ELECTROCHEMISTRY

ELECTROCHEMISTRY HAVING COMPLETED THIS TOPIC I CAN: describe the types of half cells discuss the idea of cell potential describe the use and make up of standard hydrogen electrode describe how electrode potentials are measured use the electrochemical series combine half cells use cell diagrams explain the uses of E° values

TYPES OF HALF CELL METALS IN CONTACT WITH SOLUTIONS OF THEIR IONS These are systems involving oxidation or reduction METALS IN CONTACT WITH SOLUTIONS OF THEIR IONS Reaction Cu2+(aq) + 2e¯ Cu(s) Electrode copper Solution Cu2+(aq) (1mol L-1) - 1mol L-1 copper sulfate solution Potential + 0.34V Half cell Cu2+/Cu Reaction Zn2+(aq) + 2e¯ Zn(s) Electrode zinc Solution Zn2+(aq) (1mol L-1) - 1mol L-1 zinc sulfate solution Potential - 0.76V Half cell Zn2+/Zn

TYPES OF HALF CELL GASES IN CONTACT WITH SOLUTIONS OF THEIR IONS These are systems involving oxidation or reduction GASES IN CONTACT WITH SOLUTIONS OF THEIR IONS Reaction 2H+(aq) + 2e¯ H2(g) Electrode platinum Solution H+(aq) (1mol L-1) - 1mol L-1 HCl or 0.5mol L-1 H2SO4 Gas hydrogen ( 1 atm pressure ) Potential 0.00V Half Cell 2H+(aq), H2(g) / Pt (s) Reaction Cl2(g) + 2e¯ 2Cl¯(aq) Solution Cl¯(aq) (1mol L-1) - 1mol L-1 sodium chloride Gas chorine ( 1 atm pressure ) Potential + 1.36V Half Cell Cl2(g), 2Cl-(aq) / Pt (s)

TYPES OF HALF CELL SOLUTIONS OF IONS IN TWO DIFFERENT OXIDATION STATES These are systems involving oxidation or reduction SOLUTIONS OF IONS IN TWO DIFFERENT OXIDATION STATES Reaction Fe3+(aq) + e¯ Fe2+(aq) Electrode platinum Solution Fe3+(aq) (1mol L-1) and Fe2+(aq) (1mol L-1) Potential + 0.77 V Half Cell Fe3+(aq),Fe2+(aq) / Pt (s) SOLUTIONS OF OXIDISING AGENTS IN ACID SOLUTION Reaction MnO4¯ + 8H+(aq) + 5e¯ Mn2+(aq) + 4H2O(l) Solution MnO4¯(aq)(1mol L-1) and Mn2+(aq) (1mol L-1) and H+(aq) Potential + 1.52 V Half Cell MnO4-(aq),Mn2+(aq) / Pt (s)

CELL POTENTIAL Each electrode / electrolyte combination has its own half-reaction which sets up a potential difference The value is affected by ... TEMPERATURE PRESSURE OF ANY GASES SOLUTION CONCENTRATION Measurement it is impossible to measure the potential of a single electrode… BUT... you can measure the potential difference between two electrodes it is measured relative to a reference cell under standard conditions The reference used is the STANDARD HYDROGEN ELECTRODE.

THE STANDARD HYDROGEN ELECTRODE SOLUTION OF 1 mol L-1 H+(aq) The reference used is the STANDARD HYDROGEN ELECTRODE. HYDROGEN GAS AT 1 ATMOSPHERE PRESSURE 298K (25°C) SOLUTION OF 1 mol L-1 H+(aq) e.g. 1mol L-1 HCl or 0.5 mol L-1 H2SO4 PLATINUM ELECTRODE The standard hydrogen electrode is assigned an E° value of 0.00V.

MEASUREMENT OF E° VALUES SALT BRIDGE HYDROGEN (1 ATM) PLATINUM ELECTRODE HYDROCHLORIC ACID (1mol L-1) ZINC ZINC SULFATE (1mol L-1) In the diagram the standard hydrogen electrode is shown coupled up to a zinc half cell. The voltmeter reading gives the standard electrode potential of the zinc cell. conditions temperature 298K solution conc. 1 mol L-1 with respect to ions gases 1 atmosphere pressure salt bridge filled with saturated potassium chloride solution it enables the circuit to be completed by allowing the passage of ions from one half cell to the other. This avoids a build-up of charge.

