Balancing redox reactions


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Before you balance an overall redox equation,you have to able to balance two half-equations,one for oxidation (electron loss) and one for reduction (electron gain).Sometimes that's easy, sometimes its tough, lets see first the steps to balance half equations :
(a) Balance the atoms of the element being oxidized or reduced.
(b) Balance oxidation number by adding electrons:-For a reduction half equation add the appropriate number of electrons to the left;for an oxidation half equation add electrons to the right.
(c) Balance charge by adding H+ ions in acidic solution, OH- ions in basic solution.
(d) balance hydrogen by adding H2) molecules.
(e) Check to make sure that oxygen is balanced.

For example: balance the following half equations.
(1) MnO4- (aq) ----> Mn^2+ (aq) ( acidic solution)
(2) Cr(OH)3 (s) ------> CrO4^2- (aq) (basic solution)

solution (1) because there is one atom of Mn on both the sides, no adjustment is required.
MnO4- (aq) ----> Mn^2+ (aq)
(b) Because manganese is reduced from an oxidation number of +7 to +2,five electrons must be added to left.
MnO4- (aq) + 5e- ----> Mn^2+ (aq)
(c) There is a total charge of -6 on the left versus +2 on the right.To balance, add eight H+ to the left to give a charge of +2 on both sides.
MnO4- (aq)+ 8H+ (aq) + 5e- ----> Mn^2+ (aq)
(d) To balance the eight H+ ions on the left,add four H2O molecules to the right.
MnO4- (aq)+ 8H+ (aq) + 5e- ----> Mn^2+ (aq) + 4H2O
(e) Note that there are the same number of oxygen atoms, four on both sides,as there should be.The equation shown in green is the correctly balanced reduction half -equation.

(2) (a) Again, there is one chromium atom on both sides.
Cr(OH)3 (s) ------> CrO4^2- (aq)
(b) Because the oxidation number of chromium increases from +3 to +6, add three electrons to the right.
Cr(OH)3 (s) + 5OH- (aq) -------> CrO4^2- (aq) + 3e-
(c) There is a charge zero on the left,non on the right.Add four H2O molecules to the right.
Cr(OH)3 (s) + 5OH- (aq) -------> CrO4^2- (aq) + 3e- + 4H2O
(e) There are eight oxygen atoms on both side; the oxidation half equation is properly balanced.

The process used to balance an overall redox reaction:
(1) Split the equation into two half equations, one for the reduction, the other for oxidation.
(2) Balance one of the half equations with respect to both atoms and charge as described above
(3) Balance the other half equation.
(4) Combine the two half-equation in such a way as to eliminate electrons.

For example:
balance the following redox reaction:
(a) Fe^2+ (aq) + MnO4- (aq)-----------> Fe^3+ (aq) + Mn^2+ (aq) (acidic solution)
(b) Cl2 (g) + Cr(OH)3 (s) ---------> Cl- (aq) + CrO4^2- (aq) (basic solution)

(a) oxidation: Fe^2+ (aq) ------> Fe^3+ (aq)
reduction: MnO4- (aq) ---------> Mn^2+ (aq)
Fe^2+ (aq) ------> Fe^3+ (aq) (oxidation no. Fe: +2 -----> +3)
It is clear that mass and charge balance can be achieved by adding an electron to the right:
Fe^2+ (aq) ------> Fe^3+ (aq) + e-
MnO4- (aq) ---------> Mn^2+ (aq)
this has already been balanced in the above problem
MnO4- (aq)+ 8H+ (aq) + 5e- ----> Mn^2+ (aq) + 4H2O

To eliminate the electrons,multiply the oxidation half-equation by 5 and add to the reduction half equation
5[ Fe^2+ (aq) ------> Fe^3+ (aq) + e-]
MnO4- (aq)+ 8H+ (aq) + 5e- ----> Mn^2+ (aq) + 4H2O
5Fe^2+ (aq) + MnO4- (aq) + 8H+ (aq) ----------> 5Fe^3+ (aq) + Mn^2+ (aq) + 4H2O

(b) reduction: Cl2 (g) --------> Cl- (aq)
oxidation: Cr(OH)3 (s) ------> CrO4^2- (aq)
Cl2 (g) -------> Cl- (oxid no, Cl: 0 ------> -1)
mass balance is obtained by writing a coefficent of 2 for Cl-; charge is then balanced by adding two electrons to the left.The balanced equation is
Cl2 (g) + 2e- ------->2Cl- (aq)

Cr(OH)3 (s) ------> CrO4^2- (aq) ( the equation has already been balanced in above example)
Cr(OH)3 (s) + 5OH- (aq) -------> CrO4^2- (aq) + 3e- + 4H2O
multiply the reduction half equation by 3, the oxidation half equation by 2,then add.This will produce 6e- on both sides,so they will cancel.
3 [ Cl2 (g) + 2e- ------->2Cl- (aq)]
2[ Cr(OH)3 (s) + 5OH- (aq) -------> CrO4^2- (aq) + 3e- + 4H2O]
3Cl2 (g) + 2Cr(OH)3 (s) + 10 OH- (aq) ----------> 6Cl- (aq) + 2CrO4^2- (aq) + 8H2O