Energy: Nuclear Reaction and Redox Reaction

Marcelino Dela Rama Tradio Jr., RCh, LPT, PhD-Chem

Assistant Professor
Chemistry Analytical Environmental Section
Department of Chemistry
School of Arts and Sciences
University of San Carlos
Nasipit Talamban Cebu City
6000, Philippines
Nuclear Reactions as Source of Energy

• Nuclear reactions involve the conversion of one chemical element into

another and are typically accompanied by the emission of radiation in the
form of high energy particles and photons.
• Some nuclear reactions occur naturally, whereas others can be induced by
human intervention.
• The four main nuclear reaction types which can be used to harness energy
1. Fission
2. Fusion
3. Transmutation
4. Nuclear Decay/radioactive decay

n.b. In nuclear reactions, the mass is converted into energy.

Nuclear Reactions as Source of Energy

• Nuclear Particles involve in nuclear reactions

Balancing Nuclear Reactions

• Nuclear reactions are balanced in two ways

1. The sum of the mass numbers of the reactants equals the sum of the mass numbers of the
products.
2. The sum of the charges of the reactants equals the sum of the charges of the products.

n.b. balancing nuclear reactions can be used to identify a particular nuclide in a nuclear reaction

Example 1: The reaction of an alpha particle with magnesium-25 (2512Mg) produces a proton and a
nuclide of another element. Identify the new nuclide produced.
25 Mg + 4 ɑ ⟶ 1 H + A X ; A=? , Z=? , X =?
12 2 1 z
25 Mg + 4 ɑ ⟶ 1 H + 28 X
12 2 1 13
25 Mg + 4 ɑ ⟶ 1 H + 28 Al
12 2 1 13
Balancing Nuclear Reactions

Example 2:
210 Po ⟶ 20682Pb + AzX ; A=? , Z=? , X =?
84
84 Po ⟶
210 206 Pb + 42X
82
210 Po ⟶ 206 Pb + 42He
84 82

Example 3:
7N + 42ɑ ⟶ 11H + AzX ; A=? , Z=? , X =?
14

14 N + 42ɑ ⟶ 11H + 178X

7
14 N + 42ɑ ⟶ 11H + 178O
7
Balancing Nuclear Reactions

Example 4:
9 Be + A X ⟶ 12 C + 1 n ; A=? , Z=? , X =?
4 z 6 0
9 Be + 4 X ⟶ 12 C + 1 n
4 2 6 0
9 Be + 4 He ⟶ 12 C + 1 n
4 2 6 0

Example 5: Nb-89 ejects a positron

89 Nb ⟶ A X + 0 e ; A=? , Z=? , X =?
41 z +1
89 Nb ⟶ 89 X + 0 e
41 40 +1
89 Nb ⟶ 89 Zr + 0 e
41 40 +1
Balancing Nuclear Reactions

Example 6: Tc-94 undergoes electron capture

94 Tc + 0 e ⟶ A X ; A=? , Z=? , X =?
43 -1 z
94 Tc + 0 e ⟶ 94 X
43 -1 42
94 Tc + 0 e ⟶ 94 Mo
43 -1 42
Energetics of Nuclear Reactions

• Binding Energy of a nucleus: the energy that would be released if the nucleus were formed from
a collection of free nucleons.

Example: Calculate the binding energy during the formation of He-4 atom that has an experimental
mass of 4.002603 amu.

total mass of nucleons = total mass of protons + total mass of neutrons

total mass of nucleons = 2(1.007825 amu) + 2(1.008665 amu) = 4.032980 amu
∆m = total mass of nucleons – experimental mass
∆m = 4.032980 amu – 4.002603 amu = 0.030377 amu
E = ∆mc2
E = 0.030377 amu (1.66054 x 10-27 kg/amu) (2.99792 x 108 m/s)2
E = 4.5335 x 10-12 kg m2/s2 per atom
E = 4.5335 x 10-12 J per atom
Fission Reaction

• Splitting of heavy nucleus into two or more smaller nuclei to attain greater stability
• Not all nucleus are fissionable or fissile.
• Fission reaction happens spontaneously (large nucleus simply breaks into smaller nucleus); or
through neutron bombardment.
example: 23592U + 10n ⟶ 23692U* ⟶ 14156Ba + 9236Kr + 3 10n

Sample problem:
Calculate the energy released by a nucleus of uranium-235 if it splits into a barium-141 nucleus and a
krypton-92 nucleus according to the equation above.
Fission Reaction
Sample problem 1:
Calculate the energy released by a nucleus of uranium-235 if it splits into a barium-141 nucleus and a
krypton-92 nucleus according to the equation above.

