Redox Reaction and Electrochemical Cell (Reaksi Redoks dan Sel Elektrokimia)

REDOX REACTION &
ELECTROCHEMICAL CELL

What is Redox Reaction?
An oxidation-reduction (redox) reaction
is a type of chemical reaction that involves a
transfer of electrons between two species.
An oxidation-reduction reaction is any
chemical reaction in which the oxidation
number of a molecule, atom, or ion changes
by gaining or losing an electron.
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Oxidation:
•Gain of oxygen
•Loss of electrons
Reduction:
•Loss of oxygen
•Gain of electrons
Increase in
oxidation
number
Decrease in
oxidation
number
Difference between oxidation
and reduction
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H2 + F2  2HF
Oxidised –
gains oxygen
Must be a redox!
H2  2H+
F2  2F-
+2e-
+2e-
Oxidised – loss of e-
Reduced – gain of e-
H 2 F2
-Oxidized -Reduced
-Reducing agent -Oxidizing agent
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Oxidation Number
Oxidation number (O.N) is also known as oxidation
state. It is defined as the charge the atom would have if
electrons were not shared but were transferred completely
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Example
Determine the oxidation number (O.N.) of
element in these compounds:
- CaCO3
- SO4
2-
Identify the oxidizing agent and reducing
agent in each of the following:
2H2(g) + O2(g) 2H2O (g)
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+2 -2
• CaCO3
(+2)+ x + (-2) 3 = 0
x = +4
-2
• SO4
2-
x + (-2) 4= -2
x = +4
2H2+ O2 2H2O
- O2 was reduced (O.N. of O: 0 -> -2); O2 is the
oxidizing agent
- H2 was oxidized (O.N. of H: 0 -> +1); H2 is the
reducing agent
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Practice Problem
1. Determine the oxidation number (O.N.) of
element in these compounds:
a) KNO3
b) XeF4
c) NH4
+
2. Identify the oxidizing agent and reducing
agent in each of the following:
a) Cu + 4HNO3 Cu(NO3)2+ 2NO2 + 2H2O
b) CuO + CO Cu + CO2
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BALANCING REDOX EQUATIONS
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Copyright © Pearson Education, Inc.,
or its affiliates. All Rights Reserved.
Balancing Redox Equations
Two different methods for balancing
redox equations are the oxidation-
number-change method and the
half-reaction method.
• These two methods are based on the fact
that the total number of electrons gained
in reduction must equal the total number
of electrons lost in oxidation.
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The Oxidation Number
Change Method
To use this method, start with the skeleton
equation for the redox reaction.
Fe2O3(s) + CO(g) → Fe(s) + CO2(g) (unbalanced)
The example above is for the reaction
of iron(III) oxide with carbon
monoxide. This reaction is one that
takes place in a blast furnace during
the processing of iron ore into
metallic iron
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• Using Oxidation-Number Changes
Step 1: Identify which atoms are oxidized and
which are reduced.
• Iron is reduced.
• Carbon is oxidized.
Fe2O3(s) + CO(g) → Fe(s) + CO2(g)
+3 –2 +2 –2 0 +4 –2
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Step 2: Use one bracketing line to connect the
atoms that undergo oxidation and
another such line to connect those
that undergo reduction.
–3 (reduction)
Fe2O3(s) + CO(g) → Fe(s) + CO2(g)
+3 –2 +2 –2 0 +4 –2
+2 (oxidation)
• Write the oxidation-number change at
the midpoint of each line.
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Step 3: Make the total increase in oxidation
number equal to the total decrease in
oxidation number by using appropriate
coefficients.
2 × (–3) = –6
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
+3 –2 +2 –2 0 +4 –2
3 × (+2) = +6
• The oxidation-number increase should be
multiplied by 3 and the decrease by 2.
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Step 4: Finally, make sure the equation is
balanced for both atoms and charge.
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
• If necessary, finish balancing the
equation by inspection.
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Acidic: Basic:
add H2O to the side
needing oxygen
balance as if in acidic sol’n
then add H+ to balance the
hydrogen
add enough OH- to both
sides to cancel out each
H+ (making H2O) & then
cancel out water as
appropriate
If needed, reactions that take place in acidic or
basic solutions can be balanced as follows:
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Example: Balance the following equation,
assuming it takes place in acidic solution.
ClO4
- + I-  Cl- + I2
Step 1:
Oxidation
numbers
+7 -2 -1 -1 0
Step 2: Which was
oxidized? Iodine, -1 to 0 = +1
Which was reduced?
