. Calculate the Gibbs free energy changes, AG, by using the potential differences measured in Part I andII by using the formula, AG = -nFɛ. n is the number of electrons transferred in the reaction. F is the Faraday'sconstant and equals to 96500 Coulomb.|3|Total cell reaction: Zn(s) + Cu"(aq) -Zn*(aq) + Cu(s)Part I. 1. Beaker: 0.1 M CUSO4 solution, 2. Beaker: 0.1M ZnSO4 solution.Part II. 1. Beaker: 0.1 M CuSO4 solution, 2. Beaker: 0.01M ZNSO4 solution.2. Find percent errors for part I and II. The theoretical value of standard cell potential, e, is 1.100 VCell >

Answered: Image /qna-images/answer/d3821f86-04ad-49cb-8bf6-9f164be4105b.jpg

1. Calculate the Gibbs free energy changes, AG, by using the potential differences measured in Part I and II by using the formula, AG = -nFɛ. n is the number of electrons transferred in the reaction. F is the Faraday's constant and equals to 96500 Coulomb. Total cell reaction: Zn(s) + Cu*(aq) - - Zn*(aq) + Cu(s) Part I. 1. Beaker: 0.1 M CuSO4 solution, 2. Beaker: 0.1M ZNSO4 solution.

1. Calculate the Gibbs free energy changes, AG, by using the potential differences measured in Part I and II by using the formula, AG = -nFɛ. n is the number of electrons transferred in the reaction. F is the Faraday's constant and equals to 96500 Coulomb. Total cell reaction: Zn(s) + Cu(aq) - - Zn(aq) + Cu(s) Part I. 1. Beaker: 0.1 M CuSO4 solution, 2. Beaker: 0.1M ZNSO4 solution.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

