2 Electrochemical cells

Electrochemical cells •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 ). •These two half cells will produce a small voltage if connected into a circuit. (i.e. become a Battery or cell). Salt Bridge 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 because chloride ions can form complexes with copper ions. A wire is not used because the metal wire would set up its own electrode system with the solutions. Why does a voltage form? In the cell pictured above 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. This potential difference is measured with a high resistance voltmeter, and is given the symbol E. The E for the above cell is E= +1.1V. Why use a High resistance voltmeter? 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 from flowing. What happens if current is allowed to flow? If the voltmeter is removed and replaced with a bulb or if the circuit is short circuited, a current flows. The reactions will then occur separately at each electrode. The voltage will fall to zero as the reactants are used up. The most positive electrode will always undergo reduction. Cu2+ (aq) + 2e-  Cu(s) (positive as electrons are used up) The most negative electrode will always undergo oxidation. Zn(s)  Zn2+ (aq) + 2e- (negative as electrons are given off)
 
Cell Diagrams Electrochemical cells can be represented by a cell diagram: Zn(s) | Zn2+ (aq) | | Cu2+ (aq) | Cu (s) E= +1.1V Most oxidised form is put next to the double line • 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) Systems that do not include metals. If a system does not include a metal that can act as an electrode, then a platinum electrode must be used and included in the cell diagram. It provides a conducting surface for electron transfer A platinum electrode is used because it is unreactive and can conduct electricity. e.g. for Fe2+ (aq)  Fe3+ (aq) + e- there is no solid conducting surface, a Pt electrode must be used. The cell diagram is drawn as: | | Fe3+ (aq), Fe2+ (aq) |Pt Still with more oxidised form near double line A comma separates the oxidised from the reduced species. If the system contains several species e.g. MnO4 – + 8H+ + 5e- Mn2+ + 4H2O then in the cell diagram the balancing numbers, H+ ions and H2O can be left out. | |MnO4 – , Mn2+ |Pt or if on left hand side Pt | Mn2+ , MnO4 – | | If a half equation has several physical states then the solid vertical line should be used between each different state boundary. 4e- + 2H2O (l) +O2 (g) 4OH- (aq) | | O2 | H2O, OH- | Pt Cl2 (g) + 2e-  2Cl- (aq) | | Cl2 | Cl- | Pt As the phase line also separates the oxidised and reduced terms a comma is not necessary here. Calculating the EMF of a cell Mg(s) | Mg2+ (aq) | | Cu2+ (aq) | Cu (s) E= +1.1V In order to calculate the Ecell, we must use ‘standard electrode potentials’ for the half cells. Each half cell has a standard electrode potential value Mg2+ (aq)| Mg(s) E= -2.37V Cu2+ (aq) | Cu (s) E = +0.34V For the cell diagram above Ecell = 0.34 – -2.37 = + 2.71 V use the equation Ecell= Erhs – Elhs