Electrochemical Cells

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Samuel Wiggins
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Chapter 11 Electrochemical Cells work to be done. Zn gets oxidized and its standard reduction potential is ve (076 V) while that of Cu is +ve (+034 V). The standard cell potential is therefore E =1.

1 Chapter 11 Electrochemical Cells work to be done. Zn gets oxidized and its standard reduction potential is ve (076 V) while that of Cu is +ve (+034 V). The standard cell potential is therefore E =1.10 V. Electrochemical cells can be divided into two types a) galvanic, which converts chemical energy (association sector) into electrical work and b) electrolytic, in which The former can be used to charge a battery (it is in fact a battery) and the latter uses electrical energy to drive a reaction. One can move from a) galvanic to b) electrolytic by simply changing the external potential, and by finding the balance point, determine the cell potential, which is just an electron intensive version of the free energy change. To understand how this is done, we have to understand what happens when, e.g. a metal such as Zn(s) is immersed in a ZnSO 4 solution. A tiny amount of the Zn(s) dissolves leaving the e s on the metal. Thus the metal builds up a slight ve charge and the surrounding solution builds up a slight +ve charge. While the total amount of charge is tiny, the built-up voltages can be significant ( 1 ) and thus the the work required to move charged species is not the standard chemical potential We have to work with a new quantity which accounts for theworkrequiredtomovecharge(eithere s or charged ions) in a medium with a spacially varying electrostatic potential Consider the Daniell cell: Zn(s) ZnSO 4 (aq) CuSO 4 (aq) Cu. The denotes a phase boundary and the asalt bridge. The latter is inserted to make the process quasistationary and thus allow for the maximum (non-pv) ig. Galvanic (chemical electrical) Electrochemical potential Consider the following sequence of simple experiments. 1. Bring together two dissimilar metals (A and B). The level diagrams indicate the delocalized electron quantum levels (particle in a box levels). The top most energy, the energy required to add or subtract one e,is which is also called the ermi energy, As the metals are different, it is extremely unlikely that the ermi energies are equal, as a result there will be a charge transfer. With a transfer of e from A, theasidebecomes+veandthebside ve. This decreases (increases) the (electro) chemical potential corrected for the electrostatic potential for e s on the A (B) side yielding = 81

2 ig. Dissimilar metals 3. Now separate the metals ig. Sep. diss. metals+conn.+vm + = + As = 1 = ( ) Lesson: The difference in chemical potentials induces a matter motion which induces electrical potential difference. When The (+ve) potential will be greater on side A, thus reducing the energy of e, exactly compensating for the greater chemical potential. The electrochemical potential is + (11.1) This definition amounts to subsuming the electrical work sector into an enhanced (pseudo) chemical sector. Note that if the species is neutral ( =0)or the potential is zero, =0 = The charge carried by one mole of charge is araday s constant =96485 [ ]. Now add connections ( and R via identical metal, thus they posses the same electron chemical potentials) to a high impedance voltmeter. ig. Dissimilar metals+connections+vm = ; = = = but as = + = + ( )=[ ] ( ) ( )= [ ] +( ) voltmeter = chemical + electrical. 4. It is the electrochemical difference which dictates material flow. The equilibrium condition becomes = P =0 This can be viewed as the combination of the old chemical sector n and the new electrical potential sector Of course, the new sector only is present if the charge (z) is finite. 11. Conventions, Notation and Standard states 1. The work required to move a charge = from a phase to another through a potential difference is =( )[] =( )[]. The electromotive force is defined as the potential difference measured between stationary thermodynamic states. (recalling: non-pv = ) = = = = = Lesson: Connecting different metals via similar wires to a voltmeter will read zero (unless the sides are different temperatures.) = [ ] (11.) 3. Consider the Daniell cell : Zn Zn ++ Cu ++ Cu (a) Strong electrolytes (sulfates in this case) are used to from the ions in the half cells. (b) The signifies a phase boundary across which charge can be transferred. 8

3 (c) Salt bridge ( ), between half cells, allows charge transport without mixing of the cells. It does this by having a great reservoir of electrolyte (such as KCl) suspended in a gel. This electrolyte diffuses into the cells, and the electrolyte in each half cell, only encroaches slightly into the salt bridge. (The bridge also removes the so called junction potential caused by the boundary between dissimilar solutions.) (d) The circuit is completed with an external wires (of the same material) connected to a voltmeter. (e) An equilibrium between the metal and the ion is established. In the case of Zn, some more of the Zn dissolves leaving the metal -ve and the solution +ve. This potential (of the 1/ cell) cannot be measured by sticking the second electrode in the solution. 4. Electrochemical references (a) If we use identical metals on the two sides (to connect to the voltmeter) we can take the reference electron chemical potential as zero, 0 thus (b) or the neutral metal atoms: = (11.3) = (11.4) (c) or metal ions, once eq. has been established, M + +ze thus =( +)+ =( + + ) = + (11.5) This statement (consequence of 0) is also true for the reference state, = = + 0 (d) Returning to the electrochemical potential of the ion, =0 = (11.6) 5. To set this reference (of = 0, since the absolute is unmeasureable) we choose H + under the conditions of the Standard Hydrogen Electrode. ig. SHE 6. (a) The cell is assembled by bubbling H (g), with P (really f) = 1 bar over a Pt electrode immersed in a H + solution of activity + =1. (b) The reaction and equilibrium condition are: + ()+ 1 () [ + ()] + = 1 () using our reference [ +()] + = 1 () (c) Using the standard states and the offsets for concentrations and nonidealities, the eq. condition becomes [ + ln + +] + = 1 ( ()+ ln ) (d) Thus the potential becomes + = ( + + ln + )[ 1 ( + ln )] = ln + (e) With a reference state of unit activities and fugacities, the ln term vanishes. urthermore, as H (g) is the reference state for H,thereference potential is + = + 1 = + (f) With the reference chemical potential of H + [ (+ )= + () =1 0] and thus we are consistent. + 0 (11.7) 11.3 Reactions & the Nernst Eq. Rather than working the example in the text (the Daniell cell), I will work though the thermodynamics with a different cell. 1. The half reactions are (a) Ox : ++ + ; [ ]=07618 (b) Red:()+ ()+ ; ( )=033 83

