Balance the redox equation
K2Cr2O7 + FeSO4 + H2SO4$\to$K2SO4 + Cr2(SO4)3 + Fe(SO4)3 +H2O

 

Oxidation Half Reaction: ⇒ Fe2+ – e– = Fe3+ … … … (1) Reduction Half Reaction: ⇒ Cr2O72- + 6e– + 14H+ = 2Cr3+ + 7H2O … … … (2) Now, equation (1)x6 + (2), 6Fe2+ – 6e– = 6Fe3+ Cr2O72- + 14H+ + 6e– = 7H2O + 2Cr3+ 14H+ + 6Fe2+ + Cr2O72- = 2Cr3+ + 7H2O + 6Fe3+ Now adding necessary ions and radicals we get, ⇒ K2Cr2O7 + 6FeSO4 + 7H2SO4 = Cr2(SO4)3 + 3Fe2(SO4)3 + K2SO4 + 7H2O “Answer” K2Cr2O7 + 6FeSO4 + 7H2SO4 = Cr2(SO4)3 + 3Fe2(SO4)3 + K2SO4 + 7H2O Hope it helps!!