THE ELECTROCHEMICAL SERIES E° / V F2(g) + 2e¯ 2F¯(aq) +2.87 MnO4¯(aq) + 8H+(aq) + 5e¯ Mn2+(aq) + 4H2O(l) +1.52 Cl2(g) + 2e¯ 2Cl¯(aq) +1.36 Cr2O72-(aq) + I4H+(aq) + 6e¯ 2Cr3+(aq) + 7H2O(l) +1.33 Br2(l) + 2e¯ 2Br¯(aq) +1.07 Ag+(aq) + e¯ Ag(s) +0.80 Fe3+(aq) + e¯ Fe2+(aq) +0.77 O2(g) + 2H+(aq) + 2e¯ H2O2(aq) +0.68 I2(s) + 2e¯ 2I¯(aq) +0.54 Cu+(aq) + e¯ Cu(s) +0.52 Cu2+(aq) + 2e¯ Cu(s) +0.34 Cu2+(aq) + e¯ Cu+(aq) +0.15 2H+(aq) + 2e¯ H2(g) 0.00 Sn2+(aq) + 2e¯ Sn(s) -0.14 Fe2+(aq) + 2e¯ Fe(s) -0.44 Zn2+ (aq) + 2e¯ Zn(s) -0.76 Layout If species are arranged in order of their standard electrode potentials you get a series that shows how good each substance is at gaining electrons. All equations are written as reduction processes ... i.e. gaining electrons A species with a higher E° value oxidise (reverses) one with a lower value REACTION MORE LIKELY TO WORK SPECIES ON LEFT ARE MORE POWERFUL OXIDISING AGENTS

THE ELECTROCHEMICAL SERIES E° / V F2(g) + 2e¯ 2F¯(aq) +2.87 MnO4¯(aq) + 8H+(aq) + 5e¯ Mn2+(aq) + 4H2O(l) +1.52 Cl2(g) + 2e¯ 2Cl¯(aq) +1.36 Cr2O72-(aq) + I4H+(aq) + 6e¯ 2Cr3+(aq) + 7H2O(l) +1.33 Br2(l) + 2e¯ 2Br¯(aq) +1.07 Ag+(aq) + e¯ Ag(s) +0.80 Fe3+(aq) + e¯ Fe2+(aq) +0.77 O2(g) + 2H+(aq) + 2e¯ H2O2(aq) +0.68 I2(s) + 2e¯ 2I¯(aq) +0.54 Cu+(aq) + e¯ Cu(s) +0.52 Cu2+(aq) + 2e¯ Cu(s) +0.34 Cu2+(aq) + e¯ Cu+(aq) +0.15 2H+(aq) + 2e¯ H2(g) 0.00 Sn2+(aq) + 2e¯ Sn(s) -0.14 Fe2+(aq) + 2e¯ Fe(s) -0.44 Zn2+ (aq) + 2e¯ Zn(s) -0.76 Application Chlorine is a more powerful oxidising agent - it has a higher E° Chlorine will get its electrons by reversing the iodine equation Cl2(g) + 2e¯ ——> 2Cl¯(aq) and 2I¯(aq) ——> I2(s) + 2e¯ Overall equation is Cl2(g) + 2I¯(aq) ——> I2(s) + 2Cl¯(aq) AN EQUATION WITH A HIGHER E° VALUE WILL REVERSE AN EQUATION WITH A LOWER VALUE

ELECTROCHEMICAL CELLS CELLS • electrochemical cells contain two electrodes • each electrode / electrolyte combination has its own half-reaction • the electrons produced by one half reaction are available for the other • oxidation occurs at the anode • reduction occurs at the cathode. ANODE Zn(s) ——> Zn2+(aq) + 2e¯ OXIDATION CATHODE Cu2+(aq) + 2e¯ ——> Cu(s) REDUCTION The resulting cell has a potential difference (voltage) called the cell potential which depends on the difference between the two potentials It is affected by ... • temperature • pressure of any gases • solution concentrations

A TYPICAL COMBINATION OF HALF CELLS 1.10V ZINC ZINC SULFATE (1 mol L-1) COPPER COPPER SULFATE (1mol L-1) E° = - 0.76V E° = + 0.34V zinc is more reactive - it dissolves to give ions Zn(s) ——> Zn2+(aq) + 2e¯ the electrons produced go round the external circuit to the copper electrode electrons are picked up by copper ions Cu2+(aq) + 2e¯ ——> Cu(s) As a result, copper is deposited overall reaction Zn(s) + Cu2+(aq) ——> Zn2+(aq) + Cu(s)

