Electrochem Lecture Problems Problem 1 - A mercury battery uses the following electrode half-reactions.. HgO(s) + H O(l) + e - ---> Hg(l) + OH -. E Hg 0.098V ZnO(s) + H O(l) + e - ---> Zn(s) + OH -. E Zn 1.60V Calculate the cell potential, the G o for the cell reaction and the equilibrium constant. The mercury will act as the cathode, so we can write the overall after eliminating the electrons and hydroxides as they cancel out... HgO(s) + Zn(s) ---> Hg(l) + ZnO(s) E std E Hg E Zn 1.358 V J J F 96485 R 8.3145 T molv molk o 98.15K kj ΔG o FE std 6.053 mol ΔG o K eq e RT o 8.1 10 45 Zn ZnO(s) HgO Hg Cell notation
Problem - Compute the potential of the following electrochemical cell under standard conditions and then given the concentrations provided. Ag + + Sn -> Ag + Sn +. 0.010 M 0.00 M Sn Sn + Ag + Ag Cell notation. From tables the half cell potentials are as follows: Ag + + e - -> Ag E Ag 0.799V Sn + + e - -> Sn. E Sn 0.139V E std E Ag E Sn 0.938 V Ag ion 0.010 Sn ion 0.00 0.059V E cell E std log Sn ion Ag ion 0.87 V What does the Sn + concentration need to be to produce exactly 1.00 V? Solving for the Sn ion.. E cell 1.00V Ag ion 1.00 E cell E std Sn ion Ag ion 10 0.059V 8.043 10 3
Problem 3 A cell is make of Pb in a 0.05 M Pb solution as one half cell and a 0.75 M solution of a weak acid in the other half cell with a Pt electrode. The cell voltage is measured to me 0.13 V What is the Ka of the acid? Pb + H + = Pb + + H E Pb 0.16V E H 0.00V E cell_o E H E Pb E cell_o 0.16 V E cell 0.13V n M HA 0.75 0.059V E cel = E cell_o log n Pb aq P H H Solution... E cell E cell_o n 0.059V H 10 Pbaq P H 0.8 H K a K M HA H a 0.170 If the Pb half cell contained a saturated solution of PbCl, what would the voltage be? 3 K sp_pbcl 1.710 5 K sp_pbcl Pb aq Pb 4 aq 0.016 1 0.059V Pb aq P H E cell E cell_o log E n cell 0.146 V H
Problem 4 - Compute the voltage of the following concentration cell. Zn Zn + (0.5 M) Zn + (.0 M) Zn This cell has an E std = 0. Therefore the cell potential is given as: 0.059V E cell log 0.5.0 0.018 V In this system, electrons are transferred to the concentrated side, reducing the ion concentration there and increasing it in the dilute side. Eventually the system will reach equilibrium at 1.5 M on each side. Problem 5 - The transmission of nerve impulses through the body occurs because cell membrane activity maintains the potassium ion concentration inside and outside a nerve fiber at different concentrations. Thus there is a potential. The nerve transmission is a signal that allows the ions to pass through the membrane neutralizing this potential. This causes a chain reaction effect that results in signal transmission. The potassium concentration outside the cell is maintained at approximately 150 mmol and approximately 5 mol inside the membrane. What is the cell potential across the membrane? 0.059V E cell log 150 87.446mV 1 5 which is about right for a nerve fiber. Other ions actually contribute to this to obtain a measured voltage of about - 70 mv.
