topic 14: redox II

• a cell has two half–cells • the two half cells have to be connected with a salt bridge • simple half cells will consist of a metal (acts an electrode) and a solution of a compound containing that metal (eg Cu and CuSO4) - 1 moldm-3 • these two half cells will produce a small voltage if connected into a circuit. (i.e. become a battery or cell) • standard conditions • electrodes connected to a high resistance voltmeter • LHS - oxidised • RHS - reduced

• when connected together the zinc half-cell has more of a tendency to oxidise to the Zn2+ ion and release electrons than the copper half-cell (Zn Zn2+ + 2e-) • more electrons will therefore build up on the zinc electrode than the copper electrode • a potential difference is created between the two electrodes. the zinc strip is the negative terminal and the copper strip is the positive terminal

• the voltmeter needs to be of very high resistance to stop the current from flowing in the circuit. in this state it is possible to measure the maximum possible potential difference (E). • the reactions will not be occurring because the very high resistance voltmeter stops the current fromflowing

the salt bridge is used to connect up the circuit. the free moving ions conduct the charge

• a salt bridge is usually made from a piece of filter paper (or material) soaked in a salt solution, usually potassium nitrate • the salt should be unreactive with the electrodes and electrode solutions ——-> e.g. potassium chloride would not be suitable for copper systems as chloride ions can form complexes with copper ions

because the metal wire would set up its own electrode system with the solutions and wires do not allow the flow of ions

a current flows. the reactions will then occur separately at each electrode. the voltage will fall to zero as the reactants are used up

cell diagram Zn(s) | Zn2+ (aq) | | Cu2+ (aq) | Cu (s)

• the solid vertical line represents the boundary between phases e.g. solid (electrode) and solution (electrolyte) * the double line represents the salt bridge between the two half cells • the voltage produced is indicated • the more positive half cell is written on the right if possible (but this is not essential)

• platinum electrode • provides a conducting surface for electron transfer • a platinum electrode is used because it is unreactive and can conduct electricity

• it is not possible to measure the absolute potential of a half electrode on its own. it is only possible to measure the potential difference between two electrodes • to measure it, it has to be connected to another half-cell of known potential, and the potential difference between the two half-cells measured • by convention we can assign a relative potential to each electrode by linking it to a reference electrode (hydrogen electrode), which is given a potential of zero volts

comparing their potential to that of the standard hydrogen electrode

• hydrogen gas at pressure of 100 kPa • solution containing the hydrogen ion at 1.00 mol dm-3 (solution is usually 1 mol dm-3 HCl) • temperature at 298K

• on the right hand side, the electrode getting reduced is drawn in 1 mol dm-3 of a corresponding solution - anode - positive terminal • on the left hand side, the electrode is getting oxidised in 1mol dm-3 of the corresponding solution - cathode - negative terminal • salt bridge between each beaker high resistance voltmeter drawn • the direction of current goes from LHS to RHS WRITE OUT THE STATNDARD CONDITIONS

when an electrode system is connected to the hydrogen electrode system, and standard conditions apply the potential difference measured

E(rhs) - E(lhs)

it is oxidised so it goes the opposite way

tendency for species on left to reduce, and act as oxidising agents

tendency for species on right to oxidise, and act as reducing agents

total entropy change and lnK positive Ecell will lead to a positive total entropy change

a measure of how far from equilibrium the cell reaction lies. the more positive the Ecell the more likely the reaction is to occur

as the concentration of the reactants increases so will the Ecell

most cells are exothermic in the spontaneous direction so applying Le Chatelier to a temperature rise to these would result in a decrease in Ecell because the equilibrium reactions would shift backwards

uses the energy from the reaction of a fuel with oxygen to create a voltage