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Chapter 21

Chapter 21. Electrochemistry: Chemical Change and Electrical Work. Electrochemistry: Chemical Change and Electrical Work. 21.1 Half-Reactions and Electrochemical Cells. 21.2 Voltaic Cells: Using Spontaneous Reactions to Generate Electrical Energy.

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Chapter 21

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  1. Chapter 21 Electrochemistry: Chemical Change and Electrical Work

  2. Electrochemistry: Chemical Change and Electrical Work 21.1 Half-Reactions and Electrochemical Cells 21.2 Voltaic Cells: Using Spontaneous Reactions to Generate Electrical Energy 21.3 Cell Potential: Output of a Voltaic Cell 21.4 Free Energy and Electrical Work 21.5 Electrochemical Processes in Batteries 21.6 Corrosion: A Case of Environmental Electrochemistry 21.7 Electrolytic Cells: Using Electrical Energy to Drive a Nonspontaneous Reaction

  3. Study of interchange of chemical and electrical energy. It involves oxidation –reduction reaction. Electrochemistry

  4. Key Points About Redox Reactions • Oxidation (electron loss) always accompanies reduction (electron gain). • The oxidizing agent is reduced, and the reducing agent is oxidized. • The number of electrons gained by the oxidizing agent always equals the number lost by the reducing agent.

  5. Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) Figure 21.1 A summary of redox terminology. OXIDATION One reactant loses electrons. Zn loses electrons. Reducing agent is oxidized. Zn is the reducing agent and becomes oxidized. Oxidation number increases. The oxidation number of Zn increases from 0 to +2. REDUCTION Other reactant gains electrons. Hydrogen ion gains electrons. Oxidizing agent is reduced. Hydrogen ion is the oxidizing agent and becomes reduced. Oxidation number decreases. The oxidation number of H decreases from +1 to 0.

  6. Half-Reaction Method for Balancing Redox Reactions • Summary: This method divides the overall redox reaction into oxidation and reduction half-reactions. • Each reaction is balanced for mass (atoms) and charge. • One or both are multiplied by some integer to make the number of electrons gained and lost equal. • The half-reactions are then recombined to give the balanced redox equation. • Advantages: • The separation of half-reactions reflects actual physical separations in electrochemical cells. • The half-reactions are easier to balance especially if they involve acid or base. • It is usually not necessary to assign oxidation numbers to those species not undergoing change.

  7. Cr2O72-(aq) + I-(aq) Cr3+(aq) + I2(aq) Cr2O72- Cr3+ I-I2 Cr2O72- Cr3+ 6e- + 14H+(aq) + Cr2O72- Cr3+ 2 + 7H2O(l) Balancing Redox Reactions in Acidic Solution Cr2O72-(aq) + I-(aq) Cr3+(aq) + I2(aq) 1. Divide the reaction into half-reactions - Determine the O.N.s for the species undergoing redox. +6 -1 +3 0 Cr is going from +6 to +3 I is going from -1 to 0 2. Balance atoms and charges in each half-reaction - 14H+(aq) + 2 + 7H2O(l) net: +6 Add 6e- to left. net: +12

  8. 2 + 2e- X 3 I-I2 I-I2 I-I2 6e- + 6e- + 14H+ + Cr2O72- Cr3+ 2 + 7H2O(l) 6 3 + 6e- 14H+(aq) + Cr2O72-(aq) + 6 I-(aq) 2Cr3+(aq) + 3I2(s) + 7H2O(l) 14H+(aq) + Cr2O72- Cr3+ + 7H2O(l) Balancing Redox Reactions in Acidic Solution continued 2 2 + 2e- Cr(+6) is the oxidizing agent and I(-1)is the reducing agent. 3. Multiply each half-reaction by an integer, if necessary - 4. Add the half-reactions together - Do a final check on atoms and charges.

