Study of the relationships
between electricity and
chemical reactions
Oxidation States

REVIEW: How do we determine the
oxidation state?
 What is the oxidation state of the boldfaced
element:
○ P2O5
○ NaH
○ Cr2O7-2
○ SnBr4
Redox Reactions

How do we determine if a reaction is a
redox reaction?

How would we define the following
vocabulary words:
 Oxidizing agent (oxidant): agent aquiring
electrons from another substance (reduced)
 Reducing agent (reductant): agent giving up
electrons (oxidized)
Identifying Oxidizing and
Reducing Agents

Cd (s) + NiO2 (s) + 2 H2O (l)  Cd(OH)2 (s) + Ni(OH)2 (s)

2 H2O (l) + Al (s) + MnO4-(aq)  Al(OH)4- (aq) + MnO2 (s)
Balancing Redox Reactions

How is balancing a redox reaction
different than balancing a normal
equation?

How do we split a redox reaction?

Why would using half reactions make
the process easier?
Balancing equations: acidic
solutions

In balancing reactions, we use “skeleton”
ionic equation.
 Predict why this is called the skeleton ionic
equation.

Step-by-step procedure:
 STEP 1: BALANCE EQUATION
○ Balance elements other than H and O
○ Balance O atoms by adding H2O as needed
○ Balance H atoms by adding H+ as needed
○ Finally balance charge by adding e- as needed
Balancing equations: acidic
solution cont’
 STEP 2: Multiply half-reactions by
integers as needed to make the number
of electrons lost in the oxidation halfreaction equal the number of electrons
gained in the reduction half-reaction
 STEP 3: Add half reactions (simplify if
possible by canceling species found on
both sides)
 STEP 4: Check to make sure atoms and
charges are balanced
ಅಭ್ಯಾಸ (Kannada)

Given the skeleton equation:
 MnO4- (aq) + C2O42- (aq)  Mn2+ (aq) + CO2 (g)
April 10th, 2013

DO NOW:
 Predict how balancing a redox reaction in
a basic solution is different than in an
acidic solution.
Basic Solutions

For Basic solutions:
 Follow same process except for after
balancing H+, add OH- to both reactants and
products side to “neutralize” the H+ ion.
행하다 (Korean)

See Worksheet
Voltaic Cells

Energy is released when using a voltaic
cell, how is it being used?

How can we define voltaic cell?

Draw a diagram of a voltaic cell in your
notebook.
Anodes and Cathodes!

How do we differentiate between an
anode and a cathode?

Predict how the voltaic cell conducts
electricity.

Why is using a salt bridge necessary?

Predict which direction the anions flow.
Describing a voltaic cell

Cr2O72- (aq) + 14 H+ (aq) + 6 I- (aq)  2 Cr3+ (aq) + 3 I2 (s) + 7 H2O (l)
Is a spontaneous reaction in a voltaic cell. A solution
containing K2Cr2O7 and a solution containing H2SO4 are
poured into 1 beaker and a solution containing KI is
poured into a separate beaker. 2 pieces of metal that
won’t react with either are suspended into both solutions.
Indicate the following:
○
○
○
○
○
Reaction at anode
Reaction at cathode
Direction of Electron flow
Direction of ion flow
Signs of electrodes
Electricity Flow

Describe the flow of water in a waterfall.
 Why?

Why do electrons flow spontaneously
from an anode to a cathode?

How do we measure energy between
the two?
Potential Energy

Potential Energy difference between the
two electrodes is called cell potential.
 Also known as: Electromotive force (EMF)
○ Pushes e- through circuit
 Assume 25°C  standard conditions
 Standard Cell Potential: E°cell
 Depends on particular anode and cathode
half cells

How might we measure E°cell
Reference Cell

Why is it necessary to have a reference
half reaction?
 Allows for measure of other cell potentials
directly.

Reference cell: SHE (standard hydrogen
electrode)
Calculating E°cell

How do we calculate E°cell from the
SHE?

IE: oxidation of zinc and reduction of
hydrogen ion

How would we represent this standard
reduction potential?
Reactions

How do we determine how likely a
reaction is to occur?

Given that electrons must flow
spontaneously, what assumption can we
make about the values of the E°red of
both the anode and the cathode?
April 11th, 2013

Do Now:
 Thermodyanmics Quiz.
○ Have out Calculator and Reference Sheet.
Spontaneity of Voltaic Cells

How can we use our equation for standard cell
potentials as a generalized equation?

