SCH 4U UofT Electro chemistry Questions
Surname 1Student’s Name:
Professor’s Name:
Course:
Date:
Part A
Q1
Answer: C) reduction
Q2
Answer: a) The Cr2O7^2- ion is reduced.
Q3
Answer: d) Cr2O7 2Q4
(a) a only
Answer: K2CrO4 + BaCl2 ¬ BaCrO4 + 2KCl is a redox reaction.
Q5
Answer: E) Hydrogen sulfate (H2SO4) is the reducing agent in the reaction.
Q6
Answer: E) +6
Surname 2
Q7
Answer: C) +2
Q8
Answer: E) +4
Q9
The balanced equation: BrO− + 2 Fe(OH)2 → Br− + Fe(OH)3
D) Answer: The coefficient of the bromide ion is 1.
Q10
The balanced equation is: 2CN- + 2Fe3+ ¬ CNO- + 2Fe2+
B) Answer: 2
Q11
Answer: C) two-electron
Q12
Answer: E) Fe (s) ¬ Fe2+ (aq) + 2e
Q13
Answer: E) maintain electrical neutrality in the half-cells via migration of ions.
Q14
Answer: B) anode, cathode
Surname 3
Q16
Answer: C) Cl2
Q17
Answer: E) Halogens
Q18
Answer: D) Cu2+ (aq)
Q19
Answer: E) -0.46
Q20
Answer: B) +1.94 V
Part B Short Answers
1. Electrochemical Cell: Anode: Mn/Mn2+ Cathode: Ag+/Ag Salt Bridge: NaCl Voltage
Produced: +0.80V
Surname 4
2. Mn+2
3. MnO4 –
4. a) Mg= +2, Ti= +4, O= -2
b) K= +1, Cr= +6, O= -2
c) Mn= +7, O= -2
5. A Galvanic cell uses a redox reaction to generate an electric current and is a type of battery.
An electrolytic cell uses electrical energy to drive a non-spontaneous reaction, such as
electrolysis.
6. Oxidation: Cr2O7 2- (aq) → Cr3+ (aq) + 7H+ (aq) + 7eOxidation number of Cr in Cr2O7 2- = +6
Oxidation number of Cr in Cr3+ = +3
Reduction: H2SO3 (aq) + 7e- → HSO4- (aq) + 2H+ (aq)
Oxidation number of S in H2SO3 = +4
Oxidation number of S in HSO4- = +6
Surname 5
Balanced: Cr2O7 2- (aq) + H2SO3 (aq) → Cr3+ (aq) + HSO4- (aq) + 4H+ (aq)
7. A)
b) Reduction: Ag+(aq) + e- → Ag(s)
Oxidation: Cu(s) → Cu2+(aq) + 2eNet reaction: Ag+(aq) + Cu(s) → Ag(s) + Cu2+(aq)
c)The electric potential of the cell can be calculated using the Nernst equation:
Ecell = E°cell – (RT/nF) ln Q
where E°cell is the standard cell potential, R is the ideal gas constant, T is the temperature in
Kelvin, n is the number of moles of electrons transferred, F is the Faraday constant, and Q is
the reaction quotient.
For this reaction, E°cell = +0.34 V, n = 2, R = 8.314 J/K mol, F = 96485 C/mol, and Q =
[Ag+][Cu2+]/[Ag][Cu].
Therefore, the electric potential of the cell is:
Surname 6
Ecell = 0.34 – (8.314 J/K mol x 298 K)/(2 x 96485 C/mol) ln ([Ag+][Cu2+]/[Ag][Cu])
= 0.34 – 0.0245 ln ([Ag+][Cu2+]/[Ag][Cu])
The reaction is spontaneous if the electric potential of the cell is positive. Therefore, if the
logarithm of the reaction quotient is greater than 14.02, the reaction will be spontaneous.
