Electrochemical Cells

Electrochemical Cells Volta Cell Luigi Galvani (1780) and Alessandro Volta (1800) discovered the formation of electric current from chemical reactions. The chemical reaction that occurs is a redox reaction (reduction and oxidation) and this device is called a voltaic cell. 1. Process Copper metal was dipped in CuSO4 solution (1 M) and zinc metal was dipped in ZnSO4 solution (1 M). The two solutions are onnected by a salt bridge. The salt bridge is a U tube filled with agar and KCl salt. Meanwhile, the two electrodes (Cu metal and Zn metal) are connected to a current indicating device, namely a voltmeter. Zn metal will release electrons and change to form Zn2+ ions and combine in the ZnSO4 solution . Electrons flow from the Zn electrode to the Cu electrode. Cu2+ ions in CuSO4 solution accept electrons and these ions change to form Cu metal precipitates. The reaction can be described: Oxidation reaction : Zn Zn2+ + 2 e– Reduction reaction : Cu2+ + 2 e– Cu Net reaction in cells : Zn + Cu2+ Zn2+ + Cu Writing can be abbreviated as Zn | Zn2+ || Cu2+ || Cu 2. Electrodes on Volta Cells Cathode: - The electrode where the reduction reaction occurs, means Cu metal. - In voltaic cells it is called the positive electrode. Anode: - The electrode where the oxidation reaction occurs, means Zn metal. - In voltaic cells it is called the negative electrode. 3. Function of the Salt Bridge In the ZnSO4 solution there is an increase in the number of Zn2+ ions and in the CuSO4 solution there is a decrease in the number of Cu2+ ions . Meanwhile, the number of cations (Zn2+ or Cu2+) must be equivalent to anions. To balance cations and anions, Cl– anions enter the ZnSO4 solution from the salt bridge according to the increase in Zn2+ ions. In the CuSO4 solution, there is a shortage of Cu2+ or it can be called an excess of ions, so the ions enter the salt bridge to replace Cl– which enters the ZnSO4 solution . So, the function of the salt bridge is to balance the cations and anions in solution. 4. Electrode Potential The amount of electric current generated from the two electrodes above can be determined by determining the electrode potential of Zn and Cu. It's just that the electrode potential of a substance cannot stand alone, there must be a benchmark that becomes the standard. As a standard electrode, a hydrogen electrode is used. This electrode consists of pure hydrogen gas with a pressure of 1 atm at a temperature of 25 ºC which flows through a piece of platinum immersed in a solution containing 1 mol/ liter of H+ ions. The standard hydrogen electrode potential is given the value = 0 volts (Eº = 0 volts). Reaction: 2 H+ (aq) + 2 e– H2 (g); ΔH = 0 volts Eº = 0 volts According to international agreements, if a substance turns out to be easier to reduce than hydrogen, then the electrode potential value is positive. The reduction potential is positive. Cu 2+(aq) + 2 e– Cu(s); Eº = + 0.34 volts Ag+ (aq) + e– Ag(s); Eº = + 0.80 volts But if the substance turns out to be easier to carry out oxidation reactions than hydrogen, then the electrode potential value is negative. In this case the oxidation potential is positive, but because the electrode potential must be written as reduction, it means the reduction potential is negative. Zn2+(aq) + 2 e– Zn(s); Eº = 0.76 volts A13+(aq) + 3 e– A1(s); Eº = 1.76 volts So, the electrode potential is described by a reduction reaction. 5. Cell Potential Calculation The magnitude of the cell potential of a redox reaction in a voltaic cell is Calculate the total of the electrode potentials of the elements according to the reaction. In this case, the cell potential calculation result can be positive or can be negative. If the cell potential is positive, it means the reaction can take place. whereas if the cell potential is negative, it means the reaction cannot occur taking place. E°se1 = E°(+) – E°(–) with: E°(+) = electrode potential is more positive (larger) E°(–) = electrode potential is more negative (smaller) Example. 1. Given: Cu2+(aq) + 2 e– Cu(s) E° = + 0.34 V Zn2+(aq) + 2 e– Zn(s) E° = – 0.76 V Determine the cell potential (Eo cell) of the two electrodes! Answer: Because the Zn electrode potential is more negative, Zn must oxidize and acts as an anode. Cathode: Cu2+ + 2 e– Cu(s) Eo = + 0.34 Anode: Zn2+(aq) + 2 e– Eo = + 0.76 V Cu + Zn2+ Eo cell = +1.10 V Zn(s) Cu2+ + Zn