Zn Zn2+ Cu2+ Cu CELL DIAGRAMS These give a diagrammatic representation of what is happening in a cell. • Place the cell with the more positive E° value on the RHS of the diagram. Cu2+(aq) + 2e¯ Cu(s) E° = + 0.34V put on the RHS Zn2+(aq) + 2e¯ Zn(s) E° = - 0.76V put on the LHS ZINC IS IN CONTACT THE SOLUTIONS A SOLUTION OF WITH A SOLUTION ARE JOINED VIA A COPPER IONS IN OF ZINC IONS SALT BRIDGE CONTACT WITH COPPER Zn Zn2+ Cu2+ Cu

+ _ Zn Zn2+ Cu2+ Cu CELL DIAGRAMS These give a diagrammatic representation of what is happening in a cell. • Place the cell with the more positive E° value on the RHS of the diagram. Cu2+(aq) + 2e¯ Cu(s) E° = + 0.34V put on the RHS Zn2+(aq) + 2e¯ Zn(s) E° = - 0.76V put on the LHS • Draw as shown… the cell reaction goes from left to right • the zinc metal dissolves Zn(s) ——> Zn2+(aq) + 2e¯ OXIDATION • copper is deposited Cu2+(aq) + 2e¯ ——> Cu(s) REDUCTION • oxidation takes place at the anode • reduction at the cathode _ Zn Zn2+ Cu2+ Cu +

cell voltage = +0.34V - (-0.76V) = +1.10V CELL DIAGRAMS These give a diagrammatic representation of what is happening in a cell. • Place the cell with the more positive E° value on the RHS of the diagram. Cu2+(aq) + 2e¯ Cu(s) E° = + 0.34V put on the RHS Zn2+(aq) + 2e¯ Zn(s) E° = - 0.76V put on the LHS • Draw as shown… the electrons go round the external circuit from left to right • electrons are released when zinc turns into zinc ions • the electrons produced go round the external circuit to the copper • electrons are picked up by copper ions and copper is deposited cell voltage = +0.34V - (-0.76V) = +1.10V _ Zn Zn2+ Cu2+ Cu + V

USE OF Eo VALUES - WILL IT WORK? E° values Can be used to predict the feasibility of redox and cell reactions In theory ANY REDOX REACTION WITH A POSITIVE E° VALUE WILL WORK In practice, it proceeds if the E° value of the reaction is greater than + 0.40V An equation with a more positive E° value reverse a less positive one

USE OF Eo VALUES - WILL IT WORK? An equation with a more positive E° value reverse a less positive one What happens if an Sn(s) / Sn2+(aq) and a Cu(s) / Cu2+(aq) cell are connected? Write out the equations Cu2+(aq) + 2e¯ Cu(s) ; E° = +0.34V Sn2+(aq) + 2e¯ Sn(s) ; E° = -0.14V the half reaction with the more positive E° value is more likely to work it gets the electrons by reversing the half reaction with the lower E° value therefore Cu2+(aq) ——> Cu(s) and Sn(s) ——> Sn2+(aq) the overall reaction is Cu2+(aq) + Sn(s) ——> Sn2+(aq) + Cu(s) the cell voltage is the difference in E° values... (+0.34) - (-0.14) = + 0.48V

USE OF Eo VALUES - WILL IT WORK? An equation with a more positive E° value reverse a less positive one Will this reaction be spontaneous? Sn(s) + Cu2+(aq) ——> Sn2+(aq) + Cu(s) Write out the appropriate equations Cu2+(aq) + 2e¯ Cu(s) ; E° = +0.34V as reductions with their E° values Sn2+(aq) + 2e¯ Sn(s) ; E° = - 0.14V The reaction which takes place will involve the more positive one reversing the other i.e. Cu2+(aq) ——> Cu(s) and Sn(s) ——> Sn2+(aq) The cell voltage will be the difference in E° values and will be positive... (+0.34) - (- 0.14) = + 0.48V If this is the equation you want then it will be spontaneous If it is the opposite equation (going the other way) it will not be spontaneous

USE OF Eo VALUES - WILL IT WORK? An equation with a more positive E° value reverse a less positive one Will this reaction be spontaneous? Sn(s) + Cu2+(aq) ——> Sn2+(aq) + Cu(s) Split equation into two half equations Cu2+(aq) + 2e¯ ——> Cu(s) Sn(s) ——> Sn2+(aq) + 2e¯ Find the electrode potentials Cu2+(aq) + 2e¯ Cu(s) ; E° = +0.34V and the usual equations Sn2+(aq) + 2e¯ Sn(s) ; E° = - 0.14V Reverse one equation and its sign Sn(s) ——> Sn2+(aq) + 2e¯ ; E° = +0.14V Combine the two half equations Sn(s) + Cu2+(aq) ——> Sn2+(aq) + Cu(s) Add the two numerical values (+0.34V) + (+ 0.14V) = +0.48V If the value is positive the reaction will be spontaneous

THE END