8.
a. At the zinc electrode: Zn → Zn2+ + 2e- At the silver electrode: Ag+ + e- → Ag
b. The zinc electrode is the anode and it is negative. The silver electrode is the cathode and it is
positive.
c. The overall reaction for the cell is: Zn + 2Ag+ → 2Ag + Zn2+. The standard cell potential is
E°cell = 0.04 V.
d. The electron flow is from the zinc electrode (anode) to the silver electrode (cathode).
e. The ion flow in the salt-bridge is from the zinc compartment to the silver compartment.
f. The oxidizing agent is Zn2+ and the substance oxidized is Zn.
g. The standard cell notation for the spontaneous reaction occurring in the cell is: Zn (s) | Zn2+
(1.0 M) || Ag+ (1.0 M) | Ag (s).
9. Balanced equation: 2S²- (aq) + Br2 (g) → 2Br- (aq) + SO4²- (aq)
10. Anode: 2 Cl- (aq) → Cl2 (g) + 2 eCathode: Sn2+ (aq) + 2 e- → Sn (s)
Net ionic equation: 2 Cl- (aq) → Cl2 (g) + Sn (s)
Surname 7
11.
The half-reaction equation for the copper electrode is as follows: Cu2+ (aq) → Cu (s) + 2eThis equation shows that when copper ions are reduced at the copper electrode, copper atoms are
deposited on the electrode and two electrons are released. This explains why the concentration of
copper ions in the solution has decreased and the mass of the electrode has increased.
The half-reaction equation for the other electrode is as follows: 2H+ (aq) + 2e- → H2 (g) This
equation shows that when hydrogen ions are oxidized at the other electrode, hydrogen gas is
produced and two electrons are consumed. This explains why the concentration of hydrogen ions
in the solution has increased.
12.
(a) Anode: 2Br- (aq) → Br2 (l) + 2eCathode: Ni2+ (aq) + 2e- → Ni (s)
Metal Plated: Yes
(b) Anode: 2F- (aq) → F2 (l) + 2eCathode: Al3+ (aq) + 3e- → Al (s)
Metal Plated: Yes
(c) Anode: 2I- (aq) → I2 (l) + 2eCathode: Mn2+ (aq) + 2e- → Mn (s)
Metal Plated: Yes
Surname 8
13.
(a) Cathode: Ni2+ + 2e- –> Ni
Anode: 2Br- –> Br2 + 2e(b) Cathode: 2Al3+ + 6e- –> 2Al
Anode: 3F- –> F3 + 3e(c) Cathode: Mn2+ + 2e- –> Mn
Anode: 2I- –> I2 + 2e-
Surname 9
Works Cited
10.7 Electrolysis
Learning Goals …
… explain the difference between a galvanic and an
electrolytic cell
… predict the products of the electrolysis of molten
and aqueous solution
There are two types of electrochemical cells:
1. A galvanic cell produces a current when an oxidationreduction reaction occurs spontaneously
2. An electrolytic cell uses electrical energy to produce a
chemical reaction that would not occur spontaneously
Examples of electrolytic cells:
• Recharging a battery
• Electroplating metals
Galvanic Cell
vs
Electrolytic Cell
Galvanic Cell
vs
Electrolytic Cell
Electrolysis of Water
The electrolysis of water produces hydrogen gas at the
cathode and oxygen gas at the anode.
Note:
electrolysis is not a
spontaneous reaction
and requires a battery
Cathode (reduction):
2 H2O(l) + 2e− → H2(g) + 2 OH-(aq)
Er= -0.83V
Anode (oxidation):
2 H2O(l) → O2(g) + 4 H+(aq) + 4e−
Er = 1.23V
Overall reaction:
2 H2O(l) → 2 H2(g) + O2(g)
Ecell = -2.06V
Predicting the Products of Electrolysis
of an Aqueous Solution
1. Identify the two possible electrolysis reactions that may
occur (electrolysis of salt and electrolysis of water).
2. List the four relevant half reactions and their reduction
potentials.
