Electrochemistry

ElectrochemistryIs the study of the relationship between electricityand chemical reactionChemical reactions involved in electrochemistry are :OxidationReductionREDOX REACTIONOne type of reaction cannot occur without the other.

REDUCTION gain of electron Oxidation no. decrease Reaction at cathodeRememberRED CAT = REDuction at CAThode Example:Cu2+ + 2e- CuOxidation no. OXIDATION loss of electron Oxidation no. increase Reaction at anode Example:Mg Mg2+ + 2e-Oxidation no. REDOX Reaction

Reduction :Oxidation :Cu2+(aq) + 2e- Cu(s)Zn(s) Zn2+(aq) + 2e-Half-cellreactionCu2+(aq) + Zn(s) Cu(s) + Zn2+(aq)ExampleOverall cellreaction :

Electrochemical reaction consists of reduction and oxidation.These two reactions are called half-cell reactionsThe combination of 2 half reactions are called cell reaction

CellsThere are 2 type of cellsElectrochemicalCellsElectrolyticCellswhere chemical reaction produces electricityUses electricity to produce chemical reactionAlso called; Galvanic cell or Voltaic cell

Component and Operation of Galvanic cellConsists of :1) Zn metal in an aqueous solution of Zn2+

2) Cu metal in an aqueous solution of Cu2+ The 2 metals are connected by a wire The 2 containers are connected by a salt bridge. A voltmeter is used to detect voltage generated.

Zn2+Cu2+VoltmeterGalvanic cell

What happens at the zinc electrode ? Zinc is more electropositive than copper. Tendency to release electrons: Zn > Cu. Zn (s) Zn2+ (aq) + 2e- Zinc dissolves. Oxidation occurs at the Zn electrode. Zn2+ ions enter ZnSO4 solution. Zn is the ve electrode since it is a source ofelectrons anode.

Cu2+ (aq) + 2e- Cu (s) Copper is deposited. Reduction occurs at the Cu electrode. Cu is the +ve electrode cathodeWhat happens at the copper electrode ? The electron from the Zn metal moves out through the wire enter the Cu metal Cu2+ ions from the solution accept electrons.

Reactions Involved:Anode :Cathode :Zn (s) Zn2+ (aq) + 2e-Cu2+ (aq) + 2e- Cu (s)Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s)Overall cellreaction :

Functions Salt bridge helps to maintain electrical neutrality Completes the circuit by allowing ions carrying charge to move from one half-cell to the other.Salt bridgeAn inverted U tube containing a gel permeated with solution of an inert electrolyte such as KCl, Na2SO4, NH4NO3.

What happened if there is no salt bridge?eZn2+Cu2+Veeeee As the zinc rod dissolves, the concentration of Zn2+ in the left beaker increase. The reaction stops because the nett increase in positive charge is not neutralized.This excess charge build-up can be reduced by adding a salt bridge

Salt bridge (KCl)Zn2+-+ZnCueeZn (s) Zn2+ (aq) + 2e-Cu2+ (aq) + 2e- Cu (s)ANODE (-)CATHODE (+)Cu2+E = +1.10 V

How does the cell maintains its electrical neutrality?Left CellRight CellCl- ions from salt bridge move into Zn half cellK+ ions from salt bridge move into Cu half cellElectrical neutrality is maintainedZn (s) Zn2+ (aq) + 2e-Cu2+ (aq) + 2e- Cu (s)Zn2+ ions enter the solution.Causing an overall excess of tve charge.Cu 2+ ions leave the solution.Causing an overall excess of -ve charge.

Electrochemical Cells19.2spontaneousredox reactionanodeoxidationcathodereductionhalf

Zn (s) | Zn2+ (aq) || Cu2+ (aq) | Cu (s)anodecathodeAlso can be represented as:Cell notationPhase boundarySalt bridgeCell notation

ExerciseFor the cell below, write the reaction at anode and cathode and also the overall cell reaction.Zn (s) | Zn2+(aq) || Cr3+ (aq) | Cr (s)Zn(s) Zn2+(aq) + 2e-Cr3+(aq) + 3e- Cr(s)3Zn(s) + 2Cr3+(aq) + 3Zn2+ (aq) +2Cr(s) Cathode :Anode :Overall cellreaction:X 2X 3Cell notation33226e6e-

The difference in electrical potential between the anode and cathode is called: cell voltage electromotive force (emf) cell potentialActs as electrical pressure that pushes electron through the wire.measured by a voltmeter

A measure of the ability of a half-cell to attract electrons towards it.Cu2+(aq) + 2e Cu(s) Eored = +0.34 VZn2+(aq) + 2e Zn(s) Eored = -0.76 VStandard reduction potential of copper half-cell is more positive compared to zinc. Standard reduction potentialThe more positive the half-cells electrode potential, the stronger the attraction for electrons.Tendency for reduction (cathode)Zinc half-cell becomes anode.