Mass of reactants = 235.0439231 amu + 1.0086649 amu = 236.0525880 amu

Mass of products = 141.9144064 amu + 91.9261528 amu + 3(1.0086649 amu) = 235.8665539 amu
Mass defect, ∆m = mass of products – mass of reactants
Mass defect, ∆m = 235.8665539 – 236.0525880 = − 0.1860341
E = ∆mc2
E = − 0.1860341 amu (1.66053886 x 10-27 kg / amu) (2.99792458 x 108 m/s)2
E = —2.776406 x 10-11 kg m2/s2
E = —2.776406 x 10-11 J

n.b. 1 J = 1 kg m2/s2
Fission Reaction

Nuclear Power Plant

Fusion Reaction

• Small nuclei combine to form larger, and more stable nuclei.

• Though fusion reaction is relatively safe source of energy; however, the use of
fusion to generate electricity is complicated due to the repulsion of positively
charged nuclei that requires enormous energy to force the nuclei close enough
to overcome the Coulombic forces (this require temperature at the order of 106
K)
• The energy of the sun originate in a fusion reaction.
4 11H ⟶ 42He + 2 01β + 2𝑣 + energy
Transmutation

• One nucleus of a particular element changes to another, either by natural decay

or in response to some outside intervention, such as neutron bombardment.

10 1 11 7 4
5B + 0n ⟶ 5B* ⟶ 3Li + 2He

• Reverse of nuclear decay.

• Transmutation reactions are used to produce a number of medically useful
radioisotopes.
Nuclear Decay or Radioactive Decay

• Process by which an unstable isotope of a particular element spontaneously transforms

into a new element by emission of ionizing radiation.
• Isotopes: atoms of a given element with the same atomic number but different atomic
mass; hence, they have different number of neutrons.
• Atomic mass of a given element: is equal to the weighted average atomic masses of all
the naturally-occurring isotopes
• It is classified depending upon the type of emitted or absorbed ionizing radiation as:
1. Alpha Decay
2. Beta Decay
3. Gamma Decay
4. Electron Capture
5. Positron Emission
Alpha Decay

• A nuclear decay of unstable isotopes which eject alpha particle that is

accompanied with the decrease of mass number of 4 and decrease of atomic
number of 2.
238 U ⟶ 234 Th + 4 He
92 90 2
Beta Decay

• A nuclear decay of unstable isotopes which emit beta particle, β- or 0-1β that is
accompanied with the ejection of electron from the nucleus and emission of
antineutrino, ν̄, that has no charge and no mass.
14 C ⟶ 14 N + 0 β + ν̄
6 7 -1
Gamma Decay

• A nuclear decay that emits a high-energy photon that is accompanied with other
types of decay
• In gamma decay, the nucleus will be in the excited states, and to return to its
ground state, it emits a photon (a subatomic light particle that has discrete
bundle of electromagnetic energy)
• Since, the nuclear energy level is very large, this photon will take the form of very
high energy gamma radiation (wavelength is 10-12 m and frequency is 3 x 1020/s)
• Gamma decay will neither change the mass number nor the atomic number.
14 C ⟶ 14 N + 0 β + ν̄ + 0 γ
6 7 -1 0
Electron Capture

• Nucleus captures an electron from the first (n=1) shell in the atom.
• Proton in the nucleus is converted to neutron.
1 p+0 e⟶1 n+𝝼
1 -1 0

