Reduction Half reaction : Cr2O72−+6e−→2Cr3+
Oxidation Half reaction :
SO32−→SO42−+2e−×3
Overall reaction :
Cr2O72−+3SO32−→2Cr3++3SO42−
To balance ‘O’atoms, adding H2O on LHS
Cr2O72−+3SO32−→2Cr3++3SO42−+4H2O
To balance ‘H’ atoms, adding H+ on RHS
Cr2O72−+3SO32−+8H+→2Cr3++3SO42−+4H2O
Therefore, a=1,b=3,c=8.