✔ 最佳答案
Answer :
10FeSO4 + 2KMnO4 + 8H2SO4 → 5Fe2SO4 + 2MnSO4 + K2SO4 + 8H2O
Firstly, write a completely ionic equation from two half equations.
The half equation of oxidation :
Fe^2+ → Fe^3+ + e^- ... (*)
The half equation of reduction :
MnO4^- + 8H^+ + 5e^- → Mn^2+ + 4H2O ... (#)
5(*) + (#), and cancel 5e^- on the both sides:
5Fe^2+ + MnO4^- + 8H^+ → 5Fe^3+ + Mn^2+ + 4H2O
Secondly, add dummy ions to the both sides of the equation :
5Fe^2+ + 5SO42- + K^+ + MnO4^- + 8H^+ + 4SO4^2-
→ 5Fe^3+ + (15/2)SO4^2- + Mn^2+ + SO4^2- + K^+ + (1/2)SO4^2- + 4H2O
5FeSO4 + KMnO4 + 4H2SO4 → (5/2)Fe2SO4 + MnSO4 + (1/2)K2SO4 + 4H2O
Multiply both the two sides by 2 :
10FeSO4 + 2KMnO4 + 8H2SO4 → 5Fe2SO4 + 2MnSO4 + K2SO4 + 8H2O