Cell Potential (Eocell)= Eocatode Eoanode= +0.34 (-0.76)= +1.1 VCu2+(aq) + 2e Cu(s) Eored = +0.34 VZn2+(aq) + 2e Zn(s) Eored = -0.76 V

Zn2+ (aq) + 2e- Zn (s) Cu2+ (aq) + 2e- Cu (s)E0 = -0.76VE0 = +0.34VE0cell = E0cathode - E0anode= +0.34 (-0.76)= +1.10 VorE0cell = E0red + E0ox= +0.34 + (+0.76)= +1.10 VChange the signHalf-cell equation at:Anode :Cathode :Zn (s) Zn2+ (aq) + 2e-Cu2+ (aq) + 2e- Cu (s)

Standard Electrode Potentials (Eo)at 25oC, the pressure is 1 atm (for gases), and the concentration of electrolyte is 1M.A measure of the ability of half-cell to attract electrons towards itThe sign of E0 changes when the reaction is reversedChanging the stoichiometric coefficients of a half-cell reaction does not change the value of E0

For example:

Standard Hydrogen Electrode(SHE)Made up of a platinum electrode, immersed in an aqueous solution of H+ (1 M) and bubbled with hydrogen gas at 1 atm pressure, and temperature at 25oCPtelectrodeH+ (aq)1 MH2 gasat 1 atmThe standard reduction of SHE is 0 V

Standard reduction potential of Zinc half cell is measured by setting up the electrochemicalcell as below.

Standard Electrode PotentialsZn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)Anode (oxidation):Cathode (reduction):Cell reaction

Standard reduction potential of Copper half cell is measured by setting up the electrochemicalcell as below. H2 (g) 25oC 1 atm.CuCu2CuSO4(aq) 1M H+(aq)1 M VPt E0 = 0 +-+-

Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s)Anode (oxidation):Cathode (reduction): H2 + Cu2+ Cu (s) + 2H+

The direction of half-reaction of SHE depends on the otherhalf-cell connected on it.The cell notation for SHE is either: Pt(s) | H2(g) | H+ (aq) when it is anode H+(aq) | H2(g) | Pt(s) when it is cathodeIn either case, E0 of SHE remains 0

Zn2+ (aq) + 2e- Zn (s) Cu2+ (aq) + 2e- Cu (s)E0 = -0.76VE0 = +0.34VE0cell = E0cathode - E0anode= +0.34 (-0.76)= +1.10 VorE0cell = E0red + E0ox= +0.34 + (+0.76)= +1.10 VChange the signHalf-cell equation at:Anode :Cathode :Zn (s) Zn2+ (aq) + 2e-Cu2+ (aq) + 2e- Cu (s)

At standard-state conditionE0cell = E0red + E0oxE0cell = E0cathode - E0anodeor

ExerciseCalculate the standard cell potential of the following electrochemical cell.Co(s) | Co2+(aq) || Ag+(aq) | Ag(s)AnswerE0cell = E0cathode - E0anode= +0.80 (-0.28)= +1.08 VAg+(aq) + e- Ag(aq)Co2+(aq) + 2e- Co(s) E0 = -0.28VE0 = +0.80V

Oxidation agent left of the half cell equationReduction agent right of the half cell equationExample :Oxidation agentReducing agent Refer to the list of Standard Reduction Potential:Increase strength asreducing agentThe more +ve the value of E0 the stronger the oxidizing agentThe more -ve the value of E0 the stronger the reducing agent

Arrange the 3 elements in order of increasing strength of reducing agentsX3+ + 3e- X E0 = -1.66 VY2+ + 2e- Y E0 = -2.87 VL2+ + 2e- L E0 = +0.85 VAnswer :L < X < YExercise

Calculate the E0 cell for the reaction ::Mg(p) | Mg2+(ak) || Sn4+(ak),Sn2+(ak) | Pt(p)Given :Mg2+(ak) + 2e Mg(p) E = -2.38 VSn4+(ak) + 2e Sn2+(ak) E= +0.15 V

Oxidation : Mg(p) Mg2+(ak) + 2e Eoox = +2.38 VReduction : Sn4+(ak) + 2e Sn2+(ak) Eo = +0.15 V Mg(p) + Sn4+(ak) Mg2+(ak) + Sn2+(ak) Ecell = +2.53 V

ExampleEcell = E o red + E o oxE0cell = Ecathode - Eanode=+0.15- (-2.38)=+2.53V = +2.38 + 0.15= +2.53 V

ExerciseA cell is set up between a chlorine electrode and ahydrogen electrode Draw a diagram to show the apparatus and chemicalsused. Discuss the chemical reactions occurring in theelectrochemical cell.Pt | H2(g, 1 atm) | H+(aq, 1M) || Cl2(g, 1atm) | Cl-(aq, 1M) | PtE0cell = +1.36 V

AnswerH2 (g),1 atm.E0cell =1.36V -+V-+Cl2 (g),1 atm.

