Electrochemistry Redox Chemistry

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Electrochemistry(Redox Chemistry)

• The study of chemical reactions involving
  electron transfer.

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Historical Definitions

• Reduction
• Originally referred to the reduction in mass as a
  metal is refined from its ore
• Eg.
• 2 Fe2O3 (s) 3 C (s) 4 Fe (s) 3 CO2 (g)
• The produced iron has a lower mass than the ore.
  It has been reduced.

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Historical Definitions

• Oxidation
• Originally referred to the reaction of a
  substance (often a metal) with oxygen
• Eg.
• 2 Mg (s) O2 (g) MgO (s)
• The magnesium has been oxidized.

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Modern Definitions

• Reduction - The gain or partial gain of electrons
  in a chemical reaction.
• Oxidation - The loss or partial loss of electrons
  in a chemical reaction.

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Redox Example 1

• Cu (s)2 AgNO3 (aq) Cu(NO3)2 (aq) 2Ag (s)
• Cu, 2 Ag, 2 NO3- Cu2,2 NO3-, 2Ag

Cu has lost 2e-. It has been oxidized.
Ag has gained one electron. It has been reduced.
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Redox Example 2

• 2 H2 (g) O2 (g) 2 H2O (g)

Equal sharing Of electrons Between atoms
Equal sharing Of electrons Between atoms
Oxygen has a higher electronegativity than
hydrogen (3.4 vs. 2.2). As a result, the shared
electrons are pulled more closely to the oxygen
than to the hydrogen. In effect, hydrogen has
partially lost some of the electron density it
originally had as H2. Oxygen has gained some
electron density. We can say that oxygen has been
reduced and hydrogen has been oxidized.
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Oxidizing and Reducing Agents

• Oxidizing agent (OA) - substance which causes
  oxidation
• Reducing agent (RA) - substance which causes
  reduction

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Oxidizing and Reducing Agents (contd)

• The OAs in the previous examples are Ag (aq) and
  O2 (g). Note that although AgNO3 (aq) is a
  reactant, NO3- (aq) does not appear in the net
  equation so
• Ag (aq), not AgNO3 (aq) is the OA.

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Oxidizing and Reducing Agents (contd)

• The RAs in the previous examples are Cu (s) and
  H2 (g).

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Notes about any redox reaction

• There is a reducing agent and an oxidizing agent.
• One substance is reduced and one substance is
  oxidized.
• The net reduction must equal the net oxidation.

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Redox Cats
LEO, the lion, says GER

• Loss of
• Electrons is
• Oxidation

• Gain of
• Electrons is
• Reduction

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Help, help - Im breaking up
In writing ionic and net ionic equations, use the
following rules

• Dissociate
• Soluble ionic compounds
• Strong acids

• Do not dissociate
• Insoluble ionic compounds
• Weak acids
• Molecular compounds

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• Overall Equation
• Au(NO3)3(aq) 3Ag(s)
• Au(s) 3AgNO3(aq)
• Ionic Equation
• Au3(aq) 3NO3- (aq) 3Ag(s)
• Au (s) 3 Ag(aq) 3NO3-(aq)
• Net Ionic Equation
• Au3(aq) 3Ag(s)
• Au (s) 3Ag(aq)

RA
OA
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Building Redox Tables

• One spontaneous redox reaction like the one in
  the previous slide allows an oxidizing agent (OA)
  and a reducing agent (RA) to be identified
• A series of similar reactions allows a ranking of
  several OAs and RAs to be made as the following
  example will show

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Building Redox Tables (contd)
X not tested Y reaction N no reaction
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Building Redox Tables (contd)

• The reactivity of the metal ions (the oxidizing
  agents) are
• Ag(aq) Cu2(aq) Pb2(aq) Zn2(aq)
• The reactivity of the solid metals are
• Zn(s) Pb(s) Cu (s) Ag(s)

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Building Redox Tables (contd)

• We can then build the following table to collate
  the data
• Ag(aq) 1e- Ag(s)
• Cu2(aq) 2e- Cu (s)
• Pb2(aq) 2e- Pb(s)
• Zn2(aq) 2e- Zn(s)

WRA
SOA
WOA
SRA
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Table of Standard Electrode Potentials (Redox
table)

• This table may be considered to be organized in a
  number of redundant manners

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Redox table contd

• Oxidizing agents
• Strongest on top left F2(g)
• Weakest on bottom left Li(aq)
• Reducing agents
• Strongest on bottom right Li(s)
• Weakest on top right F-(aq)

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Redox Table contd

• Reduction potential
• Highest on top left F2(g)
• Lowest on bottom left Li(aq)
• Oxidation potential
• Highest on bottom right Li(s)
• Lowest on top right F-(aq)

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Redox Table contd

• Attraction for electrons
• Strongest on top left F2(g)
• Weakest on bottom left Li(aq)

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Predicting Redox Reactions-1

• 1. Identify all entities present in the reaction
  mixture (dissociate soluble ionic compounds and
  strong acids leave insoluble ionic compounds,
  weak acids and molecular compounds in their
  molecular forms.

