Balance redox reaction through oxidation method of an acid

Cr2O7- + Cl- --> Cr3+ + Cl2

Step 1: Write the unbalanced half-reactions for oxidation and reduction

Oxidation: Cr2O7- --> Cr3+
Reduction: Cl- --> Cl2

Step 2: Balance the number of atoms for the half-reactions except for H and O

Oxidation: Cr2O7- --> 2Cr3+
Reduction: 2Cl- --> Cl2

Step 3: Balance the number of oxygen atoms by adding H2O to the half-reaction that needs it

Oxidation: Cr2O7- + 14H+ --> 2Cr3+ + 7H2O
Reduction: 2Cl- --> Cl2

Step 4: Balance the number of hydrogen atoms by adding H+ to the other half-reaction

Oxidation: Cr2O7- + 14H+ --> 2Cr3+ + 7H2O
Reduction: 2Cl- + 2H+ --> Cl2 + H2

Step 5: Balance the charges by adding electrons to the half-reactions

Oxidation: Cr2O7- + 14H+ + 6e- --> 2Cr3+ + 7H2O
Reduction: 2Cl- + 2H+ + 2e- --> Cl2 + H2

Step 6: Multiply the half-reactions by integers to make the number of electrons equal in both half-reactions

Oxidation: 6Cr2O7- + 42H+ + 18e- --> 12Cr3+ + 21H2O
Reduction: 21Cl- + 21H+ + 21e- --> 21Cl2 + 21H2

Step 7: Add the half-reactions together and cancel out the common species

6Cr2O7- + 42H+ + 18Cl- --> 12Cr3+ + 21H2O + 21Cl2

Step 8: Verify that the charges and atoms are balanced on both sides of the equation

The charges and atoms are balanced on both sides of the equation.

To balance the redox reaction using the oxidation method, follow these steps:

Step 1: Identify the change in oxidation states for each element in the reaction.

In this reaction, we can see that the oxidation state of Cr changes from +6 to +3, indicating a reduction, while the oxidation state of Cl changes from -1 to 0, indicating an oxidation.

Step 2: Write the half-reactions for oxidation and reduction.

Oxidation half-reaction: Cl- → Cl2
Reduction half-reaction: Cr2O7- → Cr3+

Step 3: Balance the atoms other than oxygen and hydrogen in each half-reaction.

Oxidation half-reaction: 2Cl- → Cl2
Reduction half-reaction: 2Cr2O7- → 2Cr3+

Step 4: Balance the oxygen atoms by adding water molecules.

Oxidation half-reaction: 2Cl- → Cl2 + 2e-
Reduction half-reaction: 2Cr2O7- + 14H+ → 2Cr3+ + 7H2O + 6e-

Step 5: Balance the hydrogen atoms by adding hydrogen ions (H+).

Oxidation half-reaction: 2Cl- + 2H+ → Cl2 + 2e-
Reduction half-reaction: 2Cr2O7- + 14H+ → 2Cr3+ + 7H2O + 6e-

Step 6: Balance the charge by multiplying the half-reactions by appropriate coefficients so that the electrons cancel out.

Oxidation half-reaction (multiplied by 6): 6Cl- + 6H+ → 3Cl2 + 6e-
Reduction half-reaction (multiplied by 2): 4Cr2O7- + 28H+ → 4Cr3+ + 14H2O + 12e-

Step 7: Add the two half-reactions together and cancel out common species.

2Cr2O7- + 14H+ + 6Cl- + 6H+ → 4Cr3+ + 7H2O + 3Cl2 + 6e-

Step 8: Simplify the equation and combine the remaining species.

2Cr2O7- + 14H+ + 6Cl- → 4Cr3+ + 7H2O + 3Cl2

And there you have it! The balanced redox equation using the oxidation method.