1. Show the process occur at anode and cathode2. Overall reaction- Half-cell reaction

Answer Reduction (cathode)Cl2 (g) + 2e- 2Cl- (aq) Oxidation(anode)H2 (g) 2H+ (aq) + 2e- Eocell =+1.36 VE0 = 0 Eocell = Eocathode - E0anode+1.36 = Eocathode 0E0cathode = +1.36 VSo the standard reduction potential for Cl2 is:Eo = +1.36 V

Spontaneous & Non-Spontaneous reactions - Redox reaction is spontaneous when Ecell is +ve.- Non spontaneous is when Ecell is ve.

E cell = 0 The reaction is at equilibrium

Sn4+/Sn2+Predict whether the following reactions occur spontaneously or non-spontaneously under standard condition. Zn + Sn4+ Sn2+ + Zn2+ The two half-cells involved are:-Anod : Zn Zn2+ + 2e Eoox = +0.76 VCathode: Sn4+ + 2e Sn2+ Eo = +0.15 VZn + Sn4+ Zn2+ + Sn2+Eocell = Eo= +0.91 VspontaneousZn/Zn2+ Eo= +0.15 (-0.76 ) Eocell= Eored + Eoox= (+0.15) + (0.76) = +0.91 VOrSn4+/Sn2+Eo =+0.15VEoZn/Zn2+= - 0.76V.

Pb2+(aq) + 2Cl-(aq) Pb(s) + Cl2(g)

Pb2+(aq) + 2Cl-(aq) Pb(s) + Cl2(g)Pb2+(aq) + 2Cl-(aq) Pb(s) + Cl2(g)cathode: Pb2+(aq) + 2e Pb(s)anode: 2Cl-(aq) Cl2(g) + 2ePb2+(aq) + 2Cl-(aq) Pb(s) + Cl2(g)Eo = -0.13 VEoox = -1.36Eocell= Eored + EooxReductionOxidation= (-1.36) + (-0.13)= -1.48 VNon-spontaneous No Reaction

Example :Answer :

ExerciseA cell consists of silver and tin in a solution of 1 Msilver ions and tin (II) ions. Determine the spontaneityof the reaction and calculate the cell voltage of thisreaction.

Nernst equation Nernst equation can be used to calculate the E cell for any chosen concentration : At 298 K and R = 8.314 J K-1 mol-1 , 1 F = 96500 C

n = no of e- that are involved Q = reaction quotient

Example 1Calculate the Ecell for the following cell Zn(s) / Zn2+ (aq, 0.02M) // Cu2+(aq, 0.40 M) / Cu(s) Answer Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)Eocell = Eored + Eoox @ = +0.34 V + 0.76 V = +1.10 V Eocell = Eocathode Eoanode = +0.34 V - (- 0.76 V) = +1.10 V

= +1.10 V (-0.0385) = +1.139 V

At equilibrium:~ No net reaction occur (Q=K)~ Ecell = 0

Example 2Calculate the equilibrium constant (K) for the following reaction.Cu(s) + 2Ag+(ak) Cu2+(ak) + 2Ag(s)At equilibrium, E cell = 0 Eocell = Eo cathode - Eo anode= +0.80 ( +0.34)= +0.46 VAnswer

log K = 15.54 K = 3.467 x 1015

ElectrolysisElectrolysis is a chemical process that uses electricity for a non-spontaneous redox reaction to occur.Such reactions take place in electrolytic cells.Electrolytic Cell It is made up of 2 electrodes immersed in an electrolyte. A direct current is passed through the electrolyte from an external source. Molten salt and aqueous ionic solution are commonly used as electrolytes.