3. Determine which reaction will occur at cathode (the
reaction with highest reduction potential will be reduced)
4. Determine which reaction will occur at the anode (the
reaction with the lowest reduction potential will be
oxidized)
e.g. Predict the products of the electrolysis of 1.0 mol/L NiCl2 (aq).
Write the net ionic equation and calculate the cell potential.
2 possible reactions
(4 half-reactions)
Cathode (reduction):
NiCl2(aq) → Ni(s) and Cl2(g)
2 H2O(l) → 2H2(g) + O2(g)
highest reduction
potential
E = – 0.23 V
Ni2+ + 2e- → Ni(s)
2 H2O(l) + 2e− → H2(g) + 2 OH-(aq)
Anode (oxidation):
r
Er = – 0.83 V
lowest reduction
2Cl- → Cl2(g) + 2eEr = 1.36 V
potential
2 H2O(l) → O2(g) + 4 H+(aq) + 4 e−
Er = 1.23 V
Net Ionic Equation:
Do not change
these values
Multiply by 2
Cathode
Ni2+ + 2e- → Ni(s)
Er = – 0.23 V
Anode
2H2O(l) → O2(g) + 4H+(aq) + 4e−
Er = 1.23 V
2Ni2+ + 4e- → 2Ni(s)
2H2O(l) → O2(g) + 4H+(aq) + 4e−
Er(cathode) = – 0.23 V
Er(anode) = 1.23 V
2Ni2+(aq)+2H2O(l) → 2Ni(s) + O2(g) + 4H+ Ecell = – 1.46 V
(Cathode – Anode)
Self Check
How prepared am I to start my homework? Can I …
… explain the difference between a galvanic and
electrolytic cell?
… predict the products of electrolysis of a molten and
an aqueous solution?
HOMEWORK
p670 #3 (molten solutions), 8
p682 #60 (aqueous solutions)
include net ionic equations and
calculate cell potential for each
9.2 Balancing Redox Reactions
Using Oxidation Numbers
Learning Goals …
… balance complex redox reactions using oxidation
numbers in aqueous solutions
… balance redox reactions in acidic and basic
solutions
• Most simple redox reactions can be balanced by
inspection.
• For more complex reactions, we can use either of
two methods:
1. Oxidation Numbers Method
2. Half Reactions Method
• For redox reactions, the charge must be balanced,
and so must each element.
Oxidation Numbers Method
1. Determine the oxidation numbers for each element in the
equation and identify the elements for which the
oxidation numbers change.
2. Adjust the values of the coefficients to balance the
electrons transferred.
3. Balance the rest of the equation by inspection. If
necessary, balance oxygen by adding H2O.
4. If necessary, balance hydrogen by adding H+ and/or OH-.
5. Check your answer.
6. Write the balanced equation.
• Redox reactions often take place in aqueous
solutions, which can be acidic, basic or
neutral.
when balancing, you may need to include
H2O, H+ or OH-.
Eg 1)
This number becomes the coefficient
required to balance the electrons
in ox #, oxidized, 7e- lost (x4)
4 NH3 + 7 O2
(-3)(+1)
(0)
4 NO2 + 6 H2O
(+4)(-2)
in ox #, reduced, 2e- gained /O
4e- gained /O2 (x7)
• Add H2O to balance the oxygens
Eg 2) A Redox Reaction in Acidic Solution
in ox #, oxidized, 1e- lost (x5)
8 H+ + MnO4- + 5 Fe2+
(+7)(-2)
(+2)
Mn2+ + 5 Fe3+ + 4 H2O
(+2)
(+3)
in ox #, reduced, 5e- gained (x1)
• Add H2O to balance the oxygens
• Add H+ to balance the hydrogens
Note: The charges must balance (+17 on each side)
Eg 3) A Redox Reaction in Basic Solution
in ox #, red., 6e- gained (x1)
I- + 6 CO2 + 3 H2O + 6 OH6 H+ + IO3- + 3C2O42(-1) (+4)(-2)
6 OH- (+5)(-2) (+3)(-2)
6 H2O
3 H2O
in ox #, ox. 1e- lost /C
2e- lost /C2
(x3)
• Add H2O to balance the oxygens
• Add H+ to balance the hydrogens
• Add same # of OH- (as H+ added) to both sides.
