Question

MnO4- + Fe2+ + H+ → Mn2+ + Fe3+ +H20 (acidic medium)​

Answer 1

To balance the equation :-

$\sf MnO_4^- + Fe^{2+} + H^+ \rightarrow Mn^{2+} + Fe^{3+} + H_2O$

Answer

$\sf \tiny{MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}}$

Solution

We weill use half-reaction method to balance the equation. Separate the oxidation half

$\sf Fe^{2+} \rightarrow Fe^{3+}$

Add one electron on right to balance the oxidation number.

$\sf Fe^{2+} \rightarrow Fe^{3+} + e^-$........eq(1)

Now the reduction half

$\sf MnO_4^- \rightarrow Mn^{2+}$

In LHS, oxidation number of Mn is +7. So add 5 electrons on left side to balance oxidation number.

$\sf MnO_4^- + 5e^- \rightarrow Mn^{2+}$

Balance number of oxygen atom by adding water on right side, and then balance hydrogen by adding H+ ions on the left side.

$\sf MnO_4^- + 5e^- + 8H^+ \rightarrow Mn^{2+} + 4H_2O$...........eq(2)

Now, balance number of electrons by multiplying with 5 in equation 1

thus, equation 1 becomes

$\sf 5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^-$

Now, add the two equations,

$\sf MnO_4^- + 5e^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+} + 5e^-$

Cancel electrons on both side.

$\sf MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}$

Answer 2

Answer:

mno4+fe2+ +H+ ___mn2+ +fe3+ +H2o