# Galvanic cell and Nernst equation

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1 Galvanic cell and Nernst equation

2 Galvanic cell Some times called Voltaic cell Spontaneous reaction redox reaction is used to provide a voltage and an electron flow through some electrical circuit When metallic zinc is dipped into a solution of copper sulfate: dark brown, spongy layer of metallic copper forms on the zinc.

3 Galvanic cell The reaction can be written as: Cu 2+ (aq) + Zn(s) Cu(s) + Zn 2+ (aq) The blue color of copper sulfate gradually disappeared and replaced by the colorless zinc sulfate. This reaction will not produce any flow of electrons as long as it occurs at the surface of the zinc.

4 The flow of the electrons can be achieved by making the oxidation and reduction halfreactions occur in separate compartments of a galvanic cell. Galvanic cell

5 Galvanic cell Each compartment called half-cell. Complete isolation of the two species would lead to an electrical imbalance Why?

6 Galvanic cell Explanation: on the left side Zn ions entering the solution give the sloution overall positive charge. This would prevent additional Zn ions to enter the solution. On the right side, Cu ions leave the solution, the SO4 ions left behind would give the solution negative charge and the electrode becomes positively charge. This would cause the electrode to repel Cu ions. continue

7 Galvanic cell The flow of current accompanied by continues electrical activity therefore the solution around the electrodes must be kept electrically neutral (even Zn leave the electrode compartment or anions enter to it; and cations enter the compartment of Cu to balance the charge of SO4 or SO4 ions must leave. How this can be achieved??? continue

8 Galvanic cell The salt bridge and the porous partition allow the slow mixing of the ions in the two solution.

9 Galvanic cell Salt bridge usually a tube filled with an electrolyte such as KNO3 or KCl in gelatin. Cations from the salt bridge can move into one compartment to compensate for the excess of negative charge while anions from the salt bridge diffuse into the other compartment to neutralize the excess of positive charge.

10 The signs of the electrodes in galvanic cells Oxidation occurs at the anode and reduction occurs at the cathode in both galvanic cell and electrolyte cell. What about the sign of the electrode at the anode and cathode??????

11 The signs of the electrodes in galvanic cells In the galvanic cell: anode has a negative charge and cathode has positive charge this is the opposite to the electrolyte solution.

12 Cell potentials The force that moves the electrons through the cell is called electromotive force (emf) and measured as volts (V). Emf is the passage of 1 coulomb is able to accomplish 1 joule of work. (1 volt = 1joule / coulomb) The emf in the galvanic cell is called cell potential (E cell )

13 Cell potentials The potential of the cell depends on the concentrations of the ions, the temperature, and the partial pressure of any gases that might be involved. Standard cell potential (E o cell) is measured when the concentrations of the ions are 1 M, the temperature is 25 C and the partial pressures of the gases are 1 atm. The cell potential is measured by a potentiomenter

14 Cell diagram This allow us the describe the galvaic cell to give informations: 1. The nature of the electrode materials 2. The nature of the solution in contact with the electrodes (including the concentrations of the ions 3. which of the half cell is anode and cathode 4. The reactants and the products in each of the half cells In the previous fig suppose the concentrations of both Zn and cu ions are 1 M Zn(s) Zn 2+ (1M) Cu 2+ (1M) Cu(s)

15 Reduction potential What is the origin of the cell potential?? In the cell we have two solution; one contains Zn ion and the other Cu ions. Each of these ions (zn and Cu) has a certain tendency to acquire electrons from its respective electrode and become reduced. Zn 2+ (aq) + 2e - Zn(s) Cu 2+ (aq) + 2e - Cu(s)

16 Reduction potential The larger reduction potential the larger tendency to undergo reduction When cell reaction takes place there is tug of war as each species attempts to pull electrons from its electrode so as to become reduced. The species with greater tendency to acquire electrons (greater reduction potential) wins the war. The loser must supply the electrons (oxidized)

17 Reduction potential The potential that we measure arises from the differences in the tendency of the two ions to become reduced and is equal to: E o cell = E o substances reduced - E o substances oxidized There fore for the previous reaction E o cell = E o Cu2+ Cu(s) - E o Zn2+ Zn

18 Reduction potential Experimentally we can measure the overall cell potentials but can be calculated. We use standard hydrogen electrode which (arbitrarily) sets its reduction potential of zero volts. It consists of a platinum wire encased in a glass sleeve through which hydrogen gas is passing at pressure of 1 atm. The platinum wire is attached to a platinum foil that is coated with a black velvety layer of finely divided platinum.

19 Reduction potential Standard Hydrogen Electrode The convention is to select a particular electrode and assign its standard reduction potential the value of V. This electrode is the Standard Hydrogen Electrode. 2H + (aq) + 2e H 2 (g) H 2 Pt The standard aspect to this cell is that the activity of H 2 (g) and that of H + (aq) are both 1. This means that the pressure of H 2 is 1 atm and the concentration of H + is 1M, given that these are our standard reference states. H +

20 Reduction potential E o H+ H2 =0.00V, means any substance that is more easily reduced than H + has a positive value of E o. What could happened if hydrogen electrode is paired with another half-cell? If the reduction potential of the species is greater than that for hydrogen electrode oxidation at hydrogen electrode H 2 (g) 2H + (aq) + 2e -

21 Reduction potential If the reduction potential of the species is less than that for hydrogen electrode reduction at hydrogen electrode 2H + (aq) + 2e - H 2 (g)

22 Reduction potential when connecting a voltmeter, connect the positive terminal to the positive electrode. If it reads a positive potential, you have correctly identified all the terminals. If you read a negative potential, then you have misidentified the reactions in the cells, and you have hooked it up backwards. Reverse your assignment of anode and cathode. in a galvanic cell the cathode is +ive in an electrolytic cell the cathode is ive.