Problem 6 - Consider the following two cells. Each will be used to produce a ph meter. For such meters, the hydrogen half cell is placed in a hydrogen sensitive glass bulb as a probe. For each cell, produce an expression that will compute the ph given a voltage change from the standard cell potential. What is the advantage of Cell # over Cell #1? Cell 1: Zn Zn + H + (? M), H (1 bar) Pt E Zn 0.763V H + (aq) + Zn (s) ----> H (g) + Zn + (aq) Cell : Pb Pb +, PbCl (sat'd) H + (? M), H (1 bar) Pt E Pb 0.16V H + (aq) + PbCl (s) ----> H (g) + Pb + (aq) For cell #1 E std 0 E Zn 0.763 V Zn ion 1.00 0.059V E cell = E std log Zn ion H ion Rearranging and then using the rule of logarithms E std E cell 0.059 log H ion = log Zn ion = log Zn ion logh ion Since the Zinc ion is at 1 M concentration, then the log (1) = 0. Finally, the last term is the ph. So we have.. ph = E std E cell 0.059V = 0.763V E cell 0.059V If E cell 0.64V 0.763V E cell ph ph.348 0.059V
For cell # E std 0 E Pb 0.16 V 0.059V E cell = E std log Pb ion H ion Rearranging as before and then using the rule of logarithms E std E cell 0.059 log H ion = log Pb ion = log Pb ion logh ion However, the Pb + concentration is not 1 M. The concentration is governed by the solubility of PbCl (s). This substance has a K sp associated with it. We will use this to compute the lead ion concentration and then proceed as before. This cell is more stable because the Pb + concentration is constant. In the Zn + cell, the concentration of 1 M will change as the cell operates. K sp 1.710 5 Preparing an ICE diagram.. PbCl (s) -----> Pb + (aq) + Cl -1 (aq) Initial 0 0 Change +x +x Equilibrium x x K sp = Pb ion Cl ion = ( x) ( x) = 4x 3 1 3 K sp x 0.016 So Pb 4 ion 0.016 Now we proceed as before...
E std E cell 0.059 logh ion = log Pb ion ph = E std E cell 0.059V logpb ion If E cell 0.045V for example.. E std 0.16 V ph E std E cell 0.059V logpb ion.61 alternate... A saturated solution of SnI as one half cell and an SHE produced a voltage of -0.018 V. What is the Ksp of the salt? E std 0.1364V E cell 0.081 V E std E cell log Sn ion ph = = 0 = 0 because [H + ) = 1 M 0.059V = log Sn ion E std E cell 0.059V E std E cell 0.059V Sn ion 10 0.01357 x Sn ion K sp ( x) 4x 3 4( 0.013571) 3 1.000 10 5 x 0.013571
Calibration against a standard. Problem 7 - In practice a ph meter is calibrated using a buffer solution of known ph. Presume that we use a ph = 7.0 buffer and zero the instrument. We then measure the voltage change when the probe is placed in a solution of unknown ph. For cell #, produce an expression that will compute the ph given a voltage change from a calibration buffer of ph = 7.0.. For a ph = 7, we know H buffer 10 7.0 1 10 7 ph buffer log H buffer For this cell we have, then 0.059V E cell_calib = E std log Pb ion H buffer Now, for the unknown acid solution.. 0.059V E cell_unk = E std log Pb ion H test_soln Subtracting, we have E cell_unk E cell_calib = 0.059 log Pb ion H test_soln H buffer Pb ion = 0.059 log H buffer H test_soln Very cool..the Pb + concentration dependence is now gone! NOTE! This only works because the lead ion concentration is stable and constant!
Finishing.. ΔE cell 0.059V = log H buffer log H test_soln = ph buffer ph test_soln ph test_soln ΔE cell ΔE cell ph buffer 0.059V Which is linear! ΔE cell 0.00V0.19V 0.00V 1 10 ph test_soln ΔE cell 8 6 4 Follow-up 0. 0.1 0 0.1 0. 0.3 ΔE cell The E was measured to be 0.84 V for a solution of a 0.74 M unknown acid. Calibration was made against a ph = 7.00 buffer. What is the K a of the acid. 0.84V ph 7.00 ph.03 H 10 ph 0.059V x x H K a 0.74 x 5.359 10 5
Problem 8 - Note that this can apply to any probe system! Old houses suffer from lead pipes and lead ions can leach into the tap water. Suppose we wish to measure the lead concentration in the tap water. We can use the same analysis subsitutinig an appropriate reference standard. We obtain using identical methods ppb test_soln ΔE cell ΔE cell ΔE cell = = ppb 0.059V std 0.059V logpb ion Suppose that the lead standard for calibration is the EPA standard for tap water before action is taken. This level is 15 ppb. The probe measured a E of 0.135 V relative to this standard when placed in tap water emerging from the lead pipe tap. Converting to Molarity... Pb ion 15gm 10 9 ml 1mol 07.gm 1000mL 7.39 10 8 M L Now the standard is.. ppb std log 7.3910 8 7.14 ppb test_soln ΔE cell ΔE cell ppb std Or transferring back to ppb.. Conc Pb_ppb ΔE cell 0.059V 10 Conc Pb_ppb ( 17.8mV) 9.974 ΔE cell ppb std 0.059V 07. gm mol M 1000 ml L 10 9 ml gm
