Question

How to separate oxidising and reducing agents

Answer 1

Oxidation-reduction (redox) reactions
• Reactions in which there are changes in oxidation state
(oxidation number) between reactants and products
2 MnO4
- + 10 Br- + 16 H+ → 2 Mn2+ + 5 Br2 + 8 H2O
• One reactant must be oxidized (lose electrons, the reductant or
reducing agent) and another must be reduced (gain electrons,
the oxidant or oxidizing agent)
– In some instances a single reactant will be both oxidized and
reduced (disproportionation)
– in other cases two species containing the same element in
different oxidation states will combine to give a single product
with an intermediate oxidation state (comproportionation).
Oxidation-reduction (redox) reactions, cont’d
• Redox reactions may be separated into oxidation and reduction
“half-reactions”.
MnO4
- + 8 H+ + 5 e- → Mn2+ + 4 H2O
2 Br- → Br2 + 2 e-
2 MnO4
- + 10 Br- + 16 H+ → 2 Mn2+ + 5 Br2 + 8 H2O
– When half-reaction reagents can be isolated into separate
compartments connected by a conduit for ion migration,
electron flow through an external circuit can be utilized to
perform work (basis for batteries and electrochemical cells).
– The potential (Eo) of an electrochemical cell is the sum of the
potentials of the reduction and oxidation half-reactions
– The potential for a cell or half-reaction is related to the free
energy change for the redox reaction through the relationship
ΔGo = -nöEo where n is the number of electrons transferred
and ö is the Faraday (9.65 x 104 J V-1
Balancing oxidation-reduction (redox) reactions
• Write separate half-reactions for oxidation and reduction
• Balance each half-reaction for mass utilizing water and protons as
necessary, i.e.,
– Get rid of oxygen as water
– Get rid of hydrogen as H+
– Get hydrogen from H+
– Get oxygen from water
• Balance each half-reaction for charge by adding electrons to the right
(oxidation) or left (reduction) as required
• Multiply half-reactions by the appropriate factors such that the electrons
lost in oxidation equal the electrons gained in reduction
• Add the two half-reactions; if the same reagent appears as both reactant
and product, subtract the smaller amount from both sides of the equation
HNO3(aq) + Cu(s) = NO2(g) + Cu2+
(aq)
Balancing oxidation-reduction (redox) reactions
• Write separate half-reactions for oxidation and reduction
• Balance each half-reaction for mass utilizing water and protons as
necessary, i.e.,
– Get rid of oxygen as water
– Get rid of hydrogen as H+
– Get hydrogen from H+
– Get oxygen from water
• Balance each half-reaction for charge by adding electrons to the right
(oxidation) or left (reduction) as required
• Multiply half-reactions by the appropriate factors such that the electrons
lost in oxidation equal the electrons gained in reduction
• Add the two half-reactions; if the same reagent appears as both reactant
and product, subtract the smaller amount from both sides of the equation
HClO = Cl2 + ClO3