• Reverse of beta emission

• Nuclear charge decreases by one

26 Al + 0-1e ⟶ 2612Mg + 𝝼
13
Positron Emission Decay

• Positron is a positively charged electron, β+ or 01β

• Collision of positron and electron results in the conversion of their combined
masses into energy (511 keV gamma-ray photons
• Positron decay result to the decrease of one nuclear charge.
15 O ⟶ 15 N + 0 β + 𝝼
8 7 1
Kinetics of Nuclear or Radioactive Decay
Recall: In a chemical reaction, the rate of reaction is expressed as the rate at
which a reactant is consumed or the rate at which the product is formed.
• In radioactive decay, the rate is measured directly based on the count of
decay in a given time period.
• Radioactive Activity (Decay Rate): the rate at which a sample decays and
it is equal to the slope of radioactive decay equation.
Ø A = # of decays/sec ; N(t) = Noe-kt
Ø A = d N(t)
dt
d N(t) = d (Noe-kt )
dt dt
dN(t) = N0 e-kt
dt
dN(t) = -kN(t)
dt
Kinetics of Nuclear or Radioactive Decay

• The SI unit of nuclear activity is Becquerel (Bq), defined as one nuclear

disintegration per second) or 1 decay/sec
• The older unit of nuclear activity is Curie (Ci), which is 3.7 x 1010 decay/sec or 3.7
x 1010 Bq
Kinetics of Nuclear or Radioactive Decay

• For a sample of N nuclei, the rate of disintegration is given by ∆N/∆t

• Since most activity is proportional to the number of nuclei present which
decreases exponentially with time, the N also decreases exponentially.
• Equation can be re-written to monitor the number a sample decayed:
ln N = ln N0 − kt k= decay constant or rate constant
ln N0 − ln N = kt
ln (N0 /N) = kt
N = N0 e-kt

• Half-life, t1/2: the time it takes for N0 to decay to one-half its original value
t1/2 = ln2/k = 0.693/k
Kinetics of Nuclear or Radioactive Decay
Sample Problem 2:
The half-life of strontium-90, 9038Sr is 28.8 years. Find the activity (decay rate) of 1.00 g of the material.

k= 0.693/t1/2
k= 0.693/28.8 year
k= 0.0240625/year (1year/31,536,000 sec)
k= 7.630 x 10-10/sec

We need to find the number of nuclei in 1.00 g of the material:

N = 1.00 g (1 mol/89.91 g) (6.022 x 1023 atom/mol)
N = 6.70 x 1021 nuclei

A = -kN
A = 7.630 x 10-10/sec (6.70 x 1021 nuclei)
A = 5.11 x 1010 decays/sec
A= 5.11 x 1012 Bq
Kinetics of Nuclear or Radioactive Decay

Sample Problem 3:
A sample of radon has an activity of 60,000 Bq. If the half-life of radon is 15 h, how long
before the sample’s activity is 3,750 Bq?
dN(t) = N0 e-kt
dt
3,750 Bq = 60,000 Bq e-kt
3,750 Bq = 60,000 Bq e-(0.693/15h)t
3,750 Bq = e-(0.693/15h)t
60,000 Bq
ln[0.0625= e-(0.693/15h)t]
-2.77259 = - (0.693/15h) t
t = 2.77259 (15h/0.693)
t= 60.0 h
Kinetics of Nuclear or Radioactive Decay

Sample Problem 4:
Cobalt-60 is used extensively in medicine as a source of gamma rays. Its half-life is 5.27
years. How long will it take a Co-60 source to decrease to 45% of its original activity?
dN(t) = N0 e-kt
dt
0.45 = 1 e-kt
0.45= 1 e-(0.693/5.27)t
ln[0.45 = e-(0.693/5.27)t]
-0.79851 = - (0.693/5.27 year) t
t= 0.79851 (5.27 year/0.693)
t= 6.07 year
Kinetics of Nuclear or Radioactive Decay

Sample Problem 5:
Cobalt-60 is used extensively in medicine as a source of gamma rays. Its half-life is
5.27 years. What % of its activity remains after 29 months or 2.417 year?
dN(t) = N0 e-kt
dt
dN(t) = 1 e-kt
dt

dN(t) = e-(0.693/5.27year)(2.417year) = 0.7277 or 72.8 %

dt
Oxidation—Reduction Reaction as Source of Energy

Recall:

• Oxidation Reaction: involves the loss of electrons and the chemical specie that
undergo oxidation are known as reducing agent.
Na ⟶ Na+1 + 1e-
• Reduction Reaction: involves the gain of electrons, and the chemical specie that
undergo reduction are known as oxidizing agent.
F + 1e- ⟶ F-1

N.B. If the Oxidation—Reduction Reaction is spontaneous, energy is released.