Chlorine, +7 to -1 = -8
+1
-8
8 4
Step 5: Balance acid soln with
water…
+ H2O+8 H+ 4
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The Half-Reaction Method
In the half-reaction method, you write and
balance the oxidation and reduction half-
reactions separately before combining them
into a balanced redox equation.
• The procedure is different, but the outcome is
the same as with the oxidation-number-change
method.
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The example is the oxidation of Fe2+ ions to
Fe3+ ions by dichromate (Cr2O7
2−) in acidic
solution. The dichromate ions are reduced to
Cr3+ ions.
Fe2+
(aq)+ Cr2O7
2-
(aq) →Fe3+
(aq)+ Cr3+
(aq)
(unbalanced)
Using Half-Reactions
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Fe2+
(aq)+ Cr2O7
2-
(aq) →Fe3+
(aq)+ Cr3+
(aq)
Step 1: Write the unbalanced equation in ionic
form.
Notice that the equation is far from balanced,
as there are no oxygen atoms on the right
side. This will be resolved by the balancing
method.
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Step 2: Write separate half-reactions for the
oxidation and reduction processes.
Oxidation half-reaction:
Reduction half-reaction:
+2 +3
+6 +3
• Determine the oxidation numbers first, if
necessary.
Fe2+(aq)→Fe3+(aq)
Cr2O7
2-(aq)→Cr3+(aq)
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Step 3: Balance the atoms in the half-reactions
other than the hydrogen and oxygen.
Cr2O7
2-(aq)→ 2Cr3+(aq)
In the oxidation half-reaction above, the iron
atoms are already balanced. The reduction
half-reaction needs to be balanced with the
chromium atoms.
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Step 4: Balance oxygen atoms by adding water
molecules to the appropriate side of
the equation.
•In acidic solutions, H2O and H+(aq) can
be used to balance oxygen and
hydrogen as needed.
•In basic solution, H2O and OH– are
used to balance these species.
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For the reduction half-reaction above, seven
H2O molecules will be added to the product
side.
Cr2O7
2-(aq)→ 2Cr3+(aq) + 7H2O (l)
Now the hydrogen atoms need to be balanced.
In an acidic medium, add hydrogen ions to
balance. In this example, fourteen H+ ions will
be added to the reactant side.
14 H+ + Cr2O7
2- (aq)→ 2Cr3+(aq) + 7H2O (l)
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Step 5: Balance the charges by adding
electrons to each half-reaction.
• For the oxidation half-reaction, the
electrons will need to be added to the
product side.
• For the reduction half-reaction, the
electrons will be added to the reactant
side.
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By adding one electron to the product side
of the oxidation half-reaction, there is a 2+
total charge on both sides.
Fe2+(aq)→Fe3+(aq) + 𝑒−
To balance the charge, six electrons need to
be added to the reactant side.
6𝑒− +14 H+ + Cr2O7
2-(aq)→ 2Cr3+(aq) + 7H2O (l)
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Now equalize the electrons by multiplying
everything in one or both equations by a
coefficient. In this example, the oxidation
half-reaction will be multiplied by six.
6Fe2+(aq)→6Fe3+(aq) + 6𝑒−
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Step 6: Add the balanced half-reactions to
show an overall equation.
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Step 7: Check the balancing
14 H+ + 6Fe2+(aq) + Cr2O7
2-(aq) →
6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)
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Exercise
Use the half-reaction method to balance the
following equations:
a. Cu + NO3
- Cu2+ + NO2 (Acidic)
b. MnO + PbO2 MnO4
- + Pb2+(Acidic)
c. Ag + Zn2+ Ag2O + Zn (Basic)
d. MnO4
- + C2O4
2- MnO2 + CO2 (Basic)
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Galvanic Cell or Voltaic Cell
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Electrochemical Cell
• A device is used to convert the chemical
energy produced in a redox reaction in to
Electric energy is called a electrochemical
cell or simply a chemical cell.