4 . How much can we get from this cell? = + P = + P where theextentofreactionisdefined by = 1 = + = +( + ) thus + P = +( + ) =( )+ P =( ) P However from the second law (..)5 0(Recall = ) Therefore P 5 = = () Lesson: the electrical work done BY a cell is equal to the decrement in the Gibbs.E. when done reversibly and less otherwise. 3. The above is true for standard and actual conditions. Therefore, = + ln{ ++ } and as X is 1 in a pure solid, = + ln{ ++ } = + ln{ ++ } = ln{ ++ } More generally, one has the Nernst Equation = ln (11.8). = ln{ } = 1 as = =1 ln{ } 3. Therefore, if we know and measure = = + =( + +)( ) = ±± 4. Thus, = ( )ln{ ±±} = ()ln{ ± } ()ln{ ± } This can be rearranged to get [ +()ln{ ± }] 1 measure =[ ()ln{±}] deduce Noting that in the limit lim 0 (with no other ions in solution), (a) ln ± ± (b) lim 0 [] = Aplotof =[ +()ln{ ± }] 1 vs ± will extrapolate to as 0 5. With and ( ) you can calculate ± ( ) = ()ln{ ±} ()ln{±} by recalling that ± =( + + ) 1 1:1 =1 1 = igs. Just an electron intensive rewrite of = + ln 4. or future reference, 98 = V Cell potentials How do you get the numbers in the table? Consider combining the AgCl Ag electrode (used above) with the SHE. Writing them out in the standard format, Oxidation reduction H (g) HCl(m ) AgCl Ag(s) 1. The half reactions are (a) Ox : 1 () + ()+ (b) Red: ()+ ()+ 1 (c) Net: ()+() ( )+ with = = 11.5 A few pointers & reminders 1. = and = +. [ ] e extensive [= ] e intensive 3. Thus when adding cells, Hess s rule becomes = + 84

5 4. This means when one is dealing with a OX-RED pair (which must be balanced in the e transfer) = + = + The E values ARE per unit e transferred NOT, like the energy functions, per unit of reaction! 11.6 The E ( ) As the cell EM s are just an e intensive measure of the increment function for E 0 are just a scaled version of those for However, one can also get and from E ( ) Lets see how. 1. = ( ) = ( ).. As = + = ( ) or = ( ) 3. = = + = + ( ) = [ ( ) ] external ( ) circuit one can measure the potential difference between the LHS and RHS. Recall that the EM is just an ( ) intensive version of the =. Therefore, if we can measure we can get Specifically, we will vary and measure against a constant LHS. Lets see how. 1. irst let me remind you how is related to Consider the for moving from the LHS =( ) ( ) =( + ln )( + ln ). In the amalgam case we have the same reference states (LHS and RHS). = ln 3. As = = = ln or = {ln +ln ln } 4. We can obtain values of or from the EM s by a simple graph. Rearranging the above equation, y +ln =ln ln ( ) for which the intercept at =0( =1 ln =0)isln and the difference from the intercept at finite is ln 11.7 *Another example Imagine two dilute solid solutions of Pb in Hg (with different mole fractions of Pb ( ) in contact with a common (but separated by a salt bridge) electrolyte. These solid solutions are called amalgams. The chemical potential of the s can be (and generally are) different in the two different solid solutions. This means that s will move from one side to the other in response to the generalized force mismatch (on the s) on one side versus the other. In addition, if the electrolyte consists of a Pb ++ salt, Pb ++ will migrate from one amalgam to the solution to the other amalgam, under the influence of the different = Of course, for this to happen, a path (external circuit) must be provided for the s. The salt bridge holds off attainmentof equilibrium and thus allows for the measurement of the. ollowing convention, we always write the half reaction (amalgam in this case) that provides the s on the LHS. This means that oxidation occurs on the left and reduction on the RHS. If one adds a galvanometer to the 85

CHM 213 (INORGANIC CHEMISTRY): Applications of Standard Reduction Potentials. Compiled by. Dr. A.O. Oladebeye

CHM 213 (INORGANIC CHEMISTRY): Applications of Standard Reduction Potentials Compiled by Dr. A.O. Oladebeye Department of Chemistry University of Medical Sciences, Ondo, Nigeria Electrochemical Cell Electrochemical