Oxidation—Reduction Reaction as Source of Energy

Rules for assigning oxidation numbers to atom

1. Atoms in their elemental state are assigned an oxidation number of zero (0).
2. Atoms in monoatomic (i.e., one atom) ions are assigned an oxidation number equal to
their charge. Oxidation numbers are usually written with the sign first, then the
magnitude, which differentiates them from charges.
3. In compounds, fluorine is assigned a -1 oxidation number; oxygen is usually assigned a
-2 oxidation number (except in peroxide compound, wherein its oxidation number is -
1) and in binary compounds with fluorine, wherein it has +1); and hydrogen is usually
assigned a +1 oxidation number (except when it exists as the hydride ion, H-, in which
case rule 2 prevails.
HF; H has +1 and F has -1
H2O; O has -2
4. In compounds, all other atoms are assigned an oxidation number so that the sum of
the oxidation numbers on all the atoms in the species equals the charge on the species
(where is zero if the species is neutral)
Oxidation—Reduction Reaction as Source of Energy

Calculating oxidation of an element in a compound (especially those elements

that has more than one oxidation number)

Problem 6: What is the oxidation number of manganese in KMnO4?

0 = 1 (K) + 1 (Mn) + 4(O)
0 = 1 (+1) + Mn + 4(-2)
0 = +1 + Mn - 8
0 = -7 + Mn
Mn = + 7
Oxidation—Reduction Reaction as Source of Energy

Calculating oxidation of an element in a compound (especially those elements

that has more than one oxidation number)

Problem 7: What is the oxidation number of Chromium in Cr2O7-2?

-2 = 2 Cr + 7(O)
-2 = 2 Cr + 7 (-2)
-2 = 2 Cr - 14
2Cr = -2+14
2Cr = +12
Cr = +6
Electrochemical Cell

Electrochemical Cell: a device that can generate electrical energy from the
chemical reactions occurring in it.
ü The chemical reactions are separated into two half reactions: the oxidation half cell
reaction, and reduction half cell reaction.
ü A wire or conducting solution is attached to both cells in order to drive the electrons
from one side to the other.
ü Electrochemical cell can be galvanic (also known as voltaic) or electrolytic.
Galvanic or Voltaic Cell

Galvanic or Voltaic Cell: an electrochemical cell that uses spontaneous redox reactions to
generate electric current (or electricity).

Terminologies in Galvanic cells:

Electrodes: electrically conducting sites at which either oxidation or reduction occur.
Anode: electrode where oxidation occurs
Cathode: electrode where reduction occurs
Cell Notation: shorthand notation representing specific electrochemical reactions that
contains the list of metals and ions involved in the reaction.
Anode l electrolyte of anode ll electrolyte of cathode l cathode
Cu(s) l Cu2+ (aq)(1 M) ll Ag+(aq)(1 M) ll Ag(s)
Salt Bridge: complete the circuit which contains strong electrolyte that allows either
cations or anions to migrate into the solution where they are needed to maintain charge
neutrality.
Atomic Perspective of Galvanic or Voltaic Cell
ü At the anode, oxidation
occurs, and cations dissolve
into solution, leaving
behind negative charge on
the anode.
ü At the cathode, reduction
occurs, brings positively
charge ions to the cathode.
ü The build-up of charge on
the anode and cathode
generate cell potential or
electromotive force (EMF)
ü EMF is related to the
maximum electrical work
that can be obtained from
an electrochemical cell.
wmax = qE
q= charge;
E=cell potential
Cell Potential of Galvanic or Voltaic Cell (under Standard Condition)

