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.
Balance redox reaction through oxidation method of an acid
Cr2O7- + Cl- --> Cr3+ + Cl2
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