D ATE PERFORMED: 24 SEPTEMBER 2010, FRIDAY
ELECTROCHEMISTRY : ELECTRODE POTENTIAL LBANO A.M.P. T A.M.P. TONGCO AND A.M.G. A A.M.G. A LBANO
I NSTITUTE OF C HEMISTRY , C OLLEGE HEMISTRY OLLEGE OF S CIENCE CIENCE THE P HILIPPINES HILIPPINES ILIMAN ITY HILIPPINES U NIVERSITY OF THE , D ILIMAN , Q , QUEZON C ITY , P HILIPPINES R ECEIVED ECEIVED SEPTEMBER 31, 2010
ABSTRACT
The experiment aimed to determine the cell potentials of certain half-reactions with Cu 2+/Cu. A beaker containing a solution of copper ions and a copper electrode was coupled, using a salt bridge made of filter paper soaked in potassium nitrate, to another beaker containing a solution of a given ion or halide. The entire cell was then connected to a multimeter. The voltmeter application of the multimeter was used to measure the cell potential (E° cell ) for both half-reactions occ urring in the set -up. Using the Ner nst equation, the experimental E° cell can be found by substituting the cell potential measured by the voltmeter. The reduction potential of the non-copper half-cell can also be computed using the Nernst equation.
INTRODUCTION
The study of electrochemistry focuses on the chemical changes produced by electric current, and the production of electricity by chemical reactions. The reactions involved in electrochemistry are, predictably, called electrochemical reactions, and always involve the transfer of electrons. Electrochemical reactions, therefore, are reductionoxidation reactions. Oxidation-reduction reactions, better known as redox reactions, are reactions which occur when the reactants undergo a change in oxidation number. Oxidation occurs when a reactant loses electrons, increasing the oxidation number. A reactant which undergoes oxidation is known as the reducing agent. On the other hand, reduction occurs when a reactant gains electrons, decreasing the oxidation number. A reactant which undergoes reduction is known as the oxidizing agent. The spontaneity of the reactions that occur in a cell can be measured based on the magnitude of the cell’s potential. Redox reactions generate certain cell potentials – denoted as E° cell – and when they occur under standard conditions, this is also known as the standard cell potential.
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A reaction with a positive E° cell value will be spontaneous, while a reaction with a negative E° cell value means that the reaction in the reverse direction will be spontaneous. The higher the value of the E°reduction – or E° oxidation – the the higher the tendency for the respective reaction to occur. The E°cell of a reaction can be calculated using the reduction potentials of the two half-reactions which comprise the cell. E°cell = E°reduction(cathode) - E°reduction (anode) However, not all electrochemical reactions will occur under standard conditions. In these cases, the Nernst equation is used. Instead of using the reduction potentials of the two half-reactions, the Nernst equation relies on several aspects such as temperature, pressure, and the number of moles of electrons transferred in the reaction. Ecell = E°cell –
log Q
The Nernst equation also utilizes the value of Q, which is the ratio between products and reactants in the reaction. R is is the gas constant (8.314 J/mol K), T is
the temperature, n is the number of moles of electrons transferred, and F is Faraday’s constant (96,500 J/V e ). In the experiment, it is assumed that the atmospheric pressure in the room is 1 atm and that the temperature is 298 K. These values will be substituted into the Nernst equation for the calculation of cell potentials. The experiment set-ups made use of beakers containing solutions of copper, zinc, iron, bromine, chlorine, and iodine ions. These solutions were connected using a salt bridge, which was made from a strip of filter paper soaked in potassium nitrate (KNO3 ). The salt bridge is necessary for maintaining electroneutrality between the two solutions being tested. The copper solution used has a given half-cell potential of 0.34 V. The beaker containing the copper solution and its respective copper electrode was connected to the other beakers containing ion solutions and their respective electrodes. The electrodes were then connected to a multimeter – first set to voltmeter settings – in order to obtain the measurements of the cell potentials. Figure 1 below illustrates the set-up used in the first pair in the experiment:
Using the given E° cell of copper and the E° cell values obtained from the experiment, the E° red of the other ion solutions could be computed for, which was the main purpose of the experiment.
