Chapter 10 Electrochemistry

Electrochemistry Reduction REDOX REACTION Oxidation Is the study of the relationship between electricity and chemical reaction Chemical reactions involved in electrochemistry are : Reduction REDOX REACTION Oxidation One type of reaction cannot occur without the other.

REDOX Reaction REDUCTION OXIDATION Remember… RED CAT Mg  Mg2+ + 2e- gain of electron loss of electron Oxidation no. decrease Oxidation no. increase Reaction at cathode Reaction at anode Example: Remember… RED CAT = REDuction at CAThode Mg  Mg2+ + 2e- Example: Oxidation no.  Cu2+ + 2e-  Cu Oxidation no. 

Example Electrochemical reaction consists of reduction and oxidation. These two reactions are called ‘half-cell reactions’ The combination of 2 half reactions are called ‘cell reaction’ Example Reduction : Cu2+(aq) + 2e-  Cu(s) Half-cell reaction Oxidation : Zn(s)  Zn2+(aq) + 2e- Overall cell reaction : Cu2+(aq) + Zn(s)  Cu(s) + Zn2+(aq)

Cells Electrochemical Electrolytic Cells Cells There are 2 type of cells Electrochemical Cells Electrolytic Cells Uses electricity to produce chemical reaction where chemical reaction produces electricity Also called; Galvanic cell or Voltaic cell Chemical Energy Electrical Electrical Energy Chemical

Component and Operation of Galvanic cell Consists 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.

Galvanic cell Voltmeter Cu electrode Zn electrode Salt bridge Zn2+ CuSO4(aq) solution ZnSO4(aq) solution

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 of electrons  anode.

What 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. Cu2+ (aq) + 2e-  Cu (s) Copper is deposited. Reduction occurs at the Cu electrode. Cu is the +ve electrode  cathode

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

Salt bridge An inverted U tube containing a gel permeated with solution of an inert electrolyte such as KCl, Na2SO4, NH4NO3. 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.

This excess charge build-up can be reduced by adding a salt bridge What happened if there is no salt bridge? V Cu Zn e e e e ZnSO4(aq) Zn2+ e e Cu2+ CuSO4(aq) 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

Cu Zn Zn (s)  Zn2+ (aq) + 2e- Cu2+ (aq) + 2e-  Cu (s) CATHODE (+) ANODE (-) E = +1.10 V Cu Zn - + e e ZnSO4(aq) Zn2+ CuSO4(aq) Cu2+ Salt bridge (KCl)

How does the cell maintains its electrical neutrality? Left Cell Right Cell Zn (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. Cl- ions from salt bridge move into Zn half cell K+ ions from salt bridge move into Cu half cell Electrical neutrality is maintained

Electrochemical Cells anode oxidation cathode reduction spontaneous redox reaction half 19.2

Cell notation Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq) Also can be represented as: Salt bridge Phase boundary Zn (s) | Zn2+ (aq) || Cu2+ (aq) | Cu (s) anode cathode Cell notation

Zn (s) | Zn2+(aq) || Cr3+ (aq) | Cr (s) Exercise For the cell below, write the reaction at anode and cathode and also the overall cell reaction. Cell notation Zn (s) | Zn2+(aq) || Cr3+ (aq) | Cr (s) X 3 Anode : 3 Zn(s) → Zn2+(aq) + 2e- 3 6e- Cathode : 2 Cr3+(aq) + 3e- → Cr(s) 6e 2 X 2 Overall cell reaction: 3Zn(s) + 2Cr3+(aq) + → 3Zn2+ (aq) +2Cr(s)

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

Electrode Potential Cu2+(aq) + 2e → Cu(s) Eored = +0.34 V A measure of the ability of a half-cell to attract electrons towards it. Cu2+(aq) + 2e → Cu(s) Eored = +0.34 V Zn2+(aq) + 2e → Zn(s) Eored = -0.76 V Standard reduction potential The more positive the half-cell’s electrode potential, the stronger the attraction for electrons. Tendency for reduction ↑ (cathode) Standard reduction potential of copper half-cell is more positive compared to zinc. Zinc half-cell becomes anode.