Electrolysis Problem 9 - Metal purification and isolation is done commonly by electrorefinement, or the coating of an electrode by running an electrochemical cell in reverse. Consider the following. A current of 1.35 A is applied over a 4 hour period to a cell containing Cu + ions, resulting in the plating out of Cu metal via Cu + (aq) + e - ---> Cu(s) What mass of copper is collected? i 1.35A Recall that 1 A = 1 C/s J F 96485 But J 1CV so F 96485 C molv mol Mass Cu 1.35A( 4hr) 3600s hr 1mol e 96485C 1mol Cu mol e 63.5gm mol Cu 38.38gm
Problem 10 - An Iron metal plate is to be coated with Nickel to form a protective barrier to corrosion. If the iron plate is 1 m in surface area and you are charged with determining the time required to produce a 0.05 mm thick plating. How long should the electroplating be done if 100.0 A are passed through a Ni + solution? First, compute the moles of Ni needed to get the desired thickness. From the periodic chart, the atomic volume of Ni is: AV Ni 6.60 cm3 mol OR AW 58.6934 gm D 8.90 gm AW AV mol Ni cm 3 D 6.595 cm3 mol Volume Ni 1m ( 0.05mm) 1cm 10mm 10 4 cm 50cm 3 m Volume Ni AV Ni 7.58mol Ni Volume Ni AW AV Ni 445gm 7.576mol Ni mol e mol Ni 100.0 C s 96485C mol e 1hr 3600s 4.061hr
Problem 11 Consider the electrolysis of water. The relevant half-reactions are: H O (l) + e - --> H (g) + OH - (aq) E o_h 0.83V O (g) + 4 H + (aq) + 4 e - --> H O(l) E o_o 1.3V We can see that going from H and O to H O E cell_o 1.3V ( 0.83V).06 V But this is incorrect. H + and OH - are not at standard 1 M states. Their concentration in water solution is 10-7 M. Thus we need the Nernst Equation. Executing for each. 0.059V E H E o_h log 10 7 0.416 V 0.059V E O E o_o log 10 7 4 4 0.816 V E cell E O E H 1.31 V Thus we need 1.31 Volts to start the hydrolysis. Note the overall reaction is: 4 H O (l) + 4 e - --> H (g) + 4 OH - (aq) H O(l) ---> O (g) + 4 H + (aq) + 4 e - Note the cathode reaction is reversed. 6 H O(l) ---> H (g) + O (g) + 4 H O H O(l) ---> H (g) + O (g)
Problem 1 Electrolysis of NaCl involves the electrolysis in molten NaCl. The following half-reactions apply. Na + (l) + e - --> Na(s) E Na.71V Cl (g) + e - --> Cl - (aq) E Cl 1.36V H O (l) + e - --> H (g) + OH - (aq) E H 0.416V O (g) + 4 H + (aq) + 4 e - --> H O(l) E O 0.816V E cell E Cl E Na 4.07 V 4.07V is larger than the 1.3 V for the electrolysis of water. So a solution of NaCl will not yield Na and Cl (g), but H (g) and O (g). Therefore, to hydrolyze NaCl, it needs to be molten. One might ask the question.. "Isn't the combination of 1.36 V - 0.816 V = 0.544 V a smaller threshold voltage and the starting point? Answer: No. O is not in the system. Only H O, Na + ions and Cl - ions are present. In electrolysis, the reaction at the cathode is reversed, not the anode because this is effectively an electrochemical cell run in reverse.
Problem 13 What will be the products of the electrolysis of NaI(aq)? The relevant half reactions are: H O (l) + e - --> H (g) + OH - (aq) O (g) + 4 H + (aq) + 4 e - --> H O(l) Na + (l) + e - --> Na(s) E H 0.416V E O 0.816V E Na.71V I (s) + e - --> I - (aq) E I 0.536V We see that the smallest combination is: E I E H 0.95 V This is less than the 1.3 V for the hydrolysis of water Thus the overall reaction is.. I - (aq) + H O (l) --> I (s) + H (g) + OH - (aq) One might ask the question.. "Isn't the combination of 0.816 V - 0.536 V = 0.80 V a smaller threshold voltage and the starting point? Answer: No. I is not in the system. Only H O, Na + ions and I - ions are present. In electrolysis, the reaction at the cathode is reversed, not the anode because this is effectively an electrochemical cell run in reverse.