RESULTS AND DISCUSSION
Determination of the standard cell potentials of different cells was performed during the first part of the experiment. Two different types of cells were used during the experiment - the electrochemical cell and the electrolytic cell. The zinc and iron half-cells were paired up with a copper half-cell which made up electrochemical cells while the bromide, chloride and iodide half-cells were paired with copper and made up the electrolytic cells. There was no problem with setting up the electrochemical cells because half-cells that composed it would react spontaneously with each other. On the other hand, electrolytic cells had to undergo electrolysis in order to generate the halogens needed for a spontaneous reaction to occur. Electrolytic cells make use of external sources of electricity like batteries for the halogens to be generated. In this particular experiment, a 1.5V dry cell was used to do so. After generating the halogens, the halide half-cells were then connected to the copper half-cell. The copper (II) half-cell was used as the reference electrode because its standard reduction potential remained unchanged when paired with any other half-cell as long as temperature was kept constant. Its standard reduction potential stayed at a constant 0.34V.
Figure 1. Zinc-copper electrochemical reaction set-up. Zn(s) | ZnSO4 (0.1 M) || CuSO 4 (0.1 M) | Cu (s)
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Table I shows the different half-cells that were paired with the copper half-cell and which halfreactions acted as the cathodes or anodes. It also shows the cell potentials (voltmeter reading = Ecell) when the above-listed half-cells were paired up with the copper half-cell.
Ionic Equation (Reduction Half-cell) Zn2+ (aq) + 2e - --> Zn(s) Fe3+ (aq) + 1e- --> Fe2+ (aq) Br2(s) + 2e- --> 2Br - (aq) Cl2(g) + 2e- --> 2Cl- (aq) I2(g) + 2e- --> 2I- (aq)
Table I. Cell Potentials Positive Electrode Reference Anode Cu2+/Cu Cathode Fe3+/Fe2+ Anode 2+ Cu /Cu Anode Cl2/Cl Cathode I2/I Cathode
Voltmeter Reading (V) 0.96 0.41 0.47 0.11 0.11
Using the values of the cell potentials observed, the standard reduction potentials of each cell was calculated. This was done with the use of the Nernst equation. The results for the standard reduction potential of the cells are found in Table II.
Table II. Standard Reduction Cell Potentials Ionic Equations Cell Notation [1] Cathode Reaction [2] Anode Reaction [3] Cell Reaction [1] + 2e- --> Cu(s) Zn(s)| Zn2+(0.1M)|| Cu 2+ (0.1M)|Cu(s) [2] Zn2+ (aq) + 2e- --> Zn(s) [3] Cu2+ (aq) + Zn(s) --> Cu(s) + Zn 2+ (aq) [1] Fe3+ (aq) + 1e- --> Fe2+ (aq) 2+ 3+ 2+ Cu(s)| Cu (0.1M)||Fe (0.5M), Fe (0.5M)| C(s) [2] Cu2+ (aq) + 2e - --> Cu(s) [3] Cu(s) + 2Fe 3+ (aq) --> Cu2+ (aq) + 2Fe 2+ (aq) [1] Br2(s) + 2e - --> 2Br - (aq) [2] Cu2+ (aq) + 2e - --> Cu(s) 2+ Cu(s)| Cu (0.1M)||Br (1M)|Br2 (1atm)|C(s) [3] Cu(s) + Br2(g) --> Cu 2+ (aq) + 2Br - (aq) [1] Cu2+ (aq) + 2e- --> Cu(s) [2]Cl2(g) + 2e - --> 2Cl- (aq) C(s)|Cl2 (1atm)|Cl – (1M)|| Cu2+ (0.1M)|Cu(s) [3] Cu2+ (aq) + 2Cl- (aq) --> Cu(s) + Cl 2(g) [1] Cu2+ (aq) + 2e - --> Cu(s) [2] I2(g) + 2e- --> 2I- (aq) 2+ C(s)|I2 (1atm)|I (1M)|| Cu (0.1M)|Cu(s) [3] Cu2+(aq) + 2I- (aq) --> Cu(s) + I 2(g)
E cell, V ᵒ
0.960
0.380
0.381
0.199
0.199
The standard reduction potential of a cell is related to the spontaneity of the reaction. A positive value of E °cell means that the reaction takes place spontaneously while a negative value of E °cell means that that certain reaction takes place non-spontaneously. All the values in the table above are positive which means that they occur spontaneously. Although in the case of the copper-halide cells, they occur spontaneously because the halogens have already been generated with the use of electricity. Having been able to calculate the standard cell potentials, the standard reduction potentials of the half -reactions were calculated as well, with the use of the equation: E°cell = E°red, cathode - E°red, anode