Cu2+(aq) + 2e → Cu(s) Eored = +0.34 V Zn2+(aq) + 2e → Zn(s) Eored = -0.76 V Cell Potential (Eocell)= Eocatode — Eoanode = +0.34 – (-0.76) = +1.1 V

or Zn2+ (aq) + 2e-  Zn (s) Cu2+ (aq) + 2e-  Cu (s) E0 = -0.76V Cu2+ (aq) + 2e-  Cu (s) E0 = +0.34V E0cell = E0cathode - E0anode = +0.34 – (-0.76) = +1.10 V or E0cell = E0red + E0ox Change the sign = +0.34 + (+0.76) = +1.10 V Half-cell equation at: Anode : Zn (s)  Zn2+ (aq) + 2e- Cu2+ (aq) + 2e-  Cu (s) Cathode :

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

For example: Cl2(g) + 2e-  2Cl-(aq) E0 = +1.36 V ½Cl2(g) + e-  Cl-(aq) E0 = +1.36 V Cl-(aq)  ½Cl2(g) + e- E0 = -1.36 V

Standard Hydrogen Electrode The standard reduction of SHE is 0 V 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 25oC H2 gas at 1 atm Pt electrode H+ (aq) 1 M The standard reduction of SHE is 0 V

Standard reduction potential of Zinc half cell is measured by setting up the electrochemical cell as below. H2 (g), 25oC,1 atm. Zn e Zn2+ ZnSO4(aq) 1 M H+(aq),1 M V Pt E0 = 0 E0 = +0.76 - +

Standard Electrode Potentials Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s) Anode (oxidation): Zn (s) Zn2+ (1 M) + 2e- Cathode (reduction): 2e- + 2H+ (aq,1 M) H2 (g,1 atm) Cell reaction Zn (s) + 2H+ (aq,1 M) Zn2+(aq) + H2 (g,1 atm) E0 = EH /H - EZn /Zn cell + 2+ 2 0.76 V = 0 - EZn /Zn 2+ EZn /Zn = -0.76 V 2+ Zn2+ + 2e- Zn E0 = -0.76 V

Standard reduction potential of Copper half cell is measured by setting up the electrochemical cell as below. V E0 = 0 + - - H2 (g) 25oC 1 atm. + Cu H+(aq)1 M Cu2 CuSO4(aq) 1M Pt

Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s) Anode (oxidation): H2 (1 atm) 2H+ + 2e- Cathode (reduction): 2e- + Cu2+ (1 M) Cu (s) H2 + Cu2+ Cu (s) + 2H+ E0 = Ecathode - Eanode cell Ecell = ECu /Cu – EH /H 2+ + 2 0.34 = ECu /Cu - 0 2+ ECu /Cu = 0.34 V 2+

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

or Zn2+ (aq) + 2e-  Zn (s) Cu2+ (aq) + 2e-  Cu (s) E0 = -0.76V Cu2+ (aq) + 2e-  Cu (s) E0 = +0.34V E0cell = E0cathode - E0anode = +0.34 – (-0.76) = +1.10 V or E0cell = E0red + E0ox Change the sign = +0.34 + (+0.76) = +1.10 V Half-cell equation at: Anode : Zn (s)  Zn2+ (aq) + 2e- Cu2+ (aq) + 2e-  Cu (s) Cathode :

E0cell = E0cathode - E0anode At standard-state condition E0cell = E0cathode - E0anode or E0cell = E0red + E0ox

Exercise Answer E0cell = E0cathode - E0anode Calculate the standard cell potential of the following electrochemical cell. Co(s) | Co2+(aq) || Ag+(aq) | Ag(s) Ag+(aq) + e-  Ag(aq) E0 = +0.80V Answer Co2+(aq) + 2e-  Co(s) E0 = -0.28V Cathode (Red) : Anode (Ox) : Co(s)  Co2+(aq) + 2e- Ag+(aq) + e-  Ag(aq) E0 = +0.80V E0ox = +0.28V E0cell = E0cathode - E0anode = +0.80 – (-0.28) = +1.08 V