  9. 14H2O + Cr2O72- + 6 I- 2Cr3+ + 3I2 + 7H2O + 14OH- 7H2O + Cr2O72- + 6 I- 2Cr3+ + 3I2 + 14OH- 14H+(aq) + Cr2O72-(aq) + 6 I-(aq) 2Cr3+(aq) + 3I2(s) + 7H2O(l) Balancing Redox Reactions in Basic Solution Balance the reaction in acid and then add OH- so as to neutralize the H+ ions. + 14OH-(aq) + 14OH-(aq) Reconcile the number of water molecules. Do a final check on atoms and charges.

  10. Cr2O72- I- Figure 21.2 The redox reaction between dichromate ion and iodide ion. Cr3+ + I2 Cr2O72- Cr3+ I-

  11. PROBLEM: Permanganate ion is a strong oxidizing agent, and its deep purple color makes it useful as an indicator in redox titrations. It reacts in basic solution with the oxalate ion to form carbonate ion and solid mangaese dioxide. Balance the skeleton ionic reaction that occurs between NaMnO4 and Na2C2O4 in basic solution: MnO4-(aq) + C2O42-(aq) MnO2(s) + CO32-(aq) C2O42- CO32- MnO4- MnO2 MnO4- MnO2 C2O42- CO32- 2 MnO4- MnO2 C2O42-2CO32- Sample Problem 21.1: Balancing Redox Reactions by the Half-Reaction Method PLAN: Proceed in acidic solution and then neutralize with base. SOLUTION: +7 +4 +3 +4 4H+ + + 2H2O + 2H2O + 4H+ +3e- +2e-

  12. 3C2O42- + 6H2O 6CO32- + 12H+ + 6e- C2O42- + 2H2O 2CO32- + 4H+ + 2e- C2O42- + 2H2O 2CO32- + 4H+ + 2e- 4H+ + MnO4- +3e- MnO2+ 2H2O 4H+ + MnO4- +3e- MnO2+ 2H2O 8H+ + 2MnO4- +6e- 2MnO2+ 4H2O 8H+ + 2MnO4- +6e- 2MnO2+ 4H2O X 3 X 2 3C2O42- + 6H2O 6CO32- + 12H+ + 6e- 2MnO2-(aq) + 3C2O42-(aq) + 2H2O(l) 2MnO2(s) + 6CO32-(aq) + 4H+(aq) 2MnO2-(aq) + 3C2O42-(aq) + 4OH-(aq) 2MnO2(s) + 6CO32-(aq) + 2H2O(l) Sample Problem 21.1: Balancing Redox Reactions by the Half-Reaction Method continued: + 4OH- + 4OH-

  13. A cell that produces electrical energy from a spontaneous chemical reaction. (Also known as a galvanic cell) System does work on the surroundings. batteries Voltaic cell

  14. An electrolysis apparatus in which electrical energy from an outside source is used to produce a chemical change. Cathode: The negative electrode. Anode: The positive electrode Surroundings do work on the system. Electroplating Electrolytic cell

  15. Electrode- Conduct electricity between cell and surroundings. Electrolyte-mix of ions. Oxi-anode, e- are lost, leave cell at anode. Red-cathode, e- are gained, enter at anode. Cells

  16. Oxidation half-reaction X X+ + e- Oxidation half-reaction A- A + e- Reduction half-reaction Y++ e- Y Reduction half-reaction B++ e- B Overall (cell) reaction X + Y+ X+ + Y; DG < 0 Overall (cell) reaction A- + B+ A + B; DG > 0 Figure 21.3 General characteristics of voltaic and electrolytic cells. VOLTAIC CELL ELECTROLYTIC CELL System does work on its surroundings Energy is released from spontaneous redox reaction Energy is absorbed to drive a nonspontaneous redox reaction Surroundings(power supply) do work on system(cell) Energy released Energy absorbed

  17. Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) The spontaneous reaction between zinc and copper(II) ion. Figure 21.4