23 Reduction potential Suppose we connect both Cu and H electrode: to obtain proper reading (+) terminal should connected to Cu electrode and (-) terminal to H electrode. Cu 2+ (aq) + 2e - Cu(s) H 2 (g) 2H + (aq) + 2e - (cathode) (anode) E o cell = E o Cu2+ Cu - E o H+ H V = E o Cu2+ Cu 0.00 V E o Cu2+ Cu = V

24 Reduction potential If both Zn and H electrode are connected then: Positive terminal is for H electrode Negative terminal is for Zn electrode 2H + (aq) + 2e - H 2 (g) (cathode) Zn(s) Zn 2+ (aq) + 2e - (anode) E o cell = E o H+ H2 - E o zn2+ zn 0.76 V = V - E o zn2+ zn V E o zn2+ zn = V E o negative means Zn is more difficult to reduce than H +

25 Reduction potential To calculate E o cell For Cu and Zn cell E o cell= E o Cu2+ Cu - E o Zn2+ Zn E o cell=+0.34 V (-0.76 V) E o cell= V This value is precisely the value that we observe experimentally Bear in mind, E o cell positive means spontaneous reaction

26 More Example Fe e Fe V e V Sn e Sn Ag + + e Ag To get a final positive cell potential, the more negative half-reaction (V) must act as the anode. More negative potential reaction is the anode. Multiply the Ag reaction by 2, but don t modify the cell potential. Fe 2+ + V Fe + V 2+ 2 Ag + + Sn 2 Ag + Sn 2+ E cell = (-1.19) = V E cell = (-0.14) = V

27 Standard Potential Tables All of the equilibrium electrochemical data is cast in Standard Reduction Potential tables. F 2 + 2e 2F Co 3+ + e Co Au + + e Au Ce 4+ + e Ce Br 2 + 2e 2Br Ag + + e Ag Cu e Cu AgCl + e Ag + Cl Sn e Sn H + + 2e H Pb e Pb Sn e Sn In e In Fe e Fe Zn e Zn V e V Cs + + e Cs Li + + e Li -3.05

28 Oxidative Strength F 2 + 2e 2F Co 3+ + e Co Au + + e Au Ce 4+ + e Ce Br 2 + 2e 2Br Ag + + e Ag Cu e Cu AgCl + e Ag + Cl Sn e Sn Consider a substance on the left of one of these equations. It will react as a reactant with something below it and on the right hand side. higher in the table means more likely to act in a reducing manner. when something is reduced, it induces oxidation in something else. it is an oxidizing agent or an oxidant. F 2 is a stronger oxidant than Ag +. Cu 2+ is a weaker oxidant than Ce 4+.

29 Reductive Strength F 2 + 2e 2F Co 3+ + e Co Au + + e Au Ce 4+ + e Ce Br 2 + 2e 2Br Ag + + e Ag Cu e Cu AgCl + e Ag + Cl Sn e Sn Substances on the right hand side of the equations will react so as to be oxidized. LOWER in the table means a greater tendency to be oxidized. when oxidized, it induces reduction in something else. It is a reducing agent or reductant. Ag is a stronger reductant than Au. Co 2+ is a weaker reductant than Sn 2+

30 Gibbs and the Cell Potential Here we can easily see how this Gibbs function relates to a potential. By convention, we identify work which is negative with work which is being done by the system on the surroundings. And negative free energy change is identified as defining a spontaneous process. G T,P w electrical n F E Note how a measurement of a cell potential directly calculates the Gibbs free energy change for the process.

31 Gibbs and the Cell Potential Consider reaction with n (number of electron) =2 F (number of coulombs per mole electrons= C/mole e - ) E (cell potential) = +1.1 V Then G o =-212 kj

32 Calculation of thermodynamic constants from standard cell potential G o = -RT ln K R=8.314 J/mole G o = RT log K G o = -nfe o = = RT log K E o = (2.303 RT/nF) log K At T = 25, the quantity RT/F is constant And equal to (2.303x8.314x298)/ =0.0592V

33 Calculation of thermodynamic constants from standard cell potential Finally we can get log K = (ne o )/ 0.059

34 Nernst Equation G = G o + RT ln Q Q is the reaction quotient G = G o RT log Q -nfe = -nfe o RT log Q Rearrangement E = E o (2.303RT/nF) log Q E = E o (0.059/n) log Q

35 Nernst Equation Consider the reaction of both Cu and Zn ar discussed earlier : [Zn 2+ ] = o.4m and [Cu 2+ ] = 0.02M, what is E? E = E o (0.059/n) log Q Q = [Zn 2+ ] / [Cu 2+ ] and E o = 1.1V Then E = 1.1 (0.059/2) log [Zn 2+ ] / [Cu 2+ ] =1.06

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