AnionCationOxidationReductionElectrolytic CellElectrolyte (M+X-)X-,OH- M+,H++-

AnodeCathode Positive electrode The electrode which is connected to thepositive terminal of the battery Oxidation takes place Negative electrode The electrode which is connected to thenegative terminal of the battery Reduction takes placeElectrons flow from anode to cathode

Electrode as circuit connectors as sites for the precipitation of insolubleproducts example: Platinum , Graphite (inert electrode)Electrolyte a liquid that conducts electricity due to the presence of +ve and ve ions must be in molten state or in aqueoussolution so that the ions can move freely example: KCl(l), HCl(aq), CH3COOH(aq)

Comparison between an electrochemical celland an electrolytic cellElectrolytic CellElectrochemical Cell

Electrolytic CellElectrochemical Cell Cathode = negative Anode = positive Cathode = positive Anode = negative Non-spontaneous redoxreaction requires energy to drive it Spontaneous redox reaction releases energy Oxidation occurs at anode, reduction occursat cathode Anions move towards anode, cations move towards cathode. Electrons flow from anode to cathode in an external circuit.Similarities:

Electrolysis of molten salt Electrolysis of molten salt requires high temp. Electrolysis of molten NaClCation : Na+Anion : Cl-Anode :Cathode :Na+ (l) + e- Na (s) Cl- (l) Cl2(g) + 2e-Overall :2Na+ (l) + 2Cl-(l) Cl2(g) + 2Na(s) 2Na+ (l) + 2e- 2Na (s)

Electrolysis of molten NaCl gives sodium metaldeposited at cathode and chlorine gas evolved atanode. Electrolysis of molten NaCl is industrially important.The industrial cell is called Downs Cell

Electrolysis of Aqueous Salt Electrolysis of aqueous salt is more complex because the presence of water. Aqueous salt solutions contains anion, cation and water. Water is an electro-active substance that may beoxidised or reduced in the process depending onthe condition of electrolysis.Reduction :Oxidation :E0 = -0.83 VE0 = -1.23 V

Predicting the products of electrolysisFactors influencing the products :1. Reduction/oxidation potential of the species in electrolyteConcentrations of ionsTypes of electrodes used active or inert

Electrolysis of Aqueous NaClThe electrolysis of aqueous NaCl depends on the concentration of electrolyte. NaCl aqueous solution contains Na+ cation, Cl- anionand water molecules On electrolysis, the cathode attracts Na+ ion and H2O molecules the anode attracts Cl- ion and H2O molecules

CathodeE0 = -0.83 VE0 = -2.71 VE0 for water molecules is more positive. H2O easier to reduce.Electrolysis of diluted NaCl solutionAnodeE0 = +1.36 VE0 = +1.23 VIn dilute solution, water will be selected for oxidation because of its lower Eo.

Reactions involvedE0 = -0.83 VCathode:Anode:Cell reaction:E0cell = -2.06 VE0 = -1.23 V4 H2O

Electrolysis of Concentrated NaCl solutionCathodeE0 = -0.83 VE0 = -2.71 V E0 for water molecules is more positive H2O easier to be reduceAnodeE0 = +1.36 VE0 = +1.23 VIn concentrated solution, chloride ions will be oxidised because of its high concentration.

Reactions involvedE0 = -0.83 VE0 = -1.36 VCell reaction:E0cell = -2.19 VCathode:Anode:

Na2SO4 aqueous solution contains Na+ ion, SO42- ionand water molecules On electrolysis, the cathode attracts Na+ ion and H2O molecules the anode attracts SO42- ion and H2O moleculesExercisePredict the electrolysis reaction when Na2SO4 solution is electrolysed using platinum electrodes.Solution

Anode E0 for water molecules is less positive H2O easier to oxidiseE0 = +2.01 VE0 = +1.23 VCathodeE0 = -0.83 VE0 = -2.71 V E0 for water molecules is more positive H2O easier to reduce

Cathode = H2 gas is produced and solution becomebasic at cathode because OH- ions are formed Anode = O2 gas is produced and solution becomeacidic at anode because H+ ions are formedEquationCathode:Anode:E0 = -1.23 VE0 = -0.83 VCell Reaction:E0cell = -2.06 V

Faradays Law of ElectrolysisDescribes the relationship between the amount ofelectricity passed through an electrolytic cell andthe amount of substances produced at electrode.

Faradays 1st Law m QQ = electric charge in coulombs (C)m = mass of substance discharged

Q = ItQ = electric charge in coulombs (C)I = current in amperes (A)t = time in second (s)Faraday constant (F)is the charge on 1 mole of electron1 F = 96 500 C

ExampleAn aqueous solution of CuSO4 is electrolysed using acurrent of 0.150 A for 5 hours. Calculate the massof copper deposited at the cathode.AnswerElectric charge, Q = Current (I) x time (t)Q = (0.150 A) x ( 5 x 60 x 60 )sQ = 2700 C1 mole of electron 1 F 96 500 CNo. of e- passed through = = 0.028 mol

Cu2+ (aq) + 2e- Cu (s)From equation:2 mol electrons 1 mol Cu0.028 mol electrons 0.014 mol CuMr for Cu = 63.5Mass of Copper deposited = 0.014 x 63.5= 0.889 g