  18. Oxidation half-reaction Zn(s) Zn2+(aq) + 2e- Reduction half-reaction Cu2+(aq) + 2e- Cu(s) Overall (cell) reaction Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) A voltaic cell based on the zinc-copper reaction. Figure 21.5

  19. Oxidation half-reaction 2I-(aq) I2(s) + 2e- Reduction half-reaction MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O(l) Overall (cell) reaction 2MnO4-(aq) + 16H+(aq) + 10I-(aq) 2Mn2+(aq) + 5I2(s) + 8H2O(l) A voltaic cell using inactive electrodes. Figure 21.6

  20. Two half cells. Consists of electrode in an electrolyte. Joined by wire and a salt bridge. Voltmeter used to measure voltage. Oxi half cell-anode- left Red half cell-cathode-right Construction and operation of a Voltaic Cell

  21. Anode conducts electrons out of the cell Cathode conducts electrons in its half cell. Electrons flow from anode to cathode thro the wire (left-right) Anode has excess of e-, -ve charge. cathode: +ve Construction and operation of a Voltaic Cell

  22. Salt bridge: A charge imbalance results because of ions. Salt bridge helps complete the circuit and avoid such imbalances. It contains soln of nonreacting ions: Na+, sulphate in gel. To maintain neutrality ions( Na+, sulfate) diffuse thro the soln and balance the charges Construction and operation of a Voltaic Cell

  23. Electrons move left-right thro’ wire Anions move from right-left thro’ salt bridge. Cations move from left-right thro’ salt bridge. Active electrodes- involved in reaction. Inactive electrode- graphite or platinum, conduct electrons into or out of half cell, but cannot take part in half reactions. Construction and operation of a Voltaic Cell

  24. Anode components listed on left, cathode components listed on right separated by double vertical lines (salt bridge) A phase difference boundary-single line. Write line notations for the following: 1. Al3+(aq) + Mg (s)→ Al (s) + Mg2+ (aq) 2. MnO4- (aq) +H+ (aq) +ClO3-(aq) → ClO4-(aq) + Mn2+ (aq) + H2O(l) Line Notation

  25. Zn(s) Zn2+(aq) + 2e- Cu2+(aq) + 2e- Cu(s) inert electrode Notation for a Voltaic Cell components of anode compartment (oxidation half-cell) components of cathode compartment (reduction half-cell) phase of lower oxidation state phase of lower oxidation state phase of higher oxidation state phase of higher oxidation state phase boundary between half-cells Examples: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu (s) graphite | I-(aq) | I2(s) || H+(aq), MnO4-(aq) | Mn2+(aq) | graphite

  26. PROBLEM: Diagram, show balanced equations, and write the notation for a voltaic cell that consists of one half-cell with a Cr bar in a Cr(NO3)3 solution, another half-cell with an Ag bar in an AgNO3 solution, and a KNO3 salt bridge. Measurement indicates that the Cr electrode is negative relative to the Ag electrode. e- Oxidation half-reaction Cr(s) Cr3+(aq) + 3e- Cr Ag K+ NO3- Reduction half-reaction Ag+(aq) + e- Ag(s) Cr3+ Ag+ Overall (cell) reaction Cr(s) + Ag+(aq) Cr3+(aq) + Ag(s) Sample Problem 21.2: Diagramming Voltaic Cells PLAN: Identify the oxidation and reduction reactions and write each half-reaction. Associate the (-)(Cr) pole with the anode (oxidation) and the (+) pole with the cathode (reduction). SOLUTION: Voltmeter salt bridge Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s)

  27. Why Does a Voltaic Cell Work? The spontaneous reaction occurs as a result of the different abilities of materials (such as metals) to give up their electrons and the ability of the electrons to flow through the circuit. Ecell > 0 for a spontaneous reaction 1 Volt (V) = 1 Joule (J)/ Coulomb (C)

