Cu loses 2 e - CCBC Faculty Web

Part I: Redox Review & How to
Balance Complex Redox Equations
Dr. C. Yau
Fall 2013
Jespersen 6/e
Chap. 6 Sec 1 & 2
Redox Rxn
Cu (s) + AgNO3 (aq)
Net ionic equation?
Review of Redox Rxns
Oxidation is an increase in oxidation number.
Reduction is a decrease in oxidation number.
Cu (s) + 2AgNO3 (aq)
Cu(NO3)2 (aq) + 2Ag
Cuo + 2Ag+
+ 2Ago
Cu is oxidized. Ag is reduced.
Cu loses 2 e- while Ag gains 1 e- each.
A total of 2 e- is transferred from Cu to Ag+
Cu = reducing agent;
Ag+ = oxidizing agent.
Alternatively, you can remember…
LEO the lion says GER
Loss of Electrons is Oxidation.
Gain of Electrons is Reduction.
Cu + Ag+
Ag+ + e-
Cu2+ + Ag
Cu2+ + 2 eAg
Oxidation (Loss of e-)
Reduction (Gain of e-)
Recognition of Redox
Redox always involve a change in oxidation
Reduction must be accompanied by
oxidation, and vice versa.
Oxidation numbers are hypothetical
charges assigned to each atom.
Each atom, even in molecular substances
that have no ions, is assigned a charge.
Note: The charge is hypothetical.
Oxidation numbers is only for "book
keeping" to keep track of electrons.
Assigning Oxidation Numbers
1. Oxidation # of any free element is zero. e.g.
H in H2, P in P4 are assigned zero.
2. Oxid # of simple monatomic ions is the
charge of the ion.
e.g. In MgCl2, Mg is +2. Cl is -1.
In SnS, Sn is +2 and S is -2.
Assigning Oxidation Numbers
3. In its cmpd, F is assigned -1.
4. In its cmpds, H is +1 unless it is bonded
to a metal where it is -1.
e.g. In HCl, H is +1 and Cl is -1.
In MgH2, Mg is +2 and H is -1.
5. In its cmpds, O is -2 unless it is a
peroxide. eg. in MgO, Mg is +2, O is -2.
In hydrogen peroxide (H2O2), H is +1 and
O is -1.
Assigning Oxidation Numbers
6. Oxidation # of others are generally calculated from
knowing that the sum of all charges must add up to charge
of the particle. For example,
In HClO, sum of charges = 0
+1 ? -2
H is assigned +1, O is assigned -2. Cl is calculated to be +1
in order for the sum to be zero.
In the chlorite ion, ClO2-, sum of charges = -1
? -2
Oxygen is assigned -2, what must Cl be in order for net
charge to be -1?
Ans. Cl = +3
Give the oxidation number:
1. N in NO3-
2. Cr in CrSO4
3. S in S2O42Identify which is the oxidizing agent:
Cr2O72− + 14H+ + 6 Cl−  3Cl2 + 2 Cr3+
+ 7 H 2O
Answer to questions on previous slide:
Give the oxidation number:
1. N in NO3- Ans. +5
Cr in CrSO4 Ans. +2 (Be sure to ask if you don’t know
how you can tell.)
S in S2O42- Ans. +3
Identify which is the oxidizing agent:
Cr2O72− + 14H+ + 6 Cl−  3Cl2 + 2 Cr3+ + 7 H2O
Ans. Cl is going from -1 to 0, so it is being oxidized and
cannot be the oxidizing agent.
Cr is going from +6 to +3, so it is being reduced.
The oxidizing agent is Cr2O72− (Note: By convention, you
name it as Cr2O72− and not just Cr.)
Balancing Redox Under Acidic
1. Verify that rxn is redox.
2. Divide skeleton eqns into half-reactions.
3. Balance atoms other than H and O.
4. Balance O by adding H2O to each side of eqn.
5. Balance H by adding H+ to side that needs H.
6. Balance charge by adding electrons.
7. Make the # of electrons gained equal to number lost
8. Add two half-reactions and cancel anything that is
the same on both sides of eqn.
Balance the following reactions which are
under acidic conditions.
Cr2O72− +
Cl− 
Mg + VO43− 
Sn2+ + IO4− 
Sn4+ + I−
Practice Example 6.6, p.225, Pract Exer 12, 13,
14, 15 on p.226; p.245 #6.37.
Balancing Redox Under BASIC
Follow the same steps as for acidic conditions
but add on the following steps at the end. It
would save you steps if you do this AFTER you
have already combined the half-reactions into
one equation:
9. Add to BOTH sides, the same number of OHas there are H+ (Under basic condition, there
cannot be any H+).
10. Combine H+ and OH- to form H2O.
11. Cancel any H2O that you can.
Balance the following reactions which are under
basic conditions.
1. SO32– + MnO4–
 SO42– + MnO2
Which is the reducing agent?
2. CN– + MnO4–
 CNO– + MnO2
Which is the oxidizing agent?
Practice Exer 6.16, 17, 18, 19on p. 227
and p.245 #6.39

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