Based on this equation and your knowledge of
thermodynamics, predict how we determine if
the reaction is spotaneous.

NOTE: E = EMF @ nonstandard conditions
E°= EMF @ standard conditions

Practice: Is this reaction spontaneous:
 Cu (s) + 2 H+ (aq)  Cu2+ (aq) + H2 (g)
Reactivity of Metals

How are the standard reduction
potentials related to the activity series of
metals?

Which is a stronger reducing agent:
silver or nickel? Why?
Gibbs Free Energy (it never
leaves you alone)

How can we related Gibbs free energy to
EMF?

Predict the sign of the Gibbs free energy if you
have a positive EMF. Why is this to be
expected?

Rewrite the equation for a reaction in
“standard conditions”

How can we relate E° to K?
April 12th, 2013
DO NOW:
 Using standard reduction potentials,
calculate standard free energy change and
k @ 298K for:
4 Ag (s) + O2 (g) + 4H+ (aq)  4 Ag+ (aq) + 2 H2O (l)

Suppose the reaction is halved, calculate the
change in gibbs free energy at standard
conditions, EMF at standard conditions, and k.
Non standard Conditions

As a cell is discharged (released all
possible electricity) reactants are
consumed and products are generated
the concentration changes until EMF
drops to 0 (dead cell)
We want to answer: How is EMF
generated under non-standard
conditions
Concentration is a changin!

Walther Nernst (1864-1941)
 Established many theoretical foundations of
electrochemistry

Effect of concentration on cell EMF can be
obtained from effect of concentration on
Gibbs Free Energy.

Which equation would we want to use to
relate effect of concentration on Gibbs Free
Energy. Why?
Practice… again.

Calculate the EMF @ 298K generated by the
voltaic cell in which the reaction is
Cr2O72- (aq) + 14H+ (aq) + 6 I- (aq)  2 Cr3+ (aq) + 3 I2 (s) + 7 H2O (l)
[Cr2O72-] = 2.0 M
[H+] = 1.0 M
[I-]= 1.0 M
[Cr3+ ] = 1.0 x 10-5 M
One More

If the potential of a Zn-H+ cell is at 0.45
V at 25°C when [Zn2+] = 1.0 M and PH2 =
1.0 atm, what is the H+ concentration?
Concentration cells

Voltaic cells constructed by the same
species in both half-cells with different
concentrations
Ni2+ - Ni reaction

Write the two half reactions and the overall
reaction.

How can we calculate the EMF of this cell?

Determine the EMF.

If E° is 0, what produces the driving force?

**NOTE: when concentrations become the
same: Q = 1, E = 0
pH meters

Generating a potential by a concentration
difference

Example: A voltaic cell with 2 hydrogen
electrodes: Electrode 1: PH2 = 1.0 atm and
unknown concentration of H+ ions.
Electrode 2 is a standard Hydrogen
electrode. At 298K, E° = 0.211 V, and the
electrical current is observed to flow from
electrode 2 to electrode 1. What is the pH
of the unknown solution?
Electrolysis (Electrolytic cells)

Non-spontaneous redox reactions driven
by outside electrical energy
 IE: decomposition of molten NaCl to Na
and Cl2

2 electrodes immersed in either molten
salt of solution
 Process still similar to voltaic cells
 Battery acts as electron pump
Electroplating

How is electroplating
different than general
electrolysis?
 Uses electrolysis to
deposit a thin layer of
metal on another metal
to improve beauty or
resistance to corrosion
 Uses active electrodes
instead of inert
Quantitative Electrolysis

How many moles of electrons are needed to
create the following from their ions?
 Sodium
 Copper
 Aluminum


Amount of substance reduced or oxidized in
electrolytic cell is proportional to number of
electrons passed through cell
Quantity of charge is measured in Coulombs
 Charge on 1 mol = 96485 C.
○ Coulomb = quantity of charge passing a point in a
circuit in 1 s when the current is 1 A (ampere)
 Coulombs = amperes x seconds
Practissimo!

Calculate the number of grams of
aluminum produced in 1.00 hour by the
electrolysis of molten AlCl3 if the
electrical current is 10.0 A.