ü Cell Potentials in an Electrochemical Cell are

measured based on standard electrode (i.e.,
hydrogen standard electrode, SHE)
ü Cell Potentials are expressed as Standard
Reduction Potentials
ü Standard Reduction Potentials can be either
positive or negative potentials
• Positive Voltage would mean that reduction occurs
o Large, positive standard reduction potential
implies that the substance is reduced readily
and a good oxidizing agent
• Negative Voltage would mean that oxidation occurs.
o Large, negative standard reduction potential
implies that the substance is oxidize readily an a
good reducing agent.
Cell Potential of Galvanic or Voltaic Cell (under Standard Condition)

ü In Galvanic or Voltaic Cell, the half-reaction with

the more positive reduction potential will be the
cathode.
ü Hence, the standard cell potential, Eocell , in Galvanic
cell for any pair of half-reactions can be determined
using the equation below.
Eocell = Eored − Eoox

ü Eocell in Galvanic Cell must be positive value

Eored – standard reduction potential of the cathode

Eoox − standard reduction potential of the anode
Cell Potential of Galvanic or Voltaic Cell (under Standard Condition)

Sample Problem 6:
Copper and iron (generally in the form of steel) are two of the
many metals used in designing machine. (a) Using standard
reduction potentials, identify the anode and the cathode. (b)
determine the cell potential for a galvanic cell, Eocell .

What if Cu is the anode and Fe is the cathode

Eocell = Eored − Eoox
Eocell = −0.44 − (0.337 V)
Eocell = − 0.78 V
Eocell = Eored − Eoox
Eocell = 0.337 V − (−0.44 V)
Eocell = 0.78 V
Fe(s) + Cu2+ (aq) ⟶ Fe2+(aq) + Cu(s)
Nernst Equation (nonstandard condition)

• The equation that describe cell potentials under nonstandard conditions.

o F – Faraday constant, 96, 485 J V-1 mol-1 or 96, 485 C/mol

o n – number of electron transferred in the redox reaction
o T –temperature in Kelvin
o R – gas constant, 8.314 J mol-1K-1
o Eo – standard cell potential
Nernst Equation

Sample Problem 7:
Assume that you have a cell that has an iron(II) concentration of 0.015 M and an H+ concentration
of 0.0010 M. The cell temperature is 38∘C, and the concentration of hydrogen gas is 0.04 M. What
would be the cell potential under these nonstandard condition?
Battery

Battery
• a cell or series of cells that generates an electric current.
• Composed of many different materials that harness electrical work of a galvanic cell.

Types of Batteries
1. Primary cells or primary batteries
2. Secondary cells or secondary batteries
Battery

Primary Cells or Primary Batteries

• Single-use batteries that cannot be recharged
• Example: alkaline battery or dry cell
Battery

Primary Cells or Primary Batteries

• Single-use batteries that cannot be recharged
• Example: mercury battery – long lasting battery, anode is zinc, while cathode is mercury(II) oxide
Battery

Secondary Cells or Secondary Batteries

• Single-use batteries that can be recharged (rechargeable batteries)
• Example: Nickel-Cadmium Batteries (anode: cadmium; cathode: nickel)
Battery

Secondary Cells or Secondary Batteries

• Single-use batteries that can be recharged (rechargeable batteries)
• Example: Nickel-Metal Hydride Batteries (anode: metal alloy; cathode: nickel)
Battery

Secondary Cells or Secondary Batteries

• Single-use batteries that can be recharged (rechargeable batteries)
• Example: Lead-Acid Storage Batteries (anode: Lead; cathode: Lead oxide)
Fuel Cells

Fuel Cells
• A voltaic cell in which the reactants can be supplied continuously and the products of the
reaction are continuously removed.
• Like batteries, fuel cells use a chemical reaction to produce electrical energy.
• But unlike batteries, it can be refueled on an ongoing basis.
• Example: reaction of hydrogen and oxygen gases to produce water.
Ø Hydrogen gas flows into the anode compartment, and oxygen gas flows into the cathode compartment.
Ø Oxygen is reduced at the cathode, while hydrogen is oxidized at the anode
END