chemistry, but am wondering if someone could help me with these.


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

I believe #1 is 2Zn+02+4---->2Zn Cl2+2H20,can someone confirm?

1st one is right, the second one cant be right because there is no H on the left side.

Certainly! I can help you with both of these redox reactions using different balancing methods.

1) To balance the redox reaction Zn + O2 + HCl -> ZnCl2 + H2O using the oxidation-number method, follow these steps:

Step 1: Assign oxidation numbers to each element on both sides of the equation.
Zn: 0 (unchanged)
O2: 0 (unchanged)
HCl: Zn goes from 0 to +2 (oxidation), so HCl contains Cl- with a -1 charge, so Cl goes from -1 to 0 (reduction)
ZnCl2: Zn goes from +2 to 0 (reduction), Cl goes from -1 to -2 (oxidation)
H2O: H goes from +1 to 0 (reduction), O goes from -2 to -2 (unchanged)

Step 2: Identify the elements that are getting oxidized and reduced.
Oxidized: Cl (from -1 to 0)
Reduced: Zn (from 0 to +2)

Step 3: Write half-reactions for each process.
Oxidation (loss of electrons): Cl- -> Cl
Reduction (gain of electrons): Zn -> Zn^2+

Step 4: Balance each half-reaction separately.
Oxidation: 2Cl- -> Cl2
Reduction: Zn -> Zn^2+

Step 5: Multiply each half-reaction by an appropriate number so that the electrons gained and lost are equal.
Oxidation: 2Cl- -> Cl2 + 2e^-
Reduction: Zn -> Zn^2+ + 2e^-

Step 6: Combine the balanced half-reactions while canceling out the electrons.
2Cl- + Zn -> Cl2 + Zn^2+

Step 7: Balance the number of atoms on both sides.
2Cl- + Zn + O2 + HCl -> Cl2 + ZnCl2 + H2O

Now the redox reaction is balanced using the oxidation-number method.

2) To balance the redox reaction Zn + MnO4- -> Zn(OH)2 + MnO2 in basic solution using the half-reaction method, follow these steps:

Step 1: Assign oxidation numbers to each element on both sides of the equation.
Zn: 0 (unchanged)
MnO4-: Mn goes from +7 to +4 (reduction), O goes from -2 to -2 (unchanged)
Zn(OH)2: Zn goes from 0 to +2 (oxidation), O goes from -2 to -2 (unchanged), H goes from +1 to +1 (unchanged)
MnO2: Mn goes from +7 to +4 (reduction), O goes from -2 to -2 (unchanged)

Step 2: Identify the elements that are getting oxidized and reduced.
Oxidized: Zn (from 0 to +2)
Reduced: Mn (from +7 to +4)

Step 3: Write half-reactions for each process, separating the reaction into reduction and oxidation half-reactions.
Oxidation (loss of electrons): Zn -> Zn^2+
Reduction (gain of electrons): MnO4- -> MnO2

Step 4: Balance each half-reaction separately.
Oxidation: Zn -> Zn^2+
Reduction: MnO4- -> MnO2

Step 5: Balance the atoms involved in the half-reaction other than hydrogen and oxygen.
Oxidation: Zn -> Zn^2+
Reduction: 4H2O + MnO4- -> MnO2 + 4OH-

Step 6: Balance the oxygen atoms by adding water molecules to the side that lacks oxygen.
Oxidation: Zn -> Zn^2+
Reduction: 4H2O + MnO4- -> MnO2 + 4OH-

Step 7: Balance the hydrogen atoms by adding hydrogen ions (H+) to the side that lacks hydrogen.
Oxidation: Zn -> Zn^2+
Reduction: 4H2O + MnO4- + 8H+ -> MnO2 + 4OH- + 4H2O

Step 8: Balance the charge by adding electrons (e-) to the side that lacks negative charge.
Oxidation: Zn -> Zn^2+ + 2e-
Reduction: 4H2O + MnO4- + 8H+ + 5e- -> MnO2 + 4OH- + 4H2O

Step 9: Multiply each half-reaction by an appropriate number so that the total number of electrons gained and lost are equal.
Oxidation: Zn -> Zn^2+ + 2e-
Reduction: 4H2O + MnO4- + 8H+ + 5e- -> MnO2 + 4OH- + 4H2O

Step 10: Combine the balanced half-reactions while canceling out the electrons.
Zn + 4H2O + MnO4- + 8H+ + 5e- -> Zn^2+ + 2e- + MnO2 + 4OH- + 4H2O

Step 11: Simplify the equation by removing species that appear on both sides.
Zn + MnO4- + 8H+ -> Zn^2+ + MnO2 + 4H2O

Now the redox reaction is balanced using the half-reaction method in basic solution.