Refer to the list of Standard Reduction Potential: Oxidation agent → left of the half cell equation Reduction agent → right of the half cell equation Example : Increase strength as reducing agent Ni2+ (aq) + 2e- → Ni (s) E0 = -0.25 V Cu2+ (aq) + 2e- → Cu (s) E0 = +0.34 V Ag+ (aq) + e- → Ag (s) E0 = +0.80 V Reducing agent Oxidation agent The more -ve the value of E0 → the stronger the reducing agent The more +ve the value of E0 → the stronger the oxidizing agent

Exercise Answer : L < X < Y Arrange the 3 elements in order of increasing strength of reducing agents X3+ + 3e-  X E0 = -1.66 V Y2+ + 2e-  Y E0 = -2.87 V L2+ + 2e-  L E0 = +0.85 V Answer : L < X < Y

E0cell = Ecathode - Eanode Example Calculate the E0 cell for the reaction :: Mg(p) | Mg2+(ak) || Sn4+(ak),Sn2+(ak) | Pt(p) Given : Mg2+(ak) + 2e→ Mg(p) Eθ = -2.38 V Sn4+(ak) + 2e → Sn2+(ak) Eθ= +0.15 V Oxidation : Mg(p) → Mg2+(ak) + 2e Eoox = +2.38 V Reduction : Sn4+(ak) + 2e → Sn2+(ak) Eo = +0.15 V Mg(p) + Sn4+(ak) → Mg2+(ak) + Sn2+(ak) Ecell = +2.53 V E0cell = Ecathode - Eanode =+0.15- (-2.38) =+2.53V Eθcell = E o red + E o ox = +2.38 + 0.15 = +2.53 V

Exercise A cell is set up between a chlorine electrode and a hydrogen electrode Pt | H2(g, 1 atm) | H+(aq, 1M) || Cl2(g, 1atm) | Cl-(aq, 1M) | Pt E0cell = +1.36 V Draw a diagram to show the apparatus and chemicals used. Discuss the chemical reactions occurring in the electrochemical cell.

Answer V E0cell =1.36V H2 (g), 1 atm. Cl2 (g), 1 atm. Pt Pt H+(aq), 1M - + - + Pt Pt H+(aq), 1M Cl-(aq), 1M

1. Show the process occur at anode and cathode - Half-cell reaction 2. Overall reaction

Answer Reduction (cathode) Cl2 (g) + 2e-  2Cl- (aq) Oxidation(anode) H2 (g)  2H+ (aq) + 2e- E0 = 0 Eocell =+1.36 V Eocell = Eocathode - E0anode +1.36 = Eocathode – 0 E0cathode = +1.36 V So the standard reduction potential for Cl2 is: Eo = +1.36 V

Eθ cell = 0 The reaction is at equilibrium Spontaneous & Non-Spontaneous reactions - Redox reaction is spontaneous when Ecell is +ve. - Non spontaneous is when Ecell is –ve. Eθ cell = 0 The reaction is at equilibrium

Eo Zn + Sn4+ → Zn2+ + Sn2+ Eocell = Eo → Eo 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 V Cathode: Sn4+ + 2e → Sn2+ Eo = +0.15 V Eo =+0.15V Sn4+/Sn2+ Eo = - 0.76V. Zn/Zn2+ Zn + Sn4+ → Zn2+ + Sn2+ Eocell = Eo → Eo Eocell= Eored + Eoox Sn4+/Sn2+ Zn/Zn2+ = (+0.15) + (0.76) = +0.15 – (-0.76 ) Or = +0.91 V = +0.91 V spontaneous

Predict : Spontaneous or non-spontaneous? Pb2+(aq) + 2Cl-(aq) → Pb(s) + Cl2(g)