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The results for the standard reduction potential of the half-cells are shown in Table III. Also shown in the table are the theoretical values of the standard reduction potentials of the half-cells. Table III. Experimental E°red vs. Theoretical E °red Half-Reaction Experimental E red Theoretical E red ᵒ
ᵒ
% Error
Zn(s) -> Zn2+(aq) + 2e-
-0.62
-0.763
18.7
Fe3+(aq) + 1e- -> Fe 2+(aq)
0.72
0.771
6.62
Br2(s) + 2e- -> 2Br-(aq)
0.72
1.08
33.3
2Cl-(aq) -> Cl2(g) + 2e-
0.14
1.36
89.7
2I-(aq) -> I2(g) + 2e-
0.14
0.535
62.6
It can be seen clearly that there are deviations between the experimental values and theoretical values. These deviations may be allotted to a number of things. One of which could be that an error was made when taking the readings from the voltmeter. Another possibility would be the errors done during the calculations. Another source of error may have been carelessness while performing the experiment. This could have allowed an inaccurate reading from the voltmeter. And there is the possibility of the use of impure reagents or faulty apparatus. It is very clear now that various sources may have been responsible for the deviation of the standard reduction potential values. The second part of the experiment started out basically with the same procedure as the first part. The only difference was that various applications of electrochemical cells were used. Table IV. Experimental Ecell vs. Theoretical Ecell
Cell Notation
Experimental Ecell
Theoretical Ecell
% Error
Zn(s)|Zn 2+ (0.1M)|| [Cu(NH3 )4 ]2+ (0.33M), NH3 (1M)||C(s) Zn(s)|Zn 2+ (0.1M)|| OH- (0.1M)|Cu(OH) 2 (s)|C(s)
0.76
0.738
2.98
0.61
0.49
24.5
The data in Table IV shows the two types of applications of the electrochemical cell – a cell containing redox reactions that involve complexes and one that has a partially soluble solid in it. The theoretical and experimental values of the cell potentials are also shown in the table and it can be seen that there are minor deviations between the two potentials. These deviations may be attributed to a few things such as the preparation of the complex and solid used. Also, it could have been that the voltmeter used was faulty. These are the most possible sources of error for this part of the experiment. With use of the the experimental cell potential, the standard reduction cell potential was calculated. With the E°cell at hand, the E°red of the half-cells were determined. In tu rn, this allowed the calculation of Gibb’s Free Energy for each half-cell to be done. The table below shows the values of ΔG obtained. ᵒ
Table V. Gibb’s Free Energy
Half-Cell
Ionic Equation
ΔG , kJ/mol
Zn(s)|Zn 2+ (0.1M)||
Zn(s) --> Zn2+ (aq) + 2e-
120
||[Cu(NH3 )4 ]2+ (0.33M), NH3 (1M)||C(s)
[Cu(NH3 )4 ]2+ (aq) + 2e- --> Cu(s) + 2NH3 (aq)
-24.1
||OH- (0.1M)|Cu(OH) 2 (s)|C(s)
Cu(OH)2 (s) + 2e- --> Cu(s) + 2OH- (aq)
19.1
ᵒ
The value of ΔG° shows the spontaneity of a certain reaction, in this case though, a half-reaction. A negative value means that the half-reaction is spontaneous while a positive value means it is non-spontaneous. This coincides
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with the fact that a positive value of E ° makes a reaction spontaneous while the negative value of E makes a reaction non-spontaneous. ᵒ
With the use of the ΔG ° of the half-cells, the values of K for the half-reactions of the second part of the experiment were calculated. The following equation was used to calculate K: K = e -ΔG/RT Where R = gas constant, 8.314 J/mol T = temperature in Kelvin Table VI. Experimental K vs. Theoretical K
Cell Notation ||[Cu(NH3 )4 ]2+ (0.33M), NH3 (1M)|C(s) ||OH- (0.1M)|Cu(OH) 2 (s)|C(s)
Cell Reaction [Cu(NH3 )4 ]2+ (aq)--> Cu(s) + 2NH3 (aq) Cu(OH)2 (s) --> Cu(s) + 2OH- (aq)
Experimental K
Theoretical K
% Error
5.9 x 10-5
1.18 x 1012
100
4.46 x 10-4
1.6 x 10-19
>100
Table VI shows the values of both the experimental and theoretical K. It also shows that the deviation of the experimental data from the theoretical data is very, very large. This may be due to the use of natural logarithm, which would have magnified the discrepancies in molar concentrations.
