Cr2O7 2 –( aq) + 3SO3 2 –( aq) 8H + → 2Cr 3 +( aq)+ 3SO4 2 –( aq)
Step 5: Balance the oxygen atom by adding water molecule. Cr2O7 2 –( aq) + 3SO3 2 –( aq) 8H + → 2Cr 3 +( aq)+ 3SO4 2 –( aq)+ 4H2O( l)
Half Reaction Method Balance the equation showing the oxidation of Fe 2 + ions to Fe 3 + ions by
dichromate ions( Cr2O7) 2 – in acidic medium, where in, Cr2O7 2 – ions are reduced to Cr 3 + ions.
Step 1: Produce unbalanced equation for the reaction in ionic form: Fe 2 +( aq) + Cr2O7 2 –( aq) → Fe 3 +( aq) + Cr 3 +( aq)
Step 2: Separate the equation into half reactions: + 2 + 3 Oxidation half: Fe 2 +( aq) → Fe 3 +( aq)
+ 6 – 2 + 3 Reduction half: Cr2O7 2 –( aq) → Cr 3 +( aq)
Step 3: Balance the atoms other than O and H in each half reaction individually.
Cr2O7 2 –( aq) → Cr 3 +( aq)
Step 4: For reactions occurring in acidic medium, add H2O to balance O atoms and H + to balance H atoms. Cr2O 7 2 –( aq) + 14 H + → Cr 3 +( aq) + 7H 2 O( l)
Step 5: Add electrons to one side of the half reaction to balance the charges. If need be, make the number of electrons equal in the two half reactions by multiplying one or both half reactions by appropriate coefficients.
Fe 2 +( aq) → Fe 3 +( aq) + e –
Cr2O7 2 –( aq) + 14H +( aq) + 6e – → 2Cr 3 +( aq) + 7H2O( l) 6Fe 2 +( aq) →6 Fe 3 +( aq) + 6 e –
Step 6: We add the two half reactions to achieve the overall reaction and cancel the electrons on each side. This gives the net ionic equation as: 6Fe 2 +( aq) + Cr2O7 2 –( aq) + 14H +( aq) → 6 Fe 3 +( aq) + 2Cr 3 +( aq) + 7H2O( l)