Presentation on theme: "ELECTROCHEMISTRY. Electrode potential and half cell Half cell 2 possible electron transfer reactions occur: M(s) M n+ (aq) + ne - (Oxidation) M n+ (aq)"— Presentation transcript:
Electrode potential and half cell Half cell 2 possible electron transfer reactions occur: M(s) M n+ (aq) + ne - (Oxidation) M n+ (aq) + ne - M(s)(Reduction) Reversible reaction occurs. When the reversible reaction reached equilibrium, there is a potential difference between the metal and aqueous solution – electrode potential. M n+ (aq) + ne - M(s), E = electrode potential Electrode potential may be positive or negative, depending on the position of equilibrium.
Factors affecting electrode potential, E: 1) Temperature 2) Pressure 3) Nature of metal (position in electrochemical series) 4) Concentration of ions. If [M n+ (aq)] increased, equilibrium shift to the right. More electrons removed from electrode electrode becomes more positive. Because electrode potential is affected by the 4 factors above, therefore electrode potential must be measured under standard condition (298K, 1 atm, 1.00 mol dm -3 ). The electrode potential measured under standard condition is called standard electrode potential, E θ.
Standard Hydrogen Electrode (S.H.E) *Consist of a platinised Pt electrode *in a solution of 1.00 mol dm -3 H + (aq) ions. *H 2 gas at the pressure of 1 atm is bubbled over the platinum electrode.
Electrode is alternately bathed in first H + (aq) and then H 2 (g). Hydrogen is adsorbed on the Pt and equilibrium is established. ½ H 2 (g) H + (aq) + e - E = 0.00 V Has a electrode potential of zero. Is used to compare the electrode potentials of other half-cells (as a reference electrode). Function of Pt electrode: 1) As an inert metal. 2) Allows H 2 gas to be adsorbed onto its surface. 3) Covered by loosely deposited layer of finely divided Pt – increase surface area equilibrium achieved faster.
Cell Diagram (-) electrode(+) electrode Oxidation More electrons at surface of electrode. M n+ (aq) + ne - M(s) Equilibrium shift to left. Reduction Electrons removed from electrode. M n+ (aq) + ne - M(s) Equilibrium shift to right. Writing a cell diagram: Metal electrode(s) metal ion(aq) metal ion(aq) Metal electrode(s) E.g :Zn(s) Zn 2+ (aq) Cu 2+ (aq) Cu(s)
Measuring E of metals or non-metals in contact of their ions in aqueous solution. Standard electrode potential, E Is the electromotive force (e.m.f) of a cell made up of a test electrode connected to a standard hydrogen electrode under standard condition (1 mol dm -3 concentration of H + at 25 C and 1 atm.) Unit : Volts
Measuring E of metals or non-metals in contact of their ions in aqueous solution. Standard electrode potential (E ) of Zn Reaction at electrodes: Oxidation : Reduction : Overall reaction: E cell = V (Zn electrode is negative – oxidation occurs and electron produced) Cell diagram :
Standard electrode potential (E ) of Mg Reaction at electrodes: Oxidation : Reduction : Overall reaction: E cell = V (Mg electrode is negative – oxidation occurs and electron produced) Cell diagram :
Standard electrode potential (E ) of Cu Reaction at electrodes: Oxidation : Reduction : Overall reaction: E cell = V (Cu electrode is positive – reduction occurs and electron removed) Cell diagram :
Standard electrode potential (E ) of Cl 2 Reaction at electrodes: Oxidation : Reduction : Overall reaction: E cell = V Cell diagram : **the least oxidised species is placed next to the inert electrode.
Measuring E of ions of the same element in different oxidation states. Standard electrode potential (E ) of Fe 3+ /Fe 2+ Reaction at electrodes: Oxidation : Reduction : Overall reaction: E cell = V Cell diagram : **the least oxidised species is placed next to the inert electrode.