CONCLUSION AND RECOMMENDATIONS
The experiment was successful in determination of cell potentials of half-reactions of zinc, chlorine, bromine, and iodine ions with copper ions. It displayed the proper use of electrochemical cells and their reactions, and utilized the Nernst equation in the determination of cell potentials. It is recommended that for future experiments, careful measurement of reagents for solution preparation should be done to avoid any deviations in experimental value. Extra care should also be taken when handling materials used for measurement of voltage, as false values can be given if the instrument is damaged or unclean.
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ANSWERS TO QUESTIONS
1. In the electrolysis of KX solution a) At which electrode, anode or cathode, is X 2 generated? Is this the positive or negative terminal of the electrolytic cell?
The X 2 will be generated at the anode of the electrolytic cell. The anode is the positive terminal of the cell.
2. Draw the experimental set-up for each of the following cells. Write the ionic equations for the anode, cathode, and the cell reactions. Calculate the cell potential.
For Cl- | AgCl/Ag E°red = 0.22 V a) Pt | H2 (1 atm) | HCl (0.1 M) | AgCl/Ag
The half-reaction in this situation is as follows: 2X -(aq) → X 2(g) + 2e The reaction shows that X - underwent oxidation in order to generate X 2. The oxidation reaction always takes place in the anode of the cell. Also, for electrolytic cells, the anode is considered to be the positive terminal. b) What will be formed at the other electrode? Write the ionic equation involved.
There are two possible formations at the cathode: solid copper, or a combination of hydrogen and oxygen gas. → Cu(s) → 2H2(g) + O2(g)
Cu2+(aq) + 2H2O +
2e-
2e-
c) It was assumed that the concentration (1.0 M) of X - is not significantly changed by passing a current of 0.1 A through the solution for 1 minute. Show mathematically that the assumption is valid.
)( )( ) ( A change of 1.036 mL in the volume is not significant enough to change the concentration of X -. It is therefore safe to assume that the concentration is not significantly changed, and therefore the assumption is valid.
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A: H2(g) → 2H+(aq) + 2eE°red = 0.00 V + C: 2(AgCl(s) + 1e → Ag (aq) + Cl (aq) ) E°red = 0.22 V H2(g) + 2AgCl(s) → 2Ag +(aq) + 2Cl-(aq) + 2H+(aq) E°cell = 0.22 V
b) Zn | ZnSO4 (0.1 M) || AgNO 3 (0.1 M) | Ag
c) Pt | Fe 2(SO4 )3 (0.005 M), FeSO 4 (0.01 M) || KCl (0.1 M) || Cl2 (1 atm) | Pt
A: Zn(s) → Zn2+(aq) + 2eE°red = -0.763 V + C: 2(Ag (aq) + 1e → Ag (s) ) E°red = 0.799 V + 2+ Zn(s) + 2Ag (aq) → 2Ag (s) + Zn (aq) E°cell = 1.562 V
A: 2(Fe3+(aq) → Fe2+(aq) + 1e-) E°red = 0.771 V C: 2Cl (aq) + 2e → Cl2(g) E°red = 1.360 V 2Fe3+(aq) + 2Cl-(aq) → 2Fe2+(aq) + Cl2(g) E°cell = 0.589 V
3. Which of the following reactions are conceivable sources of electrical energy? Write the cathode and anode for those that are. a) Cl 2 + 2Br - → Br 2 + 2Cl – b) Ag + + Cl - → AgCl c) 2Ag + Fe 2+ → Fe + 2Ag + d) H 2 + ½ O2 → H 2 O
Out of the four reactions, only reaction (a) is a conceivable source of electrical energy, because it is the only reaction with a positive E° cell. Cathode: Cl2 + 2e- → 2Cl Anode: 2Br- → Br2 + 2eCl2 + 2Br- → Br2 + 2Cl-
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E°red = 1.36 V E°oxid = -1.08 V E°cell = + 0.28 V
4. Given the cell: Pt | Cl2 (1 atm) | Cl - (0.1 M) || H + (1 M), MnO 4- (0.1M) | Pt
will increase, therefore decreasing the value of the final cell potential.
For H+, MnO4-, Mn2+ | Pt Cl- / Cl2 | Pt
iii) A few crystals of KMnO 4 are added to the cathode half-cell?