  28. Ecell/Voltage/electromotive force(emf): Difference in the electric potential of the two cells. Electrons flow from –ve to + ve electrode Ecell > 0 for a spontaneous reaction 1 Volt (V) = 1 Joule (J)/ Coulomb (C) Ecell <0- nonspontaneous Ecell=0, equilibrium. Cell Potential: Output of a Voltaic Cell

  29. Table 21.1 Voltages of Some Voltaic Cells Voltaic Cell Voltage (V) 1.5 Common alkaline battery 2.0 Lead-acid car battery (6 cells = 12V) 1.3 Calculator battery (mercury) Electric eel (~5000 cells in 6-ft eel = 750V) 0.15 Nerve of giant squid (across cell membrane) 0.070

  30. E0cell The potential measured at 298 K, with no current flowing and components in their standard states: 1 atm for gases, 1M for solutions, pure solid for electrodes. Standard cell potentials

  31. E0 values corresponding to reduction half-reactions with all solutes at 1 M and 1 atm are called standard reduction potentials E 0cell= E 0cathode(reduction) – E 0anode(oxidation) Standard Electrode( Half-Cell) potentials

  32. Hydrogen gas at 1 atm is passed over a Pt electrode in contact with 1 M H+ ions. Electrode is assigned zero volts. Standard Hydrogen Electrode

  33. Oxidation half-reaction Zn(s) Zn2+(aq) + 2e- Reduction half-reaction 2H3O+(aq) + 2e- H2(g) + 2H2O(l) Overall (cell) reaction Zn(s) + 2H3O+(aq) Zn2+(aq) + H2(g) + 2H2O(l) Determining an unknown E0half-cell with the standard reference (hydrogen) electrode. Figure 21.7

  34. PROBLEM: A voltaic cell houses the reaction between aqueous bromine and zinc metal: Br2(aq) + Zn(s) Zn2+(aq) + 2Br-(aq) E0cell = 1.83V anode: Zn(s) Zn2+(aq) + 2e- E = +0.76 E0Zn as Zn2+(aq) + 2e- Zn(s) is -0.76V Sample Problem 21.3: Calculating an Unknown E0half-cell from E0cell Calculate E0bromine given E0zinc = -0.76V PLAN: The reaction is spontaneous as written since the E0cell is (+). Zinc is being oxidized and is the anode. Therefore the E0bromine can be found using E0cell = E0cathode - E0anode. SOLUTION: E0cell = E0cathode - E0anode = 1.83 = E0bromine - (-0.76) E0bromine = 1.86 - 0.76 = 1.07 V

  35. Consider a galvanic cell:Al3+(aq) + Mg (s) →Al (s) + Mg2+ Give the balanced cell reaction and calculate Eo for the cell. 3) A galvanic cell is based on the reaction ; MnO4- (aq) +H+ (aq) +ClO3-(aq) → ClO4-(aq) + Mn2+ (aq) + H2O(l) Give the balanced cell reaction and calculate Eo for the cell. Problems

  36. F2(g) + 2e- 2F-(aq) Cl2(g) + 2e- 2Cl-(aq) MnO2(g) + 4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l) Ag+(aq) + e- Ag(s) Fe3+(g) + e- Fe2+(aq) O2(g) + 2H2O(l) + 4e- 4OH-(aq) strength of reducing agent Cu2+(aq) + 2e- Cu(s) strength of oxidizing agent N2(g) + 5H+(aq) + 4e- N2H5+(aq) Fe2+(aq) + 2e- Fe(s) 2H2O(l) + 2e- H2(g) + 2OH-(aq) Na+(aq) + e- Na(s) Li+(aq) + e- Li(s) Table 21.2 Selected Standard Electrode Potentials (298K) Half-Reaction E0(V) +2.87 +1.36 +1.23 +0.96 +0.80 +0.77 +0.40 +0.34 0.00 2H+(aq) + 2e- H2(g) -0.23 -0.44 -0.83 -2.71 -3.05