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Predicting Redox Reactions-2

• Label all possible oxidizing agents (OAs - left
  side of table) and all possible reducing agents
  (RAs - right side of table)
• Identify the strongest OA (SOA - highest on the
  left) and strongest RA (SRA - lowest on the right)

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Predicting Redox Reactions-3

• 4. Write 2 half reactions from the table and add
  them together to predict a net reaction
  (remembering to cancel out electrons the net
  oxidation must equal the net reduction).

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Predicting Redox Reactions-4

• Predict the spontaneity of the reaction
• Three redundant checks can be used

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Predicting Redox Reactions-5

• Check 1 Relative positions of the SOA and
  SRA
• SOA above SRA spontaneous
• SOA below SRA non-spont.
• This is the most common and easiest of the three
  checks

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Predicting Redox Reactions-6

• Check 2 Strengths of the forward and reverse
  OAs
• OA (forward)
• above spontaneous
• OA (reverse)
• OA (forward)
• below non-spontaneous
• OA (reverse)

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Predicting Redox Reactions-7

• Check 3 Cell Potential
• E positive spontaneous
• E negative non-spontaneous

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Eg. What is the expected reaction if a solution
of potassium permanganate is added to an acidic
solution of tin (II) chloride?
K(aq) MnO4-(aq) Sn2(aq) Cl-(aq)
H(aq) H2O(l) OA OA,SRA
RA OA OA,RA
SOA RA
2 x MnO4-(aq) 8 H(aq) 5e-
Mn2(aq) 4 H2O(l)
5 x Sn2(aq) Sn4(aq) 2e-
2 MnO4-(aq) 16 H(aq) 5 Sn2(aq)
2 Mn2(aq) 8 H2O(l) 5 Sn4(aq)
spont
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Oxidation Numbers

• An oxidation number may be assigned to each atom
  in any element, ion or compound
• Oxidation numbers are assigned by following rules
  that are based upon electronegativity values

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Oxidation Numbers Rules

• The oxidation number of uncombined elements is
  zero
• eg. O2, P4, Ca, S8
• The oxidation number of fluorine in any compound
  is -1
• eg. NaF, OF2

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Oxidation Number Rules

• The oxidation number of oxygen in compounds is -2
• eg. H2O, Na2O
• Except (a) peroxides, where it is -1
• eg. H2O2, Na2O2
• (b) Compounds, containing fluorine, in which
  rule 6 is followed

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Oxidation Number Rules

• 4. The oxidation number of hydrogen is 1
• eg. H2O, CH4
• except metal hydrides, in which the oxidation
  number is -1
• eg. NaH, BaH2

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Oxidation Number Rules

• The oxidation number of a simple monoatomic ion
  is the same as the charge on the ion.
• eg. Na, O2-, Al3

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Oxidation Number Rules

• The sum of the oxidation numbers of a molecule is
  zero
• Eg. C2H5OH 2 C 2x
• 6 H 6(1) 6
• 1 O 1(-2) -2
• 2x6-2 0
• so x-2

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Oxidation Number Rules

• The sum of the oxidation numbers of a polyatomic
  ion is the same as the charge of the ion.
• Eg. Cr2O72- 2 Cr 2x
• 7 O 7(-2) -14
• 2x -14 -2
• x 6

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Redox Reactions and Ox.s

• A redox reaction is a reaction in which there is
  a change in oxidation numbers between the
  reactants and products
• The element which is undergoes an increase in
  oxidation number has been oxidized
• The element which has decreased its oxidation
  number has been reduced

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Redox Reactions and Ox. s

• 7 -2 1
  2
• 2 MnO4-(aq) 16 H(aq) 5 Sn2(aq)
• 2 1 -2
  4
• Mn2(aq) 8 H2O(l) 5 Sn4(aq)
• Mn has been reduced from a 7 oxidation state to
  2
• Sn has been oxidized from a 2 oxidation state to
  4
• MnO4- / H is the oxidizing agent
• Sn2 is the reducing agent

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Balancing of Redox Equations

• This normally can not be done simply by
  inspection, as in the case of non-redox
  reactions. Instead, a set procedure is used.
• Two methods may be used
• Half-Reaction (Ion-Electron) Method
• Oxidation-Number Method

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Half-Reaction Method
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For Reactions in Acidic Solution

• Abundant H and H2O are present.
• In general
• First split into half-reactions.
• Balance principal atoms (non-O, non-H)
• Balance O
• Balance H
• Balance charge with electrons.
• Make e gain equal to e loss.
• Recombine and cancel as necessary.