Standard electrode potential (E ) of Cr 3+ /Cr 2+ Reaction at electrodes: Oxidation : Reduction : Overall reaction: E cell = V (Zn electrode is negative – oxidation occurs and electron produced) Cell diagram :
Uses of standard electrode potential (E ) 1) To determine strength of both oxidising and reducing agents. Electrochemical series:
2) To predict the direction of electron flow from a simple cell. Electron flow from the negative electrode (more negative E value) to the positive electrode (more positive E value)
Electrochemical cell Two half-cells joined by a salt bridge containing KCl or KNO 3. Function of salt bridge : 1) Complete the circuit. 2) Maintains electrical neutrality in the cell. Salt bridge (KCl)
Electrochemical cell Negative electrodePositive electrode Metal with larger negative E - Zn. Oxidation occurs. Metal dissolves in the solution : Zn(s) Zn 2+ (aq) + 2e - Electron flows through the wire and voltmeter to Cu electrode. Metal with larger positive E - Cu. Reduction occurs. Electron from Zn half cell taken by Cu 2+ (aq). Cu metal formed and depositied at Cu electrode. Cu 2+ (aq) + 2e - Cu(s) Cell reaction : Oxidation :Zn(s) Zn 2+ (aq) + 2e - Reduction :Cu 2+ (aq) + 2e - Cu(s) Overall :Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s)
Electrons flow from negative electrode through the wire and voltmeter to positive electrode. This cause a potential difference – electromotive force (e.m.f) e.m.f is the maximum potential difference of two metals. It gives a quantitative measure of the likelihood of the redox reaction taking place in the cell.
Standard Cell Potential (e.m.f), E cell. * Is the measure of tendency of electrons to flow through the external circuit * under standard conditions of 25 C, 1 atm and 1.00 mol dm -3 concentration. * It is the maximum potential difference between the electrodes. Example : Cell diagram :
Standard Cell Potential (e.m.f), E cell. Calculating E cell : Example 1 Cell reactions :E /V Oxidation : Reduction : Overall : Note : a) when reaction is reversed, the sign is changed. b) E θ is not dependent on the num. of electrons transferred. Formula method : E cell = E red - E ox
Example 2 A cell consisting of half-cells : Cu in CuSO 4 (aq) and Ag in AgNO 3 (aq). Write the cell diagram, reaction equation in each electrode, predict the electron flow and calculate the standard cell potential.
Uses of E θ cell To predict the feasibility of a reaction. Reactions with positive E θ cell are thermodynamically feasible. E.g: Mg 2+ (aq) + 2e - Mg(s)E θ = V Cu 2+ (aq) + 2e - Cu(s)E θ = V Half equations: Oxidation : Reduction : Overall : *E θ cell > 0, therefore reaction is feasible.
E.g: Is this reaction feasible? Pb(s) + 2Cr 3+ (aq) Pb 2+ (aq) + 2Cr 2+ (aq)
Limitations of prediction made using E θ cell value. Using E θ cell to predict feasibility of reaction might not be suitable when: a) The reaction has very high activation energy (kinetically slow). E θ value give no information about reaction rates. b) The reactions are not in standard conditions. E θ value apply to standard conditions only (298K, 1 atm, all concentrations at 1.00 mol dm -3 ).
b) Reactions are not in standard conditions E.g : Oxidation of concentrated HCl by MnO 2 MnO 2 + 4H + + 2e - Mn H 2 OE θ = V 2Cl - Cl 2 + 2e - E θ = V MnO 2 + 4H + + 2Cl - Mn 2+ + Cl 2 + 2H 2 O E θ cell = V *Reagents are heated. E θ cell is negative, reaction predicted not feasible. But the reactions is not in standard condition : [H + ] 1.00 mol dm -3, temperature 25°C. Cl 2 can be prepared by heating MnO 2 with concentrated HCl.
a) The reaction has very high activation energy E.g 1 : Cu 2+ (aq) + 2e - Cu(s) E θ = V H 2 (g) 2H + (aq) + 2e - E θ = 0.00 V Cu 2+ (aq) + Zn(s) Zn 2+ (aq) + Cu(s) E θ cell = V E θ cell value is positive, reaction predicted feasible. However, the rate of reaction is too slow (due to high E a ). Hence in practice, this reaction does not occur at standard condition. The prediction fails because E θ cell value gives no information about the kinetics of the reaction. E.g. 2 : Oxidation of methanal to methanoic acid by K 2 Cr 2 O 7. 3HCHO + Cr 2 O H+ 2Cr HCO 2 H + 4H 2 O E θ cell = +1.27V E θ cell is positive, reaction predicted feasible, but in practice no reaction occur (high E a ). Heating is required for reaction to occur.
Exercise 1. Water can be oxidised to oxygen according to the equation: 2H 2 O(l) O 2 (g) + 4H + (aq) + 4e - Given the following E values, what can be use to oxidise water in this way?
Effect of concentration of ions on E θ cell value. E.g 1 : Fe 3+ (aq) + e - Fe 2+ (aq) E θ = V If [Fe 3+ ] > 1.00 mol dm -3, equilibrium will shift to the right causing E θ to be more positive. If [Fe 3+ ] < 1.00 mol dm -3, equilibrium will shift to the left causing E θ to be more negative. If [Fe 2+ ] > 1.00 mol dm -3, equilibrium will shift to the left, E θ will be more negative. If [Fe 2+ ] < 1.00 mol dm -3, equilibrium will shift to the right, E θ will be more positive.