  37. Example: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Writing Spontaneous Redox Reactions • By convention, electrode potentials are written as reductions. • When pairing two half-cells, you must reverse one reduction half-cell to produce an oxidation half-cell. Reverse the sign of the potential. • The reduction half-cell potential and the oxidation half-cell potential are added to obtain the E0cell. • When writing a spontaneous redox reaction, the left side (reactants) must contain the stronger oxidizing and reducing agents. stronger reducing agent stronger oxidizing agent weaker oxidizing agent weaker reducing agent

  38. PROBLEM: (a) Combine the following three half-reactions into three balanced equations (A, B, and C) for spontaneous reactions, and calculate E0cell for each. (1) NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l) E0 = 0.96V (2) N2(g) + 5H+(aq) + 4e- N2H5+(aq) E0 = -0.23V (3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) E0 = 1.23V Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength (b) Rank the relative strengths of the oxidizing and reducing agents: PLAN: Put the equations together in varying combinations so as to produce (+) E0cell for the combination. Since the reactions are written as reductions, remember that as you reverse one reaction for an oxidation, reverse the sign of E0. Balance the number of electrons gained and lost without changing the E0. In ranking the strengths, compare the combinations in terms of E0cell.

  39. E0 = 0.96V Rev (2) N2H5+(aq) N2(g) + 5H+(aq) + 4e- E0 = +0.23V (1) NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l) (1) NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l) (3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) (3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) (A) 4NO3-(aq) + 3N2H5+(aq) + H+(aq) 4NO(g) + 3N2(g) + 8H2O(l) (2) N2H5+(aq) N2(g) + 5H+(aq) + 4e- Rev (1) NO(g) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3e- (1) NO(g) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3e- E0 = -0.96V E0 = 1.23V X2 (B) 2NO(g) + 3MnO2(s) + 4H+(aq) 2NO3-(aq) + 3Mn3+(aq) + 2H2O(l) Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength continued (2 of 4) SOLUTION: (a) E0cell = 1.19V X4 X3 E0cell = 0.27V X3

  40. Rev (2) N2H5+(aq) N2(g) + 5H+(aq) + 4e- (2) N2H5+(aq) N2(g) + 5H+(aq) + 4e- E0 = +0.23V E0 = 1.23V (3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) (3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) (C) N2H5+(aq) + 2MnO2(s) + 3H+(aq) N2(g) + 2Mn2+(aq) + 4H2O(l) Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength continued (3 of 4) E0cell = 1.46V X2 (b) Ranking oxidizing and reducing agents within each equation: (A): oxidizing agents: NO3- > N2 reducing agents: N2H5+ > NO (B): oxidizing agents: MnO2 >NO3- reducing agents: NO > Mn2+ (C): oxidizing agents: MnO2 > N2 reducing agents: N2H5+ > Mn2+

  41. Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength continued (4 of 4) A comparison of the relative strengths of oxidizing and reducing agents produces the overall ranking of Oxidizing agents: MnO2 > NO3- > N2 Reducing agents: N2H5+ > NO > Mn2+

  42. 1. Metals that can displace H from acid: If E0 cell for the reduction of H+ is more +ve for metal A than metal B, metal A is stronger R.A than metal B and a more active metal. They are placed below H. (+ve E0) 2. Metals that cannot displace H from acid: Higher the metal in the list more –ve is the E0 cell and less active it is. 3. Metals that can displace H from water: Grop IA, II A can displace H2 from H2 O. 4. Metals that can displace other metals from solution: Lower in list can reduce higher in list. Relative Reactivities (Activities) of Metals