Note: H+ + OHH2O (cancel out extra H2O)
The charges must balance (-7 on each side)
Self Check
How prepared am I to start my homework? Can I …
… balance complex redox reactions using oxidation
numbers in aqueous solutions?
… balance redox reactions in acidic and basic
solutions?
HOMEWORK
p613 #1-6
p617 #3,4
10.2 Standard Reduction
Potentials
Learning Goals …
… calculate the cell potential using a table of standard
reduction potentials
… predict the spontaneity of redox reactions based on
overall cell potentials
• The Cell Potential is the electric potential difference or
voltage between the anode and cathode.
• The Standard Reduction Potential is the ability of a halfcell to attract electrons in a cell that is operating under
standard conditions (1.0M, standard pressure and ambient
temperature (101.3 kPa, 25°C)
Calculating Standard Cell Potential:
E(cell) = Er(cathode) – Er(anode)
❖ If E°(cell) > 0 then the reaction is spontaneous
❖ If E°(cell) < 0 then the reaction is not spontaneous
Ex 1) Determine the cell potential and write the net ionic
equation for the following spontaneous reaction:
Mn2+ + 2e- Mn(s) Er = -1.18 V
Fe3+ + e- Fe2+
Er = 0.77 V
Lower reducing agent
OXIDIZED (anode)
Higher oxidizing agent
REDUCED (cathode)
Calculating the Standard Cell Potential:
E(cell) = Er(cathode) - Er(anode)
E(cell) = 0.77 - (-1.18)
= 1.95 V
Ex 1) Determine the cell potential and write the net ionic
equation for the following spontaneous reaction:
Mn2+ + 2e- Mn(s) Er = -1.18 V
Fe3+ + e- Fe2+
Er = 0.77 V
Lower reducing agent
OXIDIZED (anode)
Higher oxidizing agent
REDUCED (cathode)
Net Ionic Equation:
• Since the above reactions are both reduction reactions,
you must reverse the reaction occurring at the anode
• Also multiply the equations in order for the electrons to be
equal (Note: Do not multiply the reduction potential (E r)
Ex 1)
Reverse reaction
Mn2+ + 2e- Mn(s) Er = -1.18 V
Fe3+ + e- Fe2+
Er = 0.77 V
Lower reducing agent
OXIDIZED (anode)
Higher oxidizing agent
REDUCED (cathode)
Multiply reaction by 2
Net Ionic Equation:
Mn(s) → Mn2+ + 2e2Fe3+ + 2e- → 2Fe2+
Mn(s) + 2Fe3+ → Mn2+ + 2Fe2+
Ex 2) Determine the cell potential and write the net ionic
equation for the following:
Reaction at the
anode is reversed
Zn(s) Zn2+(aq) Sn2+(aq) Sn(s)
anode
cathode
Zn2+(aq) + 2e- Zn(s)
Er = -0.76 V
(anode)
Sn2+(aq) + 2e- Sn(s)
Er = -0.14 V
(cathode)
E(cell) = Er(cathode) - Er(anode)
E(cell) = -0.14-(-0.76)
= 0.62 V
Net Ionic Equation:
Half reactions:
Zn(s) → Zn2+(aq) + 2eSn2+(aq) + 2e- → Sn(s)
Zn(s) + Sn2+(aq) → Zn2+(aq) + Sn(s)
Ex 3) Using cell potential, determine if the following redox
reaction will be spontaneous:
Fe3+ + 2Cl-
Fe2+ + Cl2
Find the reactions on the standard reduction potential table
Fe3+ + e- Fe2+
Er= 0.77 V (cathode)
Cl2 + 2e- 2ClEr= 1.36 V (anode)
Note:
• The above reactions are both reduction reactions
• Since the chlorine is reversed in the net ionic equation,
this reaction is occurring at the anode (oxidation)
E(cell) = Er(cathode) - Er(anode)
E(cell) = 0.77 - 1.36
= -0.59 V
Not a spontaneous reaction
Self Check
How prepared am I to start my homework? Can I …
… calculate the cell potential using a table of standard
reduction potentials?