35. Calculate the equilibrium constant at 25 °C using standard cell potential data, Sn(s) + CuSO4(aq) = Cu(s) + SnSO4 (aq)
A certain metal M forms a soluble sulfate salt M₂SO4. Suppose the left half cell of a galvanic cell apparatus is filled with a 2.00 M solution of M₂ SO4 and the right half cell with a 10.0 mM solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them. The temperature of the apparatus is held constant at 25.0 °C. left Which electrode will be positive? x10 right X Ś ? What voltage will the voltmeter show? Assume its positive lead is connected to the positive electrode. 0 Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.
Can you break down the problem attached?
Consider the voltaic cell: Mg Mg2+ (1.0 M)||H* (1.0 M)|H₂ (1 atm)|Pt(s) Which of the following changes will cause a decrease in the cell voltage? increase the pressure of H₂ at 2.0 atm decrease the mass of the Mg electrode O decrease the concentration of Mg2+ ion lower the pH of the catholyte
Consider a voltaic cell in which the reaction is Zn (s) + 2H+ (aq) → Zn2+ (aq) + H2 (g) It is found that the cell voltage is +0.530 V when [Zn2+] = 1.0 M, PH2 = 1.0 atm. What must be the concentration of H+ in the cell? (Use the following information) (Show your work) 2H+ (aq) + 2e- → H2 (g) Eºred = 0.000 V Zn2+ (aq) + 2e- → Zn(s) Eºred = -0.762 V
Questions 20 and 21 refer to the following information. Galvanic Cell Cell Diagram E°cell (Volts) Mg | Mg2(1.0 M) || Pb*2(1.0 M) | Pb 1 2.24 Zn | Zn(1.0 M) || Pb*2(1.0 M) | Pb 2 0.63 Mg | Mg*(1.0 M) || Zn*2(1.0 M) | Zn 3 The chemical reaction occurring in first galvanic cell is: Mg + Pb*2 → Mg+2 + Pb 21. If the concentration of Mg+2 is changed from 1.0 M to 3.0 M in the galvanic cells above, what will happen to the observed cell voltages in galvanic cells 1 and 3? (A) The voltage in both galvanic cells 1 and 3 will increase. (B) The voltage will decrease in both galvanic cells 1 and 3. (C) The voltage in galvanic cell 1 will increase and will decrease in galvanic cell 3 (D) The voltage in galvanic cell 3 will increase and will decrease in galvanic cell 1. 22. Consider a voltaic cell based on these half-cells. Ag (aq) +e → Ag/s) Cd (aq) + 2e → Cd¢s) E* = +0.80 V E = -0.40 V Identify the anode and give the voltage of this cell under standard conditions. (A) Ag; Ecell = 0.40 V (B) Ag;…
Consider the following voltaic cellat 25°C: Al(s) | Al3*(0.20 M, aq) || Al3+(0.75 M, aq) | Al(s) If the standard reduction potential for the Al3+(aq) /Al(s) redox couple is -1.66 V, what is the overall cell reaction and the cell potential of the voltaic cell shown above? (Reference the cell as written above.) The overall cell reaction is Al3+(0.75 M, aq) AIS*(0.20 M, ag) and E°cell = +0.0113 V. There is no overall cell reaction and E°cell = 0.00 V. The overall cell reaction is Al3+(0.20 M, aq) Als*(0.75 M, aq) and E°cell = +0.0113 V. The overall cell reaction is Al3+(0.20 M, aq) →A13+(0.75 M, ag) and E°cell| = -0.0113 V. The overall cell reaction is Al3*(0.75 M, aq) →AI3*(0.20 M, aq) and E°cell = -0.0113 V. None of these
Calculate AG for the following cell reaction: Tl(s) | Tl+ (aq) || Ni²+ (aq) | Ni(s) From AG, determine the standard potential for the half-reaction Tl+ (aq) + e → Tl(s) AG; for Tl+ (aq) is -32.4 kJ/mol. 2+ AG; for Ni²+ (aq) is -45.60 kJ/mol. Standard Electrode (Reduction) Potentials in Aqueous Solution at 25°C Cathode (Reduction) Half-Reaction Ni²+ (aq) + 2e Ni(s) AG = J Standard Potential, E° (V) -0.23 Standard reduction potential for the half-reaction = V
Consider the following half-reactions and their standard reduction potentials then give the standard line (cell) notation for a voltaic cell built on these half reactions. Mn2*(aq) + 2 e Mn(s) E° = -1.18 V Fe(aq) +3 e Fe(s) E° = -0.036 V A. Fe(s) | Fe3t(aq, 1.0 M) || Mn (s) | Mn 2*(aq. 1.0 M) B. Mn (s) | Mn 2*(aq, 1.0 M) | | Fe(s) | Fe(aq. 1.0 M) C. Fe(s) | Fe3t(aq. 1.0 M) || Mn2*(aq, 1.0 M) | Mn(s) D. Mn (s) | Mn 2*(aq. 1.0 M) || Fe(aq, 1.0 M) | Fe(s)
Suppose the following reaction happening at 25°C Sn(s) + CuSO4(aq) SNSO4(aq) (reaction in equilibrium) Cu(s) + <----- a) Write the left (Anode) and right (Cathode) reactions. b) Calculate the standard cell potential E°Cell· c) Calculate the equilibrium constant “K" of the reaction using the standard potential data table. TABLE 17.1 Standard Reduction Potentials at 25°C (298 K) for Many Common Half-Reactions Half-Reaction e (V) Half-Reaction e° (V) F + 2e 2F Ag* +e - Ag" Co +e →Cơ H,O, + 2H + 2e 2H,0 Ce +e → Ce* PbO, + 4H + so + 2e PbSO, + 2H,0 MnO, + 4H* + 3e MnO, + 2H,O 2e + 2H + IO, 10, + H,0 MnO, + 8H + Se" → Mn* + 4H,0 Au + 3e -→ Au PbO, + 4H + 2e Pb* + 2H,O Ca, + 2e" → 2CI" Cr0,- + 14H + 6e" → 2Cr“ + 7H,0 O, + 4H + 4e 2H,O MnO, + 4H + 2e Mn + 2H,O 10, + 6H + Se → I, + 3H,0 Br, + 2e 2Br -2H +e VO + H,O 2.87 1.99 O, + 2H,0 + 4c → 40H Cu + 2e Cu Hg,Cl, + 2e -→ 2Hg + 2CI AgCI +e Ag + CI so- + 4H + 2e → H,SO, + H,O Cu* + e - Cu* 2H + 2e H; Fe + 3e Fe Pb* + 2e" -→ Pb Sn Ni* + 2e Ni…
What will be the products of the electrolysis of an 1-M aqueous aluminum chloride (AICI3) solution at 25 °C? (The electrodes themselves are platinum.) (Use the standard electrode potentials table in your online textbook's Appendix to help you answer the question.) No products will be produced. H₂(8) +20H (aq) + Cl₂(g) 4Al(s) + 30₂(g) + 12H*(aq) O 2H₂(g) + 4H₂O(l) + O₂(g) O2Al(s) + 3Cl2(g)
Consider the cell diagram: Pb(s)|Pb(NO3)2(aq)||FeCl3(aq)|Fe(s). Write the balanced net-ionic equations for the half-reactions that occur at the two electrodes. a. Anode= b. Cathode=