Introduction An important type of reaction mechanism is the transfer of electron

Question

Introduction An important type of reaction mechanism is the transfer of electrons from one species to another effecting a change in oxidation state. This gain or loss of electrons is referred to as reduction and oxidation, respectively. For example, consider the two balanced, chemical equations (1 & 2) that follow Cu (s) D Cu (aq) 2e (1) Cl2(aq) 2e D 2Cl (aq) (2) The first equation (1) shows elemental copper metal with a oxidation state of zero forming aqueous copper(Il) ions having an oxidation state of +2. This gain in oxidation number, 0 to +2, is called oxidation. The second equation (2) show elemental chlorine having an oxidation state of zero forming chloride ions in aqueous solution with an oxidation state of This decrease in oxidation number, 0 to -1, is called reduction. In any chemical reaction, the electrons gained or lost are conserved, i.e.. there is no net gain or loss of electrons. In fact, the two reactions illustrated above could occur simultaneously with the electrons lost by copper, gained by the chlorine in aqueous solution. Since the copper provides the electrons that subsequently reduce the chlorine copper is referred to as the reducing agent. Likewise, since the chlorine is accepting the electrons from copper, which subsequently oxidizes the copper, chlorine is referred to as he oxidizing agent. The two reactions above (1 & 2) are called half-reactions, as much as they each constitute one-half of the total reaction Equation 1 is the oxidation half reaction showing Cu(s) losing two electrons. In Equation 2, Cla(aq) gains two electrons and is the reduction half-reaction. If these two half-reaction are added together the result is the total oxidation-reduction reaction or "redox" reaction. Adding the two equations yields (3). Cu(s) Cl2(aq) D Cu (aq) 2CI (aq) (3) Whether or not the above reaction is spontaneous proceeds from reactants to products without input of energy into the system, may be determined by observing the reaction mixture if the reactants and products differ greatly in their appearance. Copper metal is a bright copper color and chlorine in aqueous solution will appear yellow. The Cu ion may turn the liquid part of the mixture blue, however, the Cris colorless and cannot be observed directly. Copper, like many of the transition metals, exhibits a variety of oxidation states depending upon the chemical environment in which it is found. In a strong reducing environment copper may exist as the free metal, Cu(s). An example of such a situation is illustrated in the following chemical equation (4). 4Cuo(s) CH4(g) D 4Cu(s) CO2(g) 2H2O(l) (4)

Explanation / Answer

Oxidation reduction reactions,

2. MnO4- reactions,

Ist : MnO4- + 8H+ + 5e- ---> Mn2+ + 4H2O

Equivalent mass of KMnO4 = 158.034/5 = 31.6068 g/mol

IInd : MnO4- + 8H+ + 4e- ---> Mn3+ + 4H2O

Equivalent mass of KMnO4 = 158.034/4 = 39.5085 g/mol

IIIrd : MnO4- + 8H+ + 7e- ---> Mn + 4H2O

Equivalent mass of KMnO4 = 158.034/7 = 22.5763 g/mol

when, Normality of KMnO4 solution = 0.2000 N,

Oxidation state of final Manganese = +2

molarity of KMnO4 solution = 0.2000/2 = 0.1000 M

3. Balancing in acidic medium,

BrO3- + P --> Br- + HPO3^2-

first half-reaction,

BrO3- --> Br-

balance oxygen,

BrO3- --> Br- + 3H2O

balance hydrogen and add electrons,

BrO3- + 6H+ + 6e- ---> Br- + 3H2O

second half-reaction,

P --> HPO3^2-

balance oxygen,

P + 3H2O --> HPO3^2-

balance hydrogen and add electrons,

P + 3H2O --> HPO3^2- + 6H+ + 3e-

multiply with 2 and add to first half-reaction,

BrO3- + 2P + 3H2O --> Br- + 2HPO3^2- + 6H+

is the net balanced equation.

Balancing in basic medium,

BrO3- + P --> Br- + HPO3^2-

first half-reaction,

BrO3- --> Br-

balance oxygen,

BrO3- --> Br- + 3H2O

balance hydrogen and add electrons,

BrO3- + 6H+ + 6e- ---> Br- + 3H2O

add OH- equal to H+,

BrO3- + 3H2O + 6e- ---> Br- + 6OH-

second half-reaction,

P --> HPO3^2-

balance oxygen,

P + 3H2O --> HPO3^2-

balance hydrogen and add electrons,

P + 3H2O --> HPO3^2- + 6H+ + 3e-

add OH- for basic medium

P + 6OH- --> HPO3^2- + 3H2O + 3e-

multiply with 2 and add to first half-reaction,

BrO3- + 2P + 6OH- --> Br- + 2HPO3^2- + 3H2O

is the net balanced equation.

4. Vanadium ions in solution,

VO2^+ : +5 oxidation state : yellow

VO^2+ : +4 oxidation state : blue

V(H2O)6^3+ : +3 oxidation state : green

V(H2O)6^2+ : +2 oxidation state : purple

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