… predict the spontaneity of redox reactions based on
overall cell potentials?
HOMEWORK
p647 #1
p648 #2-6, 8a
9.3 Predicting Redox Reactions
Learning Goal …
… predict the spontaneity of redox reactions based
on position in standard reduction potential table
• A redox reaction involves a transfer of valence electrons from
one substance to another.
• Most atoms, molecules, and ions are stable, and do not
readily release electrons.
• Since two particles must be involved in an electron transfer,
this transfer can be explained as a competition for electrons.
If one particle is able to pull
electrons away from the other,
a spontaneous reaction occurs.
If not, the reaction is not
spontaneous, and it will not
occur on its own.
Experiments can be run …
… and results can be summarized in a table like the one below
Standard Reduction Potentials Table
high likelihood of being reduced
Standard reduction
potential is the tendency
for a chemical species
to be reduced.
S.R.P. is measured in
volts at standard
conditions.
The more positive the
potential is, the more
likely it will be reduced.
The table to the right
lists the S.R.P.s of
common oxidizing
agents and reducing
agents (in order from
strongest to weakest).
The Spontaneity Rule
spontaneous
reaction
no spontaneous
reaction
spontaneous
reaction
no
spontaneous
reaction
*note – the oxidation half reaction will appear flipped on the reduction table
•
•
When an oxidizing agent (OA) is above the reducing agent (RA) on the
redox table, a spontaneous reaction occurs.
But when an oxidizing agent is below the reducing agent, no spontaneous
reaction occurs.
Ex)
Use the redox table to predict whether the following reactions
will be spontaneous: *again – the oxidation half reaction will appear
flipped on the reduction table
a) Cr (s) + NiSO4(aq) ?
RA
Cr (s) + Ni2+(aq) ?
Cr (s) Cr3+(aq) + 3e-
LEO, RA
Ni2+(aq) + 2e- Ni (s)
GER, OA
OA is above the RA
spontaneous reaction
b) Ag(s) + Pb(NO3)2(aq) ?
OA
Ag (s) + Pb2+(aq) ?
OA
Ag (s) Ag+(aq) + e-
LEO, RA
Pb2+(aq) + 2e- Pb (s)
GER, OA
OA is below the RA
no spontaneous reaction
RA
Ex) What will happen if an aluminum spoon is used to stir a
solution of iron (II) nitrate?
Balanced Chemical Equation:
2Al(s) + 3Fe(NO3)2(aq) → 2Al(NO3)3(aq) + 3Fe(s)
Net Ionic Equation:
2Al(s) + 3Fe2+(aq) → 2Al3+(aq) + 3Fe(s)
Half Reactions:
Al(s) → Al3+ + 3e- LEO, RA
OA
Fe2+ + 2e- → Fe(s) GER, OA
Since the oxidizing agent is
above the reducing agent, the
reaction will occur spontaneously
& the aluminum spoon will
dissolve in the solution
RA
Self Check
How prepared am I to start my homework? Can I …
… predict the spontaneity of redox reactions based on
position in standard reduction potential table?
