Balance redox reaction through oxidation method of an acid

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

1 answer

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.