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• Ex.
• H2C2O4(aq) MnO4(aq) ? Mn2(aq) CO2(g)
• Split into half-reactions.
• H2C2O4 ? CO2
• MnO4 ? Mn2
• Balance principal atoms.
• H2C2O4 ? 2 CO2
• Balance oxygen using H2O.
• MnO4 ? Mn2 4 H2O

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• Balance hydrogen using H.
• H2C2O4 ? 2 CO2 2 H
• 8 H MnO4 ? Mn2 4 H2O
• Balance charge using electrons.
• H2C2O4 ? 2 CO2 2 H 2 e
• 5 e 8 H MnO4 ? Mn2 4 H2O
• Make electron gain electron loss
• H2C2O4 ? 2 CO2 2 H 2 e x 5
• 5 e 8 H MnO4 ? Mn2 4 H2O x 2

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• Recombine and cancel as necessary.
• 5 H2C2O4 2 MnO4 16 H 10 e ?
• 10 CO2 2 Mn2 10 H 8 H2O 10 e
• The result
• 5 H2C2O4 (aq) 2 MnO4 (aq) 6 H (aq) ?
• 10 CO2 (g) 2 Mn2 (aq) 8
  H2O (l)
• When finished
• Make sure that the electrons have cancelled.
• Make sure that all atoms charges balance.

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B. Reactions in basic solution

• An abundance of OH and H2O is present.
• Ex. SnO22(aq) Bi(OH)3(s) ? Bi(s) SnO32(aq)
• Split into half-reactions.
• SnO22 ? SnO32
• Bi(OH)3 ? Bi
• 2) Balance principal elements.
• (Already done).

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• 3) Balance oxygen with 2 OH/H2O
• 2 OH SnO22 ? SnO32 H2O
• 3 H2O Bi(OH)3 ? Bi 6 OH
• Balance hydrogen with H2O/OH (no H!)
• 3 OH 3 H2O Bi(OH)3 ? Bi 6 OH 3 H2O
• Balance charge with electrons.
• 2 OH SnO22 ? SnO32 H2O 2 e
• 3 e Bi(OH)3 ? Bi 3 OH

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• Make electron gain electron loss
• 2 OH SnO22 ? SnO32 H2O 2 e
  x 3
• 3 e Bi(OH)3 ? Bi 3 OH x 2
• Recombine, then cancel.
• 6 OH 3 SnO22 6 e 2 Bi(OH)3 ?
• 3 SnO32 3 H2O 6 e 2 Bi 6
  OH
• The result
• 3 SnO22(aq) 2 Bi(OH)3 (s) ?
• 3 SnO32(aq) 2 Bi (s) 3
  H2O (l)

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Balancing Redox Reactions Using Oxidation Numbers
- 1

• The general steps are
• Assign oxidation numbers to each atom in the
  equation
• Determine which atoms have changed oxidation
  number (an increase is oxidation a decrease is
  reduction
• Determine the change in oxidation numbers for
  these elements

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Balancing Redox Reactions Using Oxidation Numbers
- 2

• Determine the oxidation per formula unit
• Find the least common multiple for the oxidation
  and reduction. These factors will become the
  coefficients on the left side of your equation.
  Once established, this ratio must be maintained
• Balance by inspection

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• Eg. to Balance the combustion reaction of
  acetylene
• C2H2 (g) O2 (g) CO2 (g) H2O (g)

-1 1 0 4-2
1-2
2
5
4
2
Ox 2(5)10
Red 2(-2)-4
x2 20
x 5 -20
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Redox Stoichiometry

• Redox stoichiometry is no different than any
  other stoichiometric problem
• A chemical equation must be established
• The molar ratios from the balanced equation are
  used to solve the given problem

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• Eg. A solution of potassium permanganate is used
  to analyze an acidified solution of tin (II)
  chloride.
• If 14.3 mL of 0.0032 mol/L potassium permangante
  solution is required to reach the endpoint in the
  titration of a 25.0 mL sample of tin (II)
  chloride, what is the concentration of tin ions
  in the unknown?

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• First, we need to identify what is going to react
  (predicting redox reactions)
• Then we need to write a balanced redox reaction
• Finally, we need to carry out the calculations
  necessary to answer the problem

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• The chemical species present in the reaction
  mixture are
• Sn2 Cl- H K MnO4- H2O
• RA,OA RA OA OA OA,RA
• SRA
• SOA

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• The two half-reactions from the table are
• MnO4-(aq) 8H(aq) 5e- Mn2 4H2O(l)
• Sn2(aq) Sn4(aq) 2e-

x2


x5
2MnO4-(aq) 16H(aq) 5Sn2(aq) Mn2
8H2O(l) 5Sn4(aq)
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• Now that we have a balanced reaction, it is a
  basic stoichiometry problem

Sn2
14.3 mL
x
x
x
4.58 x 10-3 mol/L