HOMEWORK
P623 #1, 3, 4, 6
p628 #7-9, 28-31, 59
9.1 Electron Transfer Reactions
Learning Goals …
… identify the reactant being oxidized and reduced in
a redox reaction
… assign oxidation numbers to determine if a
chemical reaction is a redox reaction
… identify the reducing and oxidizing agents in a redox
reaction
OXIDATION
• the process by which one or more electrons is lost by a chemical
entity
REDUCTION
• the process by which one or more electrons is gained by a
chemical entity
OXIDATION-REDUCTION (REDOX) REACTION
• the reaction in which one or more electrons are transferred
between chemical entities
e.g. When copper wire placed in a solution of
silver nitrate. The copper atoms in the
wire are displaced by silver ions in
solution. The blue tint of the solution is
due to copper ions, and the fuzzy coating
on the wire is the silver metal.
Cu(s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq)
Total ionic equation
Cu(s) + 2 Ag+(aq) + 2 NO3- (aq) → 2 Ag(s) + Cu2+(aq) + 2 NO3- (aq)
Net ionic equation
Cu(s) + 2 Ag+(aq) → 2 Ag(s) + Cu2+(aq)
Half-Reactions
• redox reactions and can be broken into two half-reactions:
Oxidation
Cu(s) → Cu2+(aq) + 2e-
Reduction
Ag+(aq) + e- → Ag(s)
e.g. Write the half-reaction equations for the redox reaction
that occurs when aluminum is placed in a silver nitrate
solution.
Balanced chemical equation:
Al(s) + 3 AgNO3(aq) → 3 Ag(s) + Al(NO3)3 (aq)
Ionic Equation:
Al(s) + 3Ag + (aq) + 3NO3- (aq) → 3Ag(s) + Al3+(aq) + 3NO3- (aq)
Net Ionic Equation:
Al(s) + 3Ag+ (aq) → 3Ag(s) + Al3+(aq)
Oxidation half reaction:
Al(s) → Al3+ (aq) + 3eReduction half reaction:
Ag + (aq) + e- → Ag (s)
OXIDATION NUMBERS
• a number used to keep track of electrons in redox reactions
• also known as oxidation state
Definition
• the apparent net electric charge that an atom would have
if the electron pairs in covalent bonds belonged entirely to
the more electronegative atom
-2 (2 extra e-)
+3
-3
+1 (1 less e-)
-2
Rules for Assigning Oxidation Numbers
Rule
Examples
1. A pure element has an oxidation
Na
O2
Fe
number of zero
(0)
( 0)
( 0)
2. The oxidation # of an ion equals the
Al3+
O2charge of the ion
(+3)
(-2 )
3. The oxidation # of an ion in an ionic
compound is its ionic charge
4. The oxidation # of hydrogen is +1,
except in metal hydrides, H is –1.
Na2S
(+1)(-2 )
MgO
(+2)( -2)
H2S
H2O
MgH2
(+1)( -2) (+1)( -2) (+2)(-1 )
5. The oxidation # of oxygen is –2,
unless it is a peroxide. (eg. H2O2)
6. The sum of all oxidation #s of all
elements in a compound is zero
Li2O
CO2
H2O2
(+1)( -2) (+4)( -2) (+1)( -1)
+4 -4 +2 -2
Na2SO4
KNO3
(+1)(+6)(-2 ) (+1)(+5)( -2)
+2 +6 -8 +1 +5 -6
HClO4
(+1)( +7)(-2 )
+1 +7 -8
PCl3
CS2
N2O5
(+3)( -1) (+4)( -2) (+5)(-2)
+3 -3 +4 -4 +10 -10
7. In covalent compounds, the
more electronegative element is
assigned an oxidation # that equals
the negative charge it usually has in
its ionic compounds
8. The sum of all oxidation #s of all the
NO2- SO42Cr2O72elements in a polyatomic ion equals (+3)( -2) (+6)(-2) (+6)( -2)
+3 -4 +6 -8 +12 -14
the charge of the ion
Applying Oxidation Numbers to Redox Reactions
Oxidation
• increase in oxidation number
Reduction
• decrease in oxidation number
* if there is no change in oxidation number then a redox
reaction does not take place *
e.g.
