Answered: What is the Gibbs Free energy for this…

What is the Gibbs Free energy for this galvanic cell in kJ/mol (hint: make sure your reaction is balanced correctly) E°, = 2.87 F2 (g) + 2e- F¯(aq) Al3+ + 3e- → Al(s) red E° -1.66 red

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...

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Half-reaction E° (V) Cr3+ (aq) + 3e- → Cr (s) -0.74 Fe2+ (aq) + 2e- → Fe (s) -0.440 Fe3+ (aq) + e- → Fe2+ (s) +0.771 Sn4+ (aq) + 2e- → Sn2+ (aq) +0.154 The standard cell potential (E°cell) for the voltaic cell based on the reaction below is ________ V.Sn2+ (aq) + 2Fe3+ (aq) → 2Fe2+ (aq) + Sn4+ (aq) a). +1.21 b). +0.617 c). -0.46 d). +1.39 e). +0.46
Consider a galvanic electrochemical cell constructed using Cr/Cr³+ and Zn/Zn²+ at 25 °C. The following half-reactions are provided for each metal: Cr³+(aq) + 3 e →→ Cr(s) E°red = -0.744 V Zn²+ (aq) + 2 e →→ Zn(s) Eºred = -0.763 V Write the balanced equation for the overall reaction in acidic solution. 3 Zn(s) + 2 Cr³+ (aq) → 3 Zn²+(aq) + 2 Cr (s) 1 1 + 4- Reset 2 H₂O+ 0 3_ ( ) 3 03 04 4 H H+ 5 5 11 Zn 6 6 ²+ 7 07 (s) H₂O ● 3+ (1) 8 9 0 e 4+ x H₂O 9 (g) (aq) □o Cr Delete
Galvanic cell has 2 half reactions O2(g) + 4H+(aq) + 4e- → 2H2O(l) E0red = +1.23 V Cr3+(aq) + 3e- → Cr(s) E0red = -0.74 V Write a balanced equation that describes the cell, ensuring that E0cell > 0 V and Calculate E0cell
Consider the standard reduction potentials and redox couples used in a galvanic cell. Unbalanced Half-Reaction E°(V) MnO4 (aq) + 3e MnO2(s) (at basic pH) 0.59 Fe3+(aq) + e- Fe2*(aq) 0.77 What is the balanced reaction taking place in the cell (in basic solution)? 2 H20(1) + Mn04 (aq) + 3 Fe3+(aq) → 3 Fe2*(aq) + MnO2(s) + 4 OH- MnO2(s) + 4 OH¯ + 3 Fe3+(aq) 3 Fe2+(aq) + 2 H20(1) + MnO4¯(aq) MO 3 Fe2*(aq) + 2 H2O(1) + MnO4¯(aq) MnO2(s) + 4 OH¯ + 3 Fe3+(aq) M 3 Fe2+(aq) + MnO2(s) + 4 OH → 2 H20(1) + MnO4 (aq) + 3 Fe3+(aq)
If you said it was possible to calculate the cell voltage, do so and enter your answer here. Round your Ov answer to 2 significant digits.
Consider the following standard reduction potentials in acid solution: Co2+ + 2e– ----> Co (s) E° = –0.28 V MnO4– + 8H+ + 5e– ---> Mn2+ + 4H2O E° = +1.51 V The strongest reducing agent listed above is:
Refer to the galvanic cell below (the contents of each half-cell are written beneath each compartment): Ni Ag 1.0 M Ni2+ 1.0 M Ag* The standard reduction potentials are as follow: Ni2+ + 2 e- → Ni(s), E° = -0.25 Ag* + e - Ag(s), E° = 0.7996 When current is allowed to flow, which species is oxidised? Select one: а. Ag b. Cannot be determined from the data given. C. Ni2+ O d. Ag* O e. Ni
Consider the following standard reduction potentials in acid solution: E°(V) Al⁺³ + 3e− → Al(s) −1.66 Ag⁺¹ + e− → Ag(s) +0.80 Sn⁺⁴ + 2e− → Sn⁺² +0.14 Fe⁺³ + e− → Fe⁺² +0.77 The strongest reducing agent among those shown above is A) Al⁰ B) Fe⁺³ C) Br⁻ D) Al⁺³
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
What is the voltage of the following cell? Use the Reduction Potential Table found in Chapter 17 notes. Cr(s)|Cr³*(0.25 M) || |2(s)|1-(0.20 M)|Pt(s)
Standard Reduction Potentials (Volts) at 25 °C F2 (g) + 2 e- 2 F- (aq) 2.870 O2(g) + 4 H3O+(aq) + 4 e- 6 H2O(l) 1.229 2 H2O(l) + 2 e- H2(g) + 2 OH-(aq) -0.828 Fe2+ (aq) + 2 e- Fe (s) -0.440 A 0.122 M neutral aqueous FeF2 solution is electrolyzed under 1 atm pressure using platinum electrodes.(a) Write the half-reactions expected to occur, taking into account the effect of the non-standard conditions on the electrode potentials.Half-reaction at anode: + + -> + + Half-reaction at cathode: + + -> + + (b) What is the expected decomposition potential? _______ V
What is the cell potential for the following cell at 25°C?Al ∣ Al3+(0.00212 M) Sn2+(0.435 M) ∣ SnAl3+ + 3 e- → Al(s) E° = -1.676 VSn2+ + 2 e- → Sn(s E° = -0.137 V