• Electrochemical cell are two type:
Galvanic Cell or
Voltaic Cell
Electrolytic Cell
Converts chemical energy
into electrical energy
Converts electrical energy
Into chemical energy
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Galvanic or Voltaic cell
 Spontaneous chemical reaction to generate electricity
- One reagent oxidized the other reduced
- two reagents cannot be in contact
 Electrons flow from reducing agent to oxidizing agent
- Flow through external circuit to go from one
reagent to the other
Net Reaction:
Reduction:
Oxidation:
AgCl(s) is reduced to Ag(s)
Ag deposited on electrode and Cl-
goes into solution
Electrons travel from Cd
electrode to Ag electrode
Cd(s) is oxidized to Cd2+
Cd2+ goes into solution
Galvanic Cells
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Electrodes
Cathode: electrode
where reduction takes
place
Anode: electrode
where oxidation takes
place
Galvanic Cells
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Galvanic Cells
Salt Bridge
 Connects & separates two half-cell reactions
 Prevents charge build-up and allows counter-ion
migration
Two half-cell reactions
Salt Bridge
 Contains electrolytes not
involved in redox reaction.
 K+ (and Cd2+) moves to cathode with
e- through salt bridge (counter
balances –charge build-up
 NO3
- moves to anode (counter
balances +charge build-up)
 Completes circuit
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Standard Cell Notation (line notation)
• Conventions:
Anode on Left
Single line : represent phase boundaries
Two line : represent liquid junction
Cu2+
V
NO3
-
Zn2+
NO3
-
CuZn
e-
e-
Anode anode solution cathode solution Cathode
Example: Zn Zn2+ Cu2+ Cu
Zn + Cu2+ Zn2+ + Cu
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Standard electrode potential
(E°)
Standard electrode potential (E°)
(standard reduction potentials) are
defined by measuring the potential
relative to a standard hydrogen electrode
using 1M solution at 25 °C and 1 atm.
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• So, Standard Hydrogen Electrode (SHE)
has a standard reduction potential of
zero (E ° red SHE = 0) .
2H +(1 M) + 2e- H2 (1 atm) E ° red = 0 V
• SHE can be used to determine the amount
of potential standard reduction (E ° red )
other electrodes. The arrangement of
these electrodes is known as the Volta
Series ( metal reactivity series ).
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Volta Series
Li - K - Ba - Sr - Ca - Na - Mg - Al - Mn - Zn - Cr
- Fe - Cd - Co - Ni - Sn - Pb - (H) - Cu - Ag - Hg
- Pt - Au
• The reactivity of metal elements decreases from left to
right.
• Reducing properties (reduction power) of the metal
decreases from left to right.
• The tendency of the metal to be oxidized decreases
from left to right.
• Oxidizing properties (metal oxidation power) are
increasing from left to right.
• The tendency of metal ions to be reduced increases
from left to right.
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Standard Cell Potential (𝐸° 𝑐𝑒𝑙𝑙)
Cell potential can be measured using a voltmeter
or can be calculated by the following formula:
𝐸° 𝑐𝑒𝑙𝑙 = 𝐸° 𝑐𝑎𝑡ℎ𝑜𝑑𝑒 − 𝐸° 𝑎𝑛𝑜𝑑𝑒
𝐸° 𝑐𝑒𝑙𝑙 = 𝐸° 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 − 𝐸° 𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛
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Example:
Calculate Eo for the following reaction:
Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)
Cu2+(aq)+2e-→Cu(s); E⁰red=+0.339V
Zn2+(aq)+2e-→Zn(s); E⁰red=-0.762 V
Cu2+(aq)+2e-→Cu(s); E⁰red= +0.339 V
Zn(s)→Zn2+(aq)+2e-; E⁰ox=+0.762 V
Cu2+(aq)+Zn(s)→Cu(s)+Zn2+(aq)
E⁰cell=+1.101V
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Exercise
Known standard reduction potential as follows:
Fe 3+ / Fe 2+ = + 0.77 V
Cu 2+ / Cu = + 0.34 V
Zn 2+ / Zn = -0.76 V
Mg 2+ / Mg = -2.37 V
For reaction
Mg + 2Fe 3+ → Mg 2+ + 2Fe 2+ has cell potential ..
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Exercise
From the following electrode potential data
standards:
Cu 2+ + 2e → Cu E ° = 0.3 V
Ag + + e → Ag E ° = 0.80 V
then the reaction:
Cu + 2Ag + → Cu 2+ + 2Ag
have cell potential ...
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Nernst Equation
Reduction Potential under Non-standard Conditions
 E determined using Nernst Equation
 Concentrations not-equal to 1M
aA + ne- bB Eo
For the given reaction:
The half cell reduction potential is given by:
Where:
n is the number of electrons involved in the reaction
of the anode and cathode
Qc is the result of the reaction, where
𝑄𝑐 =
[𝐶] 𝑐
[𝐷] 𝑑
[𝐴] 𝑎[𝐵] 𝑏
where A, B, C and D are chemical species; and a, b, c,
and d are coefficients in a balanced equation:
at 25oC
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Example
Pt|Fe2+(0.10 M),Fe3+(0.20 M)||Ag+(1.0 M)|Ag(s)
Applying the Nernst Equation for Determining Ecell.
