Cr + (aq) + 7H O (l) ( . ) Step : Add electrons to one side of the half reaction to balance the charges. If need be, make the number of electrons equal in the two half reactions by multiplying one or both half reactions by appropriate number. The oxidation half reaction is thus rewritten to balance the charge: Fe + (aq) → Fe + (aq) + e – ( .
) Now in the reduction half reaction there are net twelve positive charges on the left hand side and only six positive charges on the right hand side. Therefore, we add six electrons on the left side. – (aq) + 14H + (aq) + 6e – → 2Cr + (aq) + 7H O (l) ( . ) To equalise the number of electrons in both the half reactions, we multiply the oxidation half reaction by and write as : 6Fe + (aq) → 6Fe + (aq) + 6e – ( .
) Step : We add the two half reactions to achieve the overall reaction and cancel the electrons on each side. This gives the net ionic equation as : 6Fe + (aq) + Cr O – (aq) + 14H + (aq) → Fe + (aq) + 2Cr + (aq) + 7H O(l) ( . ) Step : Verify that the equation contains the same type and number of atoms and the same charges on both sides of the equation. This last check reveals that the equation is fully balanced with respect to number of atoms and the charges.
For the reaction in a basic medium, first balance the atoms as is done in acidic medium. Then for each H + ion, add an equal number of OH – ions to both sides of the equation. Where H + and OH – appear on the same side of the equation, combine these to give H O. Problem .
Permanganate(VII) ion, MnO – in basic solution oxidises iodide ion, I – to produce molecular iodine (I ) and manganese (IV) oxide (MnO ). Write a balanced ionic equation