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 )