Determine whether the following reactions are redox
reactions. If so, identify what is being oxidized and
reduced.
a) CH4 + Cl2 → CH3Cl + HCl
(-4)(+1)
(0)
(-2)(+1)(-1) (+1)(-1)
Oxidation (↑ in Ox #): Carbon: goes from -4 to -2 (↑2)
Reduction (↓ in Ox #): Chlorine: goes from 0 to -1 (↓1)
b) CaCO3 + 2HCl → CaCl2 + H2O + CO2
(+2)(+4)(-2) (+1)(-1) (+2)(-1) (+1)(-2) (+4)(-2)
*no ion undergoes a change in oxidation # therefore not a
redox reaction!
Oxidizing and Reducing Agents
OXIDIZING AGENT
• the reactant that is reduced
• gains electrons
REDUCING AGENT
• the reactant that is oxidized
• loses electrons
e.g. Identify the oxidizing and reducing agents in each of the
following.
a) 3 H2S + 2 NO3- + 2 H+ → 3 S + 2 NO + 4 H2O
+1 -2
+5 -2
+1
0
+2 -2
+1 -2
oxidation
reduction
reducing agent: H2S
oxidizing agent: NO3b) MnO2 + 4 HBr → MnBr2 + Br2 + 2 H2O
+4 -2
+1 -1
+2 -1
0
+1 -2
reduction
oxidizing agent: MnO2
oxidation
reducing agent: HBr
Self Check
How prepared am I to start my homework? Can I …
… identify the reactant being oxidized and reduced in a
redox reaction
… assign oxidation numbers to determine if a chemical reaction
is a redox reaction
… identify the reducing and oxidizing agents in a redox reaction
HOMEWORK
p601 #1a, 4 p604 #1-4 p606 #1b, 3
p607 #1, 5, 6ab, 7-10
9.2 Balancing Redox Reactions
Using the Half Reactions Method
Learning Goals …
… balance complex redox reactions using half
reactions in aqueous solutions
… balance redox reactions in acidic and basic
solutions
Balancing Redox Equations:
Half Reactions Method
Steps:
1. Write the oxidation and reduction half reactions
2. Balance elements other than O and H
3. Balance each half-reaction: add H20 to balance
oxygen; add H+ to balance the hydrogen (acidic
conditions); add H+ and OH- to balance the
hydrogen (in basic conditions)
4. Add electrons to balance charge
5. Balance the electrons (charge)
6. Add the equations together
7. Check your answer
Ex)
ClO4-
+
NO2
NO2 + H2O
→
Cl-
+ NO3-
→ NO3-
+ 2 H+
ClO4- + 8 H+ + 8 e- → Cl-
4
8NO2 + 8 H2O
→
(acidic)
+ e-
+ 4 H2O
8 NO3-
x8
x1
8
+ 16 H+ + 8 e-
ClO4- + 8 H+ + 8 e- → Cl- + 4 H2O
8NO2 + 4 H2O + ClO4- → 8 NO3- + Cl- + 8 H+
Ex)
ClO-
+
CrO2-
CrO2- + 2 H2O
+ 4 OH-
→
Cl2
→
CrO42- + 4 H+ + 3 e- x 2
+ 4 OH+ 4 H2O
2
Cl2
x3
+ 2 H2O
+ 4 OH-
2 ClO- + 4 H+ + 2 e- →
+ 4 OH+ 4 H2O
2
+ CrO42-
(basic)
2 CrO2- + 8 OH→ 2 CrO42- + 4 H2O + 6 e4
2
6 ClO + 6 H2O + 6 e- → 3 Cl2 + 12 OH2 CrO2- + 6 ClO- + 2 H2O→ 2 CrO42- + 3 Cl2 + 4 OH-
Self Check
How prepared am I to start my homework? Can I …
… balance complex redox reactions using half reactions in
aqueous solution?
… balance redox reactions in acidic and basic solutions
HOMEWORK
p616 #1,2
p617 #7-10
p630 #44, 45