Pb2+(aq) + 2Cl-(aq) → Pb(s) + Cl2(g) Reduction Pb2+(aq) + 2Cl-(aq) → Pb(s) + Cl2(g) Oxidation cathode: Pb2+(aq) + 2e → Pb(s) Eo = -0.13 V anode: 2Cl-(aq) → Cl2(g) + 2e Eoox = -1.36 Pb2+(aq) + 2Cl-(aq) → Pb(s) + Cl2(g) Eocell= Eored + Eoox = (-1.36) + (-0.13) = -1.48 V Non-spontaneous No Reaction

standard reduction potential Example : Predict whether the following reactions occur spontaneously : 2Ag(s) + Br2(aq) 2Ag+(aq) + 2Br-(aq) EAg /Ag = +0.8 V + 2 EBr /Br = +1.07 V - standard reduction potential Answer : 2Ag(s) 2Ag+(aq) + 2e Eθox = - 0.80 V Br2(aq) + 2e 2Br -(aq) Eθ = +1.07 V 2Ag(s) + Br2(aq) 2Ag+(aq) + 2Br-(aq) Esel = + 0.27 V The reaction is spontaneous

Exercise E0cell = E0cathode - E0anode A cell consists of silver and tin in a solution of 1 M silver ions and tin (II) ions. Determine the spontaneity of the reaction and calculate the cell voltage of this reaction. Ag+ (aq) + e- → Ag (s) Sn2+ (aq) + 2e- → Sn (s) E0 = +0.80 V E0 = -0.14 V (cathode) (anode) E0cell = E0cathode - E0anode = +0.80 – (-0.14) = +0.94 V E0cell = +ve ( reaction is spontaneous)

Nernst equation Nernst equation can be used to calculate the E cell for any chosen concentration : Ecell = Eocell – RT ln [ product ]x nF [ reactant]y At 298 K and R = 8.314 J K-1 mol-1 , 1 F = 96500 C Ecell = Eocell – 0.0257 [ product ]x n [ reactant]y 2.303 log

= Ecell = Eocell – 0.0592 [ product ]x n [ reactant]y log [ product ]x Q = Ecell = Eocell – 0.0592 n log Q n = no of e- that are involved Q = reaction quotient

Example 1 Calculate 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

E = Eo – 0.0592 log [ Zn2+] n [ Cu2+] E = +1.10 V – 0.0592 log (0.02 ) 2 ( 0.40) = +1.10 V – (-0.0385) = +1.139 V

At equilibrium: ~ No net reaction occur (Q=K) ~ Ecell = 0 Ecell = Eocell – 0.0592 n log K 0 = Eocell – 0.0592 n log K Eθcell = 0.0592 n log K

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

Ecell = Eocell – 0.0592 log K 2 0 = 0.46 – 0.0592 log K 2 0.0592 log K = 0.46 2 log K = 15.54 K = 3.467 x 1015

Electrolysis Electrolytic Cell Electrolysis 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.

Electrolytic Cell + - Oxidation Reduction Electrolyte (M+X-) Anion X-,OH- M+,H+ Anion Cation

Electrons flow from anode to cathode Positive electrode The electrode which is connected to the positive terminal of the battery Anode Oxidation takes place Electrons flow from anode to cathode Negative electrode The electrode which is connected to the negative terminal of the battery Cathode Reduction takes place

Electrode as circuit connectors as sites for the precipitation of insoluble products 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 aqueous solution so that the ions can move freely example: KCl(l), HCl(aq), CH3COOH(aq)

Comparison between an electrochemical cell and an electrolytic cell + - e- - + e- Anode Anode Cathode Cathode

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

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

Electrolysis of molten NaCl gives sodium metal deposited at cathode and chlorine gas evolved at anode. 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 be oxidised or reduced in the process depending on the condition of electrolysis. Reduction : 2H2O (l) + 2e- H2 (g) + 2OH- (aq) E0 = -0.83 V Oxidation : 2H2O (l) 4H+ (aq) + O2 (g) + 4e- E0 = -1.23 V

Factors influencing the products : Predicting the products of electrolysis Factors influencing the products : 1. Reduction/oxidation potential of the species in electrolyte Concentrations of ions Types of electrodes used – active or inert

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

Electrolysis of diluted NaCl solution Cathode Na+ (aq) + e- Na (s) E0 = -2.71 V 2H2O (l) + 2e- H2 (g) + 2OH- (aq) E0 = -0.83 V E0 for water molecules is more positive. H2O easier to reduce. Anode Cl2 (g) + 2e- 2Cl- (aq) E0 = +1.36 V O2 (g) + 4H+ (aq) + 4e- 2H2O (l) E0 = +1.23 V In dilute solution, water will be selected for oxidation because of its lower Eo.

