balance the equation by ion electron method.
Fe^++ + MnO4- = Fe3+ +Mn2+ (acidic medium) please help me..!!
Sure! I can help you balance the equation using the ion-electron method.
Step 1: Write down the unbalanced equation:
Fe^2+ + MnO4^- = Fe^3+ + Mn^2+
Step 2: Determine the changes in oxidation states for each element:
Fe goes from +2 to +3 (an increase of 1)
Mn goes from +7 to +2 (a decrease of 5)
Step 3: Divide the equation into two half-reactions based on the changes in oxidation states:
a) Oxidation half-reaction: Fe^2+ → Fe^3+
b) Reduction half-reaction: MnO4^- → Mn^2+
Step 4: Balance the atoms and charge in each half-reaction:
a) Oxidation half-reaction:
Fe^2+ → Fe^3+
To balance the Fe atoms, we need to add one Fe^3+ on the right side.
Fe^2+ → Fe^3+ + e^-
b) Reduction half-reaction:
MnO4^- → Mn^2+
To balance the Mn atoms, we need to add five Mn^2+ ions on the left side.
MnO4^- + 5e^- → Mn^2+
Step 5: Balance the electrons:
Since we have 5 electrons on the left side in the reduction half-reaction and only 1 electron on the right side in the oxidation half-reaction, we need to multiply the oxidation half-reaction by 5 to balance the electrons.
5Fe^2+ → 5Fe^3+ + 5e^-
MnO4^- + 5e^- → Mn^2+
Now the electrons are balanced.
Step 6: Balance the remaining elements and charges:
To balance the number of Fe atoms, we need to multiply the reduction half-reaction by 5:
5MnO4^- + 25e^- → 5Mn^2+
Now we can add the half-reactions together:
5Fe^2+ + 5MnO4^- + 25e^- → 5Fe^3+ + 5Mn^2+ + 5e^-
Step 7: Cancel out the electrons in the equation:
5Fe^2+ + 5MnO4^- → 5Fe^3+ + 5Mn^2+
Step 8: Balance the hydrogen and oxygen atoms and any remaining charges:
In an acidic medium, you can add H2O molecules to balance the oxygen atoms and H+ ions to balance the hydrogen atoms. In this case, we need to add H+ ions to the left side to balance the hydrogen atoms:
5Fe^2+ + 5MnO4^- + 25H+ → 5Fe^3+ + 5Mn^2+ + 15H2O
Now the equation is balanced.
Therefore, the balanced equation by the ion-electron method in acidic medium is:
5Fe^2+ + 5MnO4^- + 25H+ → 5Fe^3+ + 5Mn^2+ + 15H2O