Electrochemistry
• One of the most important areas of Applied Thermodynamics
• Study of the relationship between chemical change and electrical work. This
relationship is investigated through the use of electrochemical cells
• The electrochemical cells incorporate a redox reaction to produce or to utilize
electrical energy
Electron-transfer reactions: Cu (s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag (s)
Redox Reactions
Cu (s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag (s)
Cu (s): from 0 to +2 oxidized reducing agent
Ag (s): from +1 to 0 reduced oxidizing agent
• Oxidation (electron loss) is always accompanied by reduction (electron gain)
• Oxidizing agent is reduced the reducing agent is oxidized
• The total number of e − gained by the oxidizing agent is always equal to the total
number lost by the reducing agent
Half-reaction method for balancing redox reactions
• It divides the overall redox reaction into oxidation and reduction half-rxns
• Each half-rxn is balanced for mass (atoms) and charge
This method is very useful because
a) it separates the oxidation and reduction steps, which reflect their actual physical
separation in electrochemical cells
b) it is readily applied to redox reactions that take place in acidic or basic solutions
(common in cells)
Note: Review Oxidation Numbers (see Lecture Notes in Electrochemistry)
Example Balance the reaction in acidic solution
CuS (s) + NO 3− (aq) → Cu2+ (aq) + SO 24− (aq) + NO (g)
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1. Write two half-rxns:
CuS (s) → Cu2+ (aq) + SO 24− (aq)
NO 3− (aq) → NO (g)
2. Balance elements other than H and O:
Balanced
3. Balance oxygen by inserting H2O
CuS (s) + 4H2O (l) → Cu2+ (aq) + SO 24− (aq)
NO 3− (aq) → NO (g) + 2H2O (l)
4. Balance hydrogen by inserting H+
CuS (s) + 4H2O (l) → Cu2+ (aq) + SO 24− (aq) + 8H+ (aq)
NO 3− (aq) + 4H+ (aq) → NO (g) + 2H2O (l)
5. Balance charge by inserting e −
CuS (s) + 4H2O (l) → Cu2+ (aq) + SO 24− (aq) + 8H+ (aq) + 8e − (oxidation)
NO 3− (aq) + 4H+ (aq) + 3e − → NO (g) + 2H2O (l) (reduction)
6. Make the number of e − in the two half-rxns the same
3CuS (s) + 12H2O (l) → 3Cu2+ (aq) + 3SO 24− (aq) + 24H+ (aq) + 24e −
8NO 3− (aq) + 32H+ (aq) + 24e − → 8NO (g) + 16H2O (l)
7. Addition of the two half-rxns
3CuS (s) + 12H2O (l) + 8NO 3− (aq) + 32H+ (aq) + 24e − → 3Cu2+ (aq) + 3SO 24− (aq) +
24H+ (aq) + 24e − + 8NO (g) + 16H2O (l)
3CuS (s) + 8NO 3− (aq) + 8H+ (aq) → 3Cu2+ (aq) + 8NO (g) + 3SO 24− (aq) + 4H2O (l)
Check: Mass balance Cu: 3 3, S: 3 3, N: 8 8, H: 8 8, O: 24 24
Charge balance -8+8=0 +6-6=0
Example Balance the reaction in basic solution
Ag (s) + HS − (aq) + CrO 24− (aq) → Ag2S (s) + Cr(OH)3 (s)
1. Write the two half-rxns
Ag (s) + HS − (aq) → Ag2S (s)
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CrO 24− (aq) → Cr(OH)3 (s)
2. Balance the elements other than H and O
2Ag (s) + HS − (aq) → Ag2S (s)
CrO 24− (aq) → Cr(OH)3 (s)
3. Balance O by inserting H2O
2Ag (s) + HS − (aq) → Ag2S (s)
CrO 24− (aq) → Cr(OH)3 (s) + H2O (l)
4. Balance H by inserting H2O on the side that is deficient in hydrogen and an equal
amount of OH − on the other side (basic solution)
2Ag (s) + HS − (aq) + OH − (aq) → Ag2S (s) + H2O (l)
CrO 24− (aq) + 5H2O (l) → Cr(OH)3 (s) + H2O (l) + 5OH − (aq)
5. Balance charge by inserting e −
2Ag (s) + HS − (aq) + OH − (aq) → Ag2S (s) + H2O (l) +2e − (oxidation)
CrO 24− (aq) + 5H2O (l) +3e − → Cr(OH)3 (s) + H2O (l) + 5OH − (aq) (reduction)
6. Make the number of e − in the two half-rxns the same
6Ag (s) + 3HS − (aq) + 3OH − (aq) → 3Ag2S (s) + 3H2O (l) +6e −
2CrO 24− (aq) + 10H2O (l) +6e − → 2Cr(OH)3 (s) + 2H2O (l) + 10OH − (aq)
7. Addition of the two half-rxns
6Ag (s) + 3HS − (aq) + 3OH − (aq) + 2CrO 24− (aq) + 10H2O (l) +6e − → 3Ag2S (s)
+6e − +2Cr(OH)3 (s) + 2H2O (l) + 2OH − (aq)
6Ag (s) + 3HS − (aq) + 2CrO 24− (aq) + 5H2O (l) → 3Ag2S (s) + 2Cr(OH)3 (s) +
7OH − (aq)
Check: Mass balance Ag: 6 6, S: 3 3, Cr: 2 2, H: 13 13, O: 13 13
Charge balance -7 -7
Disproportionation Reactions
In this kind of reactions the same chemical species is both oxidized and reduced; it
reacts with itself. Balancing this type of reactions is straightforward by the half-rxn
method once it is realized that the same species may appear on the left in both half –rxns.
