Question

k2cr2o7+feso4+h2so4 =cr2(so4)3+fe2(so4)3+k2so4+h2o​

Answer 1

Answer:

Potassium dichromate iron(II) sulfuric acid (K2Cr2O7 FeSO4 H2SO4) is accessible in nearly every chemical laboratory worldwide. In the H2SO4 medium, K2Cr2O7 and FeSO4 respond mainly. To balance the redox reaction we use the ion-electron technique.

Answer 2

Explanation:

The Oxidizing agent: K2Cr2O7 or Cr2O72-

The Reducing agent: FeSO4 or, Fe2+

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+