Effect of concentration of ions on E θ cell value. E.g 2 : Cr 2 O 7 2- (aq) + 14H + (aq) + 6e - (aq) 2Cr 3+ (aq) + 7H 2 O(l) E θ = V. What will happen to the E θ value and the strength of the Cr 2 O 7 2- solution as an oxidising agent if : a) [Cr 2 O 7 2- (aq) ] were to be increased. b) [H + (aq)] were to be decreased. c) [Cr 3+ (aq) ] were to be increased.
Electrolysis Is a process in which current is passed through the electrolyte (in molten or aqueous solution). Anode(+) : oxidation occurs. Anions go the anode and are discharged. X - X + e - Cathode (-) : reduction occurs. Cations go to the cathode and are discharged. M + + e - M Only 1 element at a time can be deposited at each electrode. E.g : electrolysis of molten NaCl. At anode : 2Cl - (l) Cl 2 (g) + 2e - (oxidation) At cathode : Na + (l) + e - Na(s)(reduction)
Selective discharge of ions. Factors affecting selective discharge during electrolysis : 1. Position in the Redox series. Cations with the most positive (or less negative) E θ value is discharged first. Anions with the most negative (or less positive) E θ value is discharged first. E.g : What is discharged during electrolysis of a mixture of CuSO 4 (aq) and NaI(aq)?
2. Relative concentration of ions. Ions of higher concentration will be discharged. E.g 1 : Electrolysis of brine (concentrated NaCl). At the anode: Concentration of Cl - ion is much higher. Hence Cl - ion is discharged at the anode instead of OH - ions. Hydrogen ions are discharged at the cathode instead of Na. The E θ value of the forward reaction is negative, therefore the reaction is not feasible. In electrolysis of dilute NaCl, the OH - ion is discharged instead (OH - ion is more negative) 3. Nature of the electrodes.
Quantitative Electrolysis Faradays first law :Q = I x t Q = amount of electric charge that flows through an electric circuit, unit : C. I = current, unit : A T = time, unit : s The mass of substance discharged at an electrode in electrolysis is directly proportional to the amount of electric charge. m α Qm = mass of substance liberated, unit : g m α I x t
Example When a current of A is passed through an electrolyte for 30 minutes at 25°C, 2.00 g of metal is produced at the cathode. What mass of metal would be produced at the cathode if: a) A current of A is passes through the electrolyte for 30 minutes at 25°C. b) A current of A is passes through the electrolyte for 2.0 hours at 25°C. c) A current of A is passes through the electrolyte for 30 minutes at 60°C.
Faraday constant, F, is the amount of charge on 1 mole of electrons. F = L x e where L = Avogadro constant e = charge on the electron (1.60 x C) 1 F = C mol -1 Faradays second law :M n+ (aq) + ne+ M(s) nF of charge is needed to form 1 mole of M(s) Questions: g of metal samarium (Sm) was discharged by electrolysis with a current of 2.4 A flowing for s. What is the formula of samarium ions? [Sm = 150] Hint : I mole of electron gives 1 F.
2. A current of 4.00 A was passed through copper(II) sulphate solution 965 s. what mass of copper was discharged at the cathode? [F = C mol-1; Cu = 63.5]
3. Calculate the volume of oxygen gas produced in cm 3 (measured at s.t.p) when 2.00 A is passed through dilute H 2 SO 4, for 1830 s. using platinum electrodes.
4. A current was passed through the circuit shown in figure above. In the experiment, 5.4 g of silver metal was discharged at the cathode in electrolysis cell B. What mass of copper was discharged at the cathode in cell A? [Cu = 63.5; Ag = 108]
The Avogadros constant (L) can be found from the results of an electrolysis experiment using the relationship, F = L x e e.g : Nov 2008, paper 5.
Method : The cathode is cleaned and weighed before being placed in the copper(II) sulphate solution. The circuit is completed and the current set at 0.3 A by adjusting the variable resistor. The current is maintained at 0.3 A for exactly 40 minutes at which point the circuit is broken. The cathode is removed from the solution and carefully washed with distilled water to remove any copper(II) sulphate solution. Distilled water is removed from the cathode by rinsing it with propanone in which the water dissolves. The cathode is finally dried by allowing the propanone to evaporate from its surface. The cathode is reweighed and placed back in the solution. A constant current of 0.3 A is passed for a further 40 minutes when the rinsing, drying and weighing are repeated.
Initial mass of cathode : g Mass of cathode after 40 min : g Mass of copper metal discharged :_______g Calculate a value for the Avogadro constant (L). [charge on the electron = 1.60 x C; Cu = 63.5]