There are two methods for balancing redox reaction
First method is oxidation method and second half reaction method or Ion electron method.
Both these methods are in use and the choice of their use rests with the individual using them
Oxidation number method
The steps are
(i) Write skeleton equation and indicate oxidation number of each element so that we can identify which elements are going change in oxidation number
(ii)Determine the increase and decrease of oxidation numbers per atom .Multiply the increase or decrease of oxidation number with number of atoms undergoing the change
(iii)Equalize increasing total oxidation number and decrease in total oxidation numbers on the reactant side by multiplying the respective formula with suitable integers
(iv) Balance the equation with respect to atoms other than oxygen and hydrogen atoms
(v) balance oxygen by adding equal number of water molecules to the sites for your short oxygen atoms
(vi) Balancing of hydrogen atom will depend on the medium acidic or basic as explain below
(a) if you are working in acidic medium then deficiency in hydrogen atom can be made by adding hydrogen ions on the side that is deficient in it
(b) On the other hand in basic medium first balance the atom as it is done in acidic medium then find each $H^+$ Ion add an equal number of $OH^-$ ions to both side of equation where $H^+$ and $OH^-$ appear on the same side equation combined is to give $H_2O$
Example
$Fe^{2+} + Cr_{2}O_7^{2-} + H^{+} -> Cr^{3+} + Fe^{3+} + H_{2}O$ Solution
(i) The skeleton equation along with oxidation number of each atom
$Fe^{2+} + Cr_{2}O_7^{2-} + H^{+} -> Cr^{3+} + Fe^{3+} + H_{2}O$
+2 +6 -2 +1 +3 +3
(ii) The oxidation number for chromium decrease from +6 to +3. So total decrease for two Cr atom is 6. On the other hand, the oxidation number of iron. increases from +2 (in $Fe^{2+}$ ) to +3 (in $Fe^{3+}$ )
(iii)To balance increase and decrease of oxidation numbers, multiply Fe by 6 and $Cr_{2}O_7^{2-}$ by 1.
(iv) On counting and equating the atoms on both sides
$6Fe^{2+} + Cr_{2}O_7^{2-} + 14H^{+} -> 2Cr^{3+} + 6Fe^{3+} + 7H_{2}O$
Half reaction method or Ion electron method
In this method first of all we identify the oxidant and reductant nn The Skeleton equation and then we split The Redox reaction into half reaction corresponding to oxidation and reduction. The balancing of is carried out systematically using the below outlined method
(i)Write a skeleton equation
(ii)Write half reaction separately for oxidation and reduction now .
(iii)Balance the atoms other than O and H in each half reaction individually
(iv) Balance the half reaction with respect to Oxygen and hydrogen atom ,for balancing for oxygen atom add water molecules to the side the deficient in it .Balancing of hydrogen atom will depend on the medium acidic or basic as explain below
(a) if you are working in acidic medium then deficiency in hydrogen atom can be made by adding hydrogen ions on the side that is deficient in it
(b) On the other hand in basic medium first balance the atom as it is done in acidic medium then find each H+ Ion add an equal number of OH- ions to both side of equation where H+ and OH- appear on the same side equation combined is to give $H_2O$
(v) Now balance the charge on both the equation .if necessary we can add the electrons on either side of the equation
(vi)The last and final step to multiply the half deflection equations by appropriate Coefficient so that number of electrons involved in both the equations are same and then add the equation and cancel the electrons are the common species on both the sides Example
Permanganate(VII) ion, $MnO_4^-$ in basic solution oxidizes iodide ion, $I^-$ to produce molecular iodine ($I_2$) and manganese (IV) oxide ($MnO_2$). Write a balanced ionic equation to represent this redox reaction
Solution
(i)First we write the skeletal ionic equation, which is
$MnO_4^-(aq) + I^-(aq) -> MnO_2(s) + I_2(s)$
(ii) The two half-reactions are:
Oxidation half :
$I^- (aq) -> I_2 (s)$
-1 0
Reduction half:
$MnO_4^-(aq) -> MnO_2(s)$
+7 +4
(iii)To balance the I atoms in the oxidation half reaction, we rewrite it as:
$2I^- (aq) -> I_2 (s)$
(iv) To balance the O atoms in the reduction half reaction, we add two water molecules on the right:
$MnO_4^- (aq) -> MnO_2 (s) + 2 H_2O (l)$
To balance the H atoms, we add four H+ ions on the left:
$MnO_4^- (aq) + 4H^+(aq) -> MnO_2 (s) + 2 H_2O (l)$
As the reaction takes place in a basic solution, therefore, for four H+ ions, we add four OH– ions to both sides of the equation:
$MnO_4^- (aq) + 4H^+(aq) + 4OH^- -> MnO_2 (s) + 2 H_2O (l) + 4OH^-$
Replacing the H+ and OH– ions with water
$MnO_4^- (aq) + 4H_2O -> MnO_2 (s) + 2 H_2O (l) + 4OH^-$
or
$MnO_4^- (aq) + 2H_2O -> MnO_2 (s) + 4OH^-$
(v) Now balance the charge
$2I^- (aq) -> I_2 (s) + 2e^–$
$MnO_4^–(aq) + 2H_2O(l) + 3e^– -> MnO_2(s)+ 4OH– (aq)$
(vi) Now to equalize the number of electrons, we multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2.
$6I^- (aq) -> 3I_2 (s) + 6e^–$
$2MnO_4^–(aq) + 4H_2O(l) + 6e^– -> 2MnO_2(s)+ 8OH– (aq)$
(vii): Adding both the reactions
$2MnO_4^–(aq) + 6I^- (aq) + 4H_2O(l) -> 2MnO_2(s)+ 3I_2 (s) + 8OH– (aq)$