Example: Cl2 (aq) → ClO 3− (aq) + Cl − (aq) (unbalanced)
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Step 1 gives:
Cl2 (aq) → ClO 3− (aq)
Cl2 (aq) → Cl − (aq)
Proceeding with steps 2 to 5 yields
Cl2 (aq) + 6H2O (l) → 2ClO 3− (aq) + 12H+ (aq) + 10e − (oxidation)
Cl2 (aq) +2e − → 2Cl − (aq) (reduction)
Make the number of e − in the two half-rxns the same
6Cl2 (aq) + 6H2O (l) → 2ClO 3− (aq) + 12H+ (aq) + 10Cl − (aq)
Divide the coefficients by 2:
3Cl2 (aq) + 3H2O (l) → ClO 3− (aq) + 6H+ (aq) + 5Cl − (aq)
Electrochemical Cells
Electric charge (C) coulombs
Charge on e − 1.602 x 10-19 C
1 mol of e − : (1.602 x 10-19 C)(6.022 x 1023 mol-1) = 96, 485 C
Faraday’s constant (F)
Relation between charge and mole:
q = ne F
Electric current: the quantity of charge flowing each second through a circuit. The unit
is called Ampère (A): charge of 1C per second flowing past a point in a circuit.
The difference in electric potential, E, between two points is the work needed (or that
can be done) when moving an electric charge from one point to another. It is measured
in volts (V)
Volt = Joule/Coulomb (J/C)
Ohm’s Law: I =
E
R
(R is called resistance, units ohms (Ω))
1 Ω is the current of 1 A that flows through a circuit with a potential difference of 1 volt
if the resistance of the circuit is 1 Ω
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Anode: oxidation
Cathode: reduction
E cell = E cathode − E anode
Line notation: | (phase boundary)
|| (salt bridge)
Cd(s)|Cd(NO3)2(aq)||AgNO3(aq)|Ag(s)
Galvanic (voltaic) cell: Electrons flow
spontaneously from the negative to the
positive electrode. Ecell > 0
Conversion of chemical potential
energy into electrical energy that can
perform work
Electrical energy is proportional to Ecell
An electrolytic cell uses electrical energy to carry out a chemical rxn that would
otherwise not occur. (non-spontaneous cells)
Difference between galvanic and electrolytic cells
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Fundamentals of Electrolysis
Michael Faraday, the father of electrochemistry
Faraday’s Laws
• The quantities of substances produced and consumed at the electrodes are directly
proportional to the amount of electric charge passing through the cell
• When a given amount of electric charge passes through a cell, the quantity of a
substance produced or consumed at an electrode is proportional to its molecular
mass divided by the number of moles of electrons required to produce or consume
one mole of the substance.
If a current I flows for time t, the charge q passing any point in the circuit is:
q=It
q Coulombs, I Ampères, t seconds
coulombs
It
=
coulombs / mole F
If a reaction requires n electrons per molecule, the quantity reacting in time t is:
It
Moles reacted =
ne − F
Example
If a current of 0.17A flows for 16 min through the cell in the
adjacent figure how many grams of Cu will be deposited?
The moles of e − are:
From the diagram we conclude that:
Cathode: Cu2+ + 2e − • Cu (s)
Anode:
H2O • ½ O2 (g) + 2H+ + 2e −
Net rxn: H2O + Cu2+ • Cu (s) + ½ O2 (g) + 2H+
moles of e − :
It (0.17C / s )(16 min)(60s / min)
=
= 1.69 x10− 3 mol
F
96,485C / mol
The cathode half-reaction requires 2e − for each Cu(s)
deposited.
Moles of Cu(s) = ½ (moles of e − ) = 8.45 x 10-4 mol, or (8.45 x
10-4 mol)(63.546 g mol-1) = 0.054 g.
You may use also this equation for mass of solid deposited or
1 AtomicMass
It
mass of gas evolved: m =
F
ne −
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Example
Draw a diagram, show balanced equations, and write the notation for a voltaic cell that
consists of one half-cell with a Cr bar in a Cr(NO3)3 solution, another half-cell with an
Ag bar in an AgNO3 solution, and a KNO3 salt bridge. Measurement indicates that the
Cr electrode is negative relative to the Ag electrode.
Identify the oxidation and reduction reactions and write each half-reaction. Associate
the (-)(Cr) pole with the anode (oxidation) and the (+) pole with the cathode (reduction).
Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s)
Molten salts: refer to a salt that is in the liquid phase that is normally a solid at standard
temperature and pressure.
Molten salts have a variety of uses. Molten chloride salt mixtures are commonly used as
baths for various alloy heat treatments such as annealing of steel. Cyanide and chloride
salt mixtures are used for surface modification of alloys, fluoride, chloride, and nitrate
can also be used as heat transfer fluids as well as for thermal storage.
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Example: Electrolysis of molten MgCl2
Mg2+ (melt) + 2Cl − (melt) → Mg (l) + Cl2 (g)
Half-rxns: 2Cl − (melt) → Cl2 (g) + 2e − (oxidation)
Mg2+ (melt) + 2e − → Mg (l) (reduction)
Cell: graphite | Cl2 (g) | Cl − (melt), Mg2+ (melt) | Mg (l) | steel
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