What is the value of Ecell for the voltaic cell pictured below and
diagrammed as follows?
Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag (s)
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Example
Ecell = Ecell° - log Q
n
0.0592 V
Pt|Fe2+(0.10 M),Fe3+(0.20 M)||Ag+(1.0 M)|Ag(s)
Ecell = Ecell° - log
n
0.0592 V [Fe3+]
[Fe2+] [Ag+]
Ecell = 0.03 V – 0.0178 V = 0.0122 V
Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag (s)
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Batteries and Corrosion
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Batteries: Producing Electricity
Through Chemical Reactions
• Primary Cells (or batteries).
– Cell reaction is not reversible.
• Secondary Cells.
– Cell reaction can be reversed by passing
electricity through the cell (charging).
• Flow Batteries and Fuel Cells.
– Materials pass through the battery which
converts chemical energy to electric
energy.
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The Leclanché (Dry) Cell
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Dry Cell
Zn(s) → Zn2+(aq) + 2 e-Oxidation:
2 MnO2(s) + H2O(l) + 2 e- → Mn2O3(s) + 2 OH-Reduction:
NH4
+ + OH- → NH3(g) + H2O(l)Acid-base reaction:
NH3 + Zn2+(aq) + Cl- → [Zn(NH3)2]Cl2(s)Precipitation reaction:
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Alkaline Dry Cell
Zn2+(aq) + 2 OH- → Zn (OH)2(s)
Zn(s) → Zn2+(aq) + 2 e-
Oxidation reaction can be thought of in two steps:
2 MnO2(s) + H2O(l) + 2 e- → Mn2O3(s) + 2 OH-Reduction:
Zn (s) + 2 OH- → Zn (OH)2(s) + 2 e-
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Lead-Acid (Storage) Battery
• The most common secondary battery
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Lead-Acid Battery
PbO2(s) + 3 H+(aq) + HSO4
-(aq) + 2 e- → PbSO4(s) + 2 H2O(l)
Oxidation:
Reduction:
Pb (s) + HSO4
-(aq) → PbSO4(s) + H+(aq) + 2 e-
PbO2(s) + Pb(s) + 2 H+(aq) + HSO4
-(aq) → 2 PbSO4(s) + 2 H2O(l)
E°cell = E°PbO2/PbSO4
- E°PbSO4/Pb = 1.74 V – (-0.28 V) = 2.02 V
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The Silver-Zinc Cell: A Button
Battery
Zn(s),ZnO(s)|KOH(sat’d)|Ag2O(s),Ag(s)
Zn(s) + Ag2O(s) → ZnO(s) + 2 Ag(s) Ecell = 1.8 V
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The Nickel-Cadmium Cell
Cd(s) + 2 NiO(OH)(s) + 2 H2O(L) → 2 Ni(OH)2(s) + Cd(OH)2(s)
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Fuel Cells
O2(g) + 2 H2O(l) + 4 e- → 4 OH-(aq)
2{H2(g) + 2 OH-(aq) → 2 H2O(l) + 2 e-}
2H2(g) + O2(g) → 2 H2O(l)
E°cell = E°O2/OH- - E°H2O/H2
= 0.401 V – (-0.828 V) = 1.229 V
 = ΔG°/ ΔH° = 0.83
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Air Batteries
4 Al(s) + 3 O2(g) + 6 H2O(l) + 4 OH- → 4 [Al(OH)4](aq)
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Corrosion
Corrosion is a galvanic process by which
metals deteriorate through oxidation usually
but not always to their oxides.

Corrosion: Unwanted Voltaic Cells
O2(g) + 2 H2O(l) + 4 e- → 4 OH-(aq)
2 Fe(s) → 2 Fe2+(aq) + 4 e-
2 Fe(s) + O2(g) + 2 H2O(l) → 2 Fe2+(aq) + 4 OH-(aq)
Ecell = 0.841 V
EO2/OH- = 0.401 V
EFe/Fe2+ = -0.440 V
In neutral solution:
In acidic solution:
O2(g) + 4 H+(aq) + 4 e- → 4 H2O (aq) EO2/OH- = 1.229 V
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Corrosion prevention
Corrosion prevention is based on the
following two principles:
1. Prevent contact with oxygen and /
or water
2. Cathode protection (sacrifice
anode)
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Prevent contact with oxygen or
water
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Cathode protection
• The Use of a Sacrificial Electrode to Protect
Against Corrosion. Connecting a magnesium rod
to an underground steel pipeline protects the
pipeline from corrosion.