  43. F2(g) + 2e- 2F-(aq) Cl2(g) + 2e- 2Cl-(aq) MnO2(g) + 4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l) Ag+(aq) + e- Ag(s) Fe3+(g) + e- Fe2+(aq) O2(g) + 2H2O(l) + 4e- 4OH-(aq) strength of reducing agent Cu2+(aq) + 2e- Cu(s) strength of oxidizing agent N2(g) + 5H+(aq) + 4e- N2H5+(aq) Fe2+(aq) + 2e- Fe(s) 2H2O(l) + 2e- H2(g) + 2OH-(aq) Na+(aq) + e- Na(s) Li+(aq) + e- Li(s) Table 21.2 Selected Standard Electrode Potentials (298K) Half-Reaction E0(V) +2.87 +1.36 +1.23 +0.96 +0.80 +0.77 +0.40 +0.34 0.00 2H+(aq) + 2e- H2(g) -0.23 -0.44 -0.83 -2.71 -3.05

  44. Oxidation half-reaction Ca(s) Ca2+(aq) + 2e- Reduction half-reaction 2H2O(l) + 2e- H2(g) + 2OH-(aq) Overall (cell) reaction Ca(s) + 2H2O(l) Ca2+(aq) + H2(g) + 2OH-(aq) The reaction of calcium in water. Figure 21.8

  45. -wmax -Ecell = charge In the standard state - charge = n F DG0 = -n F E0cell n = #mols e- DG0 = - RT ln K F = Faraday constant E0cell = - (RT/n F) ln K Free Energy and Electrical Work DG a -Ecell DG = wmax = charge x (-Ecell) DG = -n F Ecell F = 96,485 C/mol e- 1V = 1J/C F = 9.65x104J/V*mol e-

  46. DG0 K E0cell Reaction at standard-state conditions spontaneous at equilibrium nonspontaneous By substituting standard state values into E0cell, we get E0cell = (0.0592V/n) log K (at 298 K) E0cell = -RT lnK nF The interrelationship of DG0, E0, and K. Figure 21.9 DG0 < 0 > 1 > 0 0 1 0 > 0 < 1 < 0 DG0 = -nFEocell DG0 = -RT lnK E0cell K

  47. 5. Calculate ∆G0 for the reaction:Cu2+ (aq) + Fe(s) →Cu (s) + Fe2+ (aq) Is this reaction spontaneous? Problem

  48. PROBLEM: Lead can displace silver from solution: Pb(s) + 2Ag+(aq) Pb2+(aq) + 2Ag(s) As a consequence, silver is a valuable by-product in the industrial extraction of lead from its ore. Calculate K and DG0 at 298 K for this reaction. E0 = 0.13V Pb2+(aq) + 2e- Pb(s) E0 = 0.80V Ag+(aq) + e- Ag(s) E0cell = 0.93V 2X Pb(s) Pb2+(aq) + 2e- Ag+(aq) + e- Ag(s) E0cell = 0.592V log K n n x E0cell (2)(0.93V) = 0.592V 0.592V Sample Problem 21.5: Calculating K and DG0 from E0cell PLAN: Break the reaction into half-reactions, find the E0 for each half-reaction and then the E0cell. Substitute into the equations found on slide SOLUTION: E0 = -0.13V E0 = 0.80V DG0 = -nFE0cell = -(2)(96.5kJ/mol*V)(0.93V) DG0 = -1.8x102kJ log K = K = 2.6x1031

  49. RT Ecell = E0cell - ln Q nF 0.0592 log Q Ecell = E0cell - n The Effect of Concentration on Cell Potential DG = DG0 + RT ln Q -nF Ecell = -nF Ecell + RT ln Q • When Q < 1 and thus [reactant] > [product], lnQ < 0, so Ecell > E0cell • When Q = 1 and thus [reactant] = [product], lnQ = 0, so Ecell = E0cell • When Q >1 and thus [reactant] < [product], lnQ > 0, so Ecell < E0cell

  50. 7. Consider a voltaic cell based on the reaction: Fe(s) + Cu2+(aq) →Fe2+(aq) + Cu(s) If [Cu2+] =0.30 M, what must [Fe2+] be to increase Ecell by 0.25 V above E0cell at 25 C? Problem:

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