E°red = +1.51 V E°red = +1.36 V
a) Calculate the i. cell potential
2(MnO 4- + 8H+ + 5e- → Mn2+ + 4H2O) +1.51 V 5(Cl → Cl2 + 2e ) - 1.36 V + 2+ 2MnO 4 + 5Cl + 16H → 2Mn + 5Cl2 + 8H2O E°cell = 0.15 V
[ ] ii. free energy change per mole of electrons transferred
b) Will the cell potential be greater, lesser, or the same if: i) 0.1 M KCl is added to the anode half-cell?
The cell potential will remain the same, because KCl is not involved in the anode half-reaction. ii) A few crystals of AgNO3 are added to the anode half-cell?
The cell potential will be lesser than the original, because the added AgNO 3 will release Ag + ions which can react with the Cl - ions to form solid AgCl crystals. This will lower the Cl- concentration. Because Cl - is a reactant involved in the cell reaction, the value of log Q
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The cell potential will be greater than the original, because the addition of the crystals will increase the concentration of MnO 4-, a reactant involved in the cell reaction. This increase in concentration will cause the value of log Q to decrease, therefore increasing the value of the final cell potential.
c) If NaOH is added to the cathode half-cell to make the H + concentration of 0.001 M, and if the concentrations of the other ions remain essentially unchanged, write the ionic equation for the spontaneous cell reaction. Show calculations to support your answer.
2Mn2+ + 5Cl2 + 8H2O → 2MnO4- + 10Cl- + 16H+
5. Give the effect (increase, decrease, none) of the following on the experimental emf obtained in Part B, and explain: a) Addition of acid in the entire cathodic chamber
There will be no change in the value of emf, because the addition of acid will not react with any of the reactant species present in the cathodic chamber (e.g. Cu2+ ). b) Addition of excess NH 3 in the entire anodic chamber
There will be a decrease in the value of emf, because the Cu2+ ions will react with ammonia, forming an ammonia complex. This will lower the Cu 2+ concentration. Because Cu 2+ is a reactant involved in the cell reaction, the value of log Q will increase, therefore decreasing the value of the experimental emf. c) Reversal of anode and cathodes used
There will be a no change in the magnitude of emf, but there will be a reversal of signs from positive to negative. So, technically, the emf decreases because it is now negative. This is because when the anode and cathode are reversed, the polarities are reversed as well.
6. Account for the percent error in Part B. Suggest ways wherein experimental deviation from the theoretical may be minimized.
A possible source of error is the inability to measure and keep the temperature constant, as the temperature used in the calculations – 298 K – was just assumed. Another possible source of error is in the preparation of solutions, because precise measurement of reagents to be mixed could not be ensured. Mishandling of instruments is another possible source of error.
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APPLICATIONS
REFERENCES
Electrochemistry has several applications in industrial processes. The most common use is electroplating, which is the process of applying an electric current in a reaction cell containing a metal cation in order to coat a certain object with a layer of that metal. Electroplating is used to provide a material with desired qualities – such as protection from corrosion, abrasion resistance, aesthetic quality, etc. – onto a surface which lacks these qualities. Another application of electroplating is to build a layer of thickness on undersized parts. It is also used to provide a “sacrificial” surface on a material to protect it from degradation.
[1] Brown, T.L., LeMay Jr H.E., Bursten, B.E. Chemistry: The Central Science 7th ed. Prentice Hall, New Jersey, 1998
The concepts employed in electrochemistry are also utilized for the electrolytic purification of metals, which is the process of removing impurities from a metal by oxidizing metal ions from an impure metal anode and reducing these ions onto the pure metal cathode. Impurities in the anode would settle to the bottom of the cell set-up to be disposed. Electrolytic purification has branched out to be used in many industries including water treatment and the handling of biological waste. Detection of foreign substances, such as ethanol, is also being done through the use of electrochemistry. An electrochemical cell is used in devices utilized to identify drunk drivers. The redox reaction of ethanol creates electricity within the electrochemical cell, which is detected by the device. Electrochemistry is also the main basis for the production of batteries. Chemical reactions with a positive reduction potential create electricity, which can be harnessed for many practical purposes. The mobility of modern batteries and the simplicity behind their use has allowed batteries to be employed in a very large range of appliances and applications.