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ELECTROCHEMICAL
CELL
Electrolytic Cell
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ELECTROLYTIC CELL
Electrolysis:
the use of electric energy to drive a
nonspontaneous chemical reaction
The anode is where oxidation occurs.
Anions migrate to the anode and lose electrons.
The cathode is where reduction occurs.
Cations migrate to the cathode and gain electrons.
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Electrolysis
Using electrical energy to produce chemical
change.
Sn2+(aq) + 2 Cl-(aq) ---> Sn(s) + Cl2(g)
SnCl2
SnCl2(aq)
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Electrolytic Cell Usage
• Electrolytic cells are typically used to:
• - plate metals on other metals
• - obtain a pure metal from it’s compounds
• - recharge batteries

Differences between
Electrochemical Cells
• Electrolytic Cells are different from Voltaic Cells because …
• They need Electricity to force a redox reaction to occur
– There is an external power source
– They don’t produce electricity
• The polarities are reversed
• The Anode is positive
• The Cathode is negative
• Electrolytic cell reactions are also different because they
usually take place in one solution/one cell
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Reaction in Electrolysis
• At anode
1. If the anode is made of other metals (not Pt, C, or Au),
the anode will oxidize to the ion.
For example, if an anode is made of Ni, Ni will be oxidized
to Ni 2+
•Ni Ni2+ + 2e-
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2. Inert Anode (Pt, Au, and C)
The hydroxide ion (OH - ): 4OH-
(aq) 2H2O(l) + O2(g) + 4e-
The remaining acidic ion:
a. The remaining oxygenated acids, such as NO3 -, PO4
3 -
are not oxidized. Instead the water is oxidized.
2H2O(l) 4H+
(aq)+O2(g)+4e-
b. The remaining ions of non-oxygenated acids, such as
Cl - , Br - , I - are oxidized to Cl 2 , Br 2 , I 2 gas.
2X- X2 + 2e-
Reaction in electrolysis
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• At Cathode
1. Hydrogen ion (H+)
Hydrogen ions are reduced to hydrogen gas
molecules.
Reaction: 2H+
(aq) + 2e- H2(g)
REACTION IN ELECTROLYSIS
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Electrolysis of Molten Sodium
Cloride
Cathode:
Na+ + e- Na
Anode:
2 Cl-  Cl2 + 2e-
Cathode: 2Na+ + 2e  2Na(l)
Anode: 2 Cl  Cl2(g) + 2e
___________________________
overall 2 Na+ + 2 Cl  2Na(l) + Cl2(g)
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2. Metal ions (Li, K, Ba, Ca, Na, Mg, Al, Mn, etc.)
a. Metal ions in solution
Reaction: H2O(l) + 2e- H2(g) + 2OH-
(aq)
b. Metal ions in molten
Reaction: LX+ + xe- L
Reaction in electrolysis
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79
Electrolysis of Sulphate Acid
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EXERCISE
• Determine the reaction that occurs at the anode and at
the cathode on the following electrolysis.
• 1. Electrolysis of HCl solution with electrode Pt.
• 2. Electrolysis of NaBr solution with C. electrode
• 3. Electrolysis of CuSO4 solution with C. electrode
• 4. Electrolysis of KNO3 solution with electrode Pt.
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Electrolysis Calculations
• Faraday Laws I
“The mass of substances produced
in electrolysis is directly proportional
to the electrical charge supplied.”
W = Q cause Q = i t
W = i t
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• Faraday Laws II
“The mass of substances produced by the
same electrical charge is proportional to the
equivalent weight of these substances.”
W = e
e =
𝐴𝑟
𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟
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In the electrolysis cell, the amount of substance (mass)
deposited or which dissolves at the electrode is directly
proportional to the amount of current passing through the
electrolyte (Faraday's Law).
W =
e i t
F
W =
e i t
96.500
w = mass of substance (g)
e = equivalent mass
i = current strength (A)
t = time (s)
F = constant Faraday = 96,500 coulombs
1 F = 1 mole of electrons
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For 2 or more electrolytes electrolyzed with
the same amount of current
WA
WB
=
eA
eB
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Redox Reaction and Electrochemical Cell (Reaksi Redoks dan Sel Elektrokimia)

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