Reactions involved 2H2O (l) + 2e- H2 (g) + 2OH- (aq) E0 = -0.83 V Cathode: 4H2O (l) + 4e- 2H2 (g) + 4OH- (aq) Anode: 2H2O (l) O2 (g) + 4H+ (aq) + 4e- E0 = -1.23 V Cell reaction: 6H2O(l) O2(g) + 2H2(g) + 4OH-(aq) + 4H+(aq) 4 H2O 2H2O(l) O2(g) + 2H2(g) E0cell = -2.06 V

Electrolysis of Concentrated NaCl solution Cathode Na+ (aq) + e- Na (s) E0 = -2.71 V 2H2O (l) + 2e- H2 (g) + 2OH- (aq) E0 = -0.83 V E0 for water molecules is more positive H2O easier to be reduce Anode Cl2 (g) + 2e- 2Cl- (aq) E0 = +1.36 V O2 (g) + 4H+ (aq) + 4e- 2H2O (l) E0 = +1.23 V In concentrated solution, chloride ions will be oxidised because of its high concentration.

Reactions involved 2H2O (l) + 2e- H2 (g) + 2OH- (aq) E0 = -0.83 V Cathode: 2H2O (l) + 2e- H2 (g) + 2OH- (aq) E0 = -0.83 V Anode: 2Cl- (aq) Cl2 (g) + 2e- E0 = -1.36 V Cell reaction: 2H2O(l) + 2Cl- Cl2(g) + H2(g) + 2OH-(aq) E0cell = -2.19 V

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

Cathode Anode Na+ (aq) + e- Na (s) E0 = -2.71 V 2H2O (l) + 2e- H2 (g) + 2OH- (aq) E0 = -0.83 V E0 for water molecules is more positive H2O easier to reduce Anode S2O82- (aq) + 2e- 2SO42- (aq) E0 = +2.01 V O2 (g) + 4H+ (aq) + 4e- 2H2O (l) E0 = +1.23 V E0 for water molecules is less positive H2O easier to oxidise

Equation 2H2O (l) + 2e- H2 (g) + 2OH- (aq) E0 = -0.83 V Cathode: 2H2O (l) + 2e- H2 (g) + 2OH- (aq) E0 = -0.83 V Anode: 2H2O (l) O2 (g) + 4H+ (aq) + 4e- E0 = -1.23 V Cell Reaction: 2H2O(l) O2(g) + 2H2(g) E0cell = -2.06 V Cathode = H2 gas is produced and solution become basic at cathode because OH- ions are formed Anode = O2 gas is produced and solution become acidic at anode because H+ ions are formed

Faraday’s Law of Electrolysis Describes the relationship between the amount of electricity passed through an electrolytic cell and the amount of substances produced at electrode. Faraday’s First Law States that the quantity of substance formed at an electrode is directly proportional to the quantity of electric charge supplied.

m α Q Faraday’s 1st Law Q = electric charge in coulombs (C) m = mass of substance discharged

The End

is the charge on 1 mole of electron Q = It Q = 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 electron 1 F = 96 500 C

Example Answer An aqueous solution of CuSO4 is electrolysed using a current of 0.150 A for 5 hours. Calculate the mass of copper deposited at the cathode. Answer Electric charge, Q = Current (I) x time (t) Q = (0.150 A) x ( 5 x 60 x 60 )s Q = 2700 C 1 mole of electron Ξ 1 F Ξ 96 500 C 2700 96 500 No. of e- passed through = = 0.028 mol

Cu2+ (aq) + 2e-  Cu (s) From equation: 2 mol electrons  1 mol Cu 0.028 mol electrons  0.014 mol Cu Mr for Cu = 63.5 Mass of Copper deposited = 0.014 x 63.5 = 0.889 g