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[2] Lower, S. “All About Electrochemistry.” Chemistry 1 Virtual Textbook. Simon Fraser University. Accessed 28 September 2010. < http://www.chem1.com/acad/webtext/elchem/> [3] Masterton, W.L., Hurley, C.N. Chemistry: Principles and Reactions, 5 th Edition. Thomson Learning Asia, Singapore, 2005. [4] Morgan, N. Chemistry in Action. Oxford University Press, New York, 1995. [5] Ohzuku, T.,Kato, J., Sawai K., Hirai, T., Electrochemistry of Manganese Dioxide in Lithium Nonaqueous Cells. Osaka, Japan. Japan Electrochemical Society, Volume 138, Issue 9, pp. 2556-2560 (1991). [5] Skoog, D.A., West, D.M., et al. Fundamentals of Analytical Chemistry 8 th edition. Brooks/Cole, Singapore. 2004. [6] Whitten, K.W., Davis, R.E., Peck, M.L., Stanley, G.G. Chemistry. Eighth Edition. Thomson Brooks/Cole, California, 2007. [7] Zumdahl, S. S. Chemical Principles: Third Edition. Houghton Mifflin Company, Boston, 1998.
APPENDIX A. WORKING EQUATIONS
1.
Ecell = E°cell – log Q Where Ecell = cell potential in volts, V E˚cell = standard cell potential in volts, V R = gas constant = 8.314 J/mol K T = absolute temperature = 298 K n = moles of electrons transferred in the balanced equation in moles F = faraday = 96, 485 J/ V mol eQ = reaction quotient (can include pressure)
Or Ecell = E°cell – log the standard temperature, 298K 2.
B. SAMPLE CALCULATIONS
For systems under
E°cell = E°red, cathode - E°red, anode Where E°cell = standard cell potential, V E°red, cathode = reduction potential of the cathode, V E°red, anode= reduction potential of the anode, V
3. ∆G° = -RT lnK ∆G° = -nFE°cell lnK = nFE° cell / RT Where ∆G° = Gibbs free energy under standard conditions R= 8.314 J/ mol K T= temperature in Kelvin, K K = equilibrium constant n= number of moles electrons in the balanced reaction F= 96, 500 J/ V mol e E°cell= cell potential in Volts, V 4. % error = |
| x 100%
Cu2+(aq) + 2e-→ Cu(s) . Zn(s) → Zn2+(aq) + 2e Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)
1. Anode: Cathode: Cell rxn:
Experimental Ecell: 0.96 V
] [ E = E° log log [] E° = 0.96V + E° = 0.96V log E° = 0.96 V cell
cell
cell cell cell
E°cell = E°red, cathode - E°red, anode E° red, anode = E°red, cathode - E°cell = 0.34V – 0.96V E°red, anode = -0.62V experimental E° red, Zn2+/Zn(s) Theoretical E°red = -0.763 V
| x 100% =| | x 100% = 18.74%
% error = |
2. Anode: Zn(s) → Zn2+(aq) + 2e Cathode: [Cu(NH3 )4 ]2++2e- Cu(s) + 4NH3(aq) Cell rxn: Zn(s) + [Cu(NH3 )4 ]2+ Cu(s) + 2+ 4NH3(aq) + Zn (aq) ↔
→
Experimental E cell = 0.76 V
log = 0.76 V + log
E°cell = Ecell +
E°cell = 0.745 V
E°cell = E°red, cathode - E°red, anode E°red, cathode = E°cell + E°red, anode = 0.805 V – 0.62 V E°red, cathode = 0.185V Theoretical E°cell= E°red(cathode)- Eored(anode) = -0.4 -(-0.763) = 0.723V
log = 0.723
Theoretical Ecell = E°cell -
= 0.738 V % error = |
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| x 100%
| x 100%
% error = | % error = 2.98 %
∆G° of cathode: ∆G° = -nFE°cell ∆G° = -(2 mol e )(96,500 J/V·mol e )(0.125 V) ∆G° = -24, 125 J = -24.1 kJ ∆G° of anode: ∆G° = -nFE°cell ∆G° = -(2 mol e )(96,485 J/V·mol e )(-0.62 V) ∆G° = 119,641.4 J = 119.641 kJ [Cu(NH3 )4 ]2+
↔
Cu(s) + 4NH3(aq) + 2e-
∆G° = -RTlnK
lnK = lnK = D
lnK D = 9.74 K D = 1.7 x 104 K f = = 5.9 X 10-5
Theoretical K f = 1.18 x 1012
| x 100% – =| | x 100% = 100%
% error = |
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