1. Complex ion formation: Potential, E Potential, E cell cell before adding 6 M NH 3 2+ after Cu(NH3)4 formed 2+ Use Equation (10) to calculate the residual concentration of free Cu² ion in equilibrium in the solution. Take [Ag*] = 1M. Show your calculations 2+ with Cu(NH3)4 Equation 10: E = E cell cell 0.0592 2 ● log 0.4089 V 0.8012 V [Cu²+] [Ag]²

1. Complex ion formation: Potential, E Potential, E cell cell before adding 6 M NH 3 2+ after Cu(NH3)4 formed 2+ Use Equation (10) to calculate the residual concentration of free Cu² ion in equilibrium in the solution. Take [Ag*] = 1M. Show your calculations 2+ with Cu(NH3)4 Equation 10: E = E cell cell 0.0592 2 ● log 0.4089 V 0.8012 V [Cu²+] [Ag]²

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

Eºcell calculation, reaction quotient, e cell calculation
What is E° for the cell Mn | Mn²*(aq) 2+ || Ag*(aq) | Ag?
How many Faradays are required to convert one mole of Cr₂O7-² to Cr+³?
A 100.0 mL solution of 0.0200 M Fe3+ in 1 M HClO4 is titrated with 0.100 M Cu+, resulting in the formation of Fe2+ and Cu2+. A Pt indicator electrode and a saturated Ag∣∣AgCl electrode are used to monitor the titration. Calculate the values of ? for the cell after each of the given volumes of the Cu+Cu+ titrant have been added. 10.0 ml E = ? V 19.5 ml E = ? V
The reduction potential to the metal of Ba 2+ is -1.57 V and that of Cu 2+ is +0.34 V. The cell potential for the reaction Ba(s) + Cu 2+ (aq) → Ba 2+ (aq) + Cu(s) is therefore: a. (+1.57) + (+0.34) b. (+1.57) + (-0.34) c. (-1.57) + (+0.34) d. 2 x (-1.57) + 2 x (0.34) e. (-1.57) + (0.34)
Which of the statements below are true for the titration of Fe2t with standard Ce? I. The indicator electrode potential before the addition of Cet solution is calculated using the Fea+ half-reaction after calculating the equilibrium concentrations of Fe3+ and Fe2+ in solution. II. Before the equivalence point, the indicator electrode potential is calculated using the Fe3+ half-reaction because only [Fe3] and [Fe2*] are known. III. At the equivalence point, the indicator potential is the average of E" for Fet and Cet. IV. After the equivalence point, the indicator electrode potential is calculated using the Ce** half-reaction because only [Ce"] and [Ce2*] are known. 1, II, II, and IV I, II, and IV II, III, and IV 1, I1. and IV
Calculate the potential at the equivalence point for titration of 0.100MFe^2+ with 0.100M Ce^4+.Given the formal potential for the reduction E^0 of Fe^3+ to Fe^2+ is +0.77V, and the formal potential for the reduction of E^0 Ce^4+ to Ce^3+ is 1.45V.
What is the standard electrode potential of these redox couple at 25°C? 1) Fe/Fe3+2) Cu/Cu2+3) Zn/Zn2+4)Al/Al3+
A solution prepared by mixing 60.0 mL of 0.360 M AGNO, and 60.0 mL of 0.360 M TINO, was titrated with 0.720 M NaBr in a cell containing a silver indicator electrode and a reference electrode with a constant potential of 0.175 V. The reference electrode is attached to the positive terminal of the potentiometer, and the silver electrode is attached to the negative terminal. The solubility constant of TIBr is Kp = 3.6 × 10-6 and the solubility constant of AgBr is = 5.0 x 10-13. Ksp Which precipitate forms first? O TIBI O AgBr Which of the expressions shows how the cell potential, E, depends on [Ag*]? E = 0.175 V – [0.799 V – 0.05916log ([Ag+])] 1 E = |0.799 V – 0.05916log - 0.175 V [Ag*) E = [0.799 V – 0.05916log ([Ag*]D] – 0.175 V E = 0.175 V - |0.799 V – 0.05916log [Ag* * (ল) Calculate the first and second equivalence points of the titration. first equivalence point: mL second equivalence point: mL What is the cell potential after each of the given volumes of 0.720 M NaBr have been…
Please solve questions 2, 3, 4, 5
A 30.0 mL solution of 0.0180 M Rh3+ in 1 M HClO4 was titrated with 0.0900 M Ce4+ to give Rh6+ and Ce3+. Calculate the potential (vs. Ag | AgCl) at the equivalence point. At what volume of titrant will the potential be equal to the standard potential of the Ce half-reaction (ignoring the Ag | AgCl electrode)?
What is the standard electrode potential for the reaction 2 Cr + 3 Pb2* →3 Pb + 2 Cr+? Standard Reduction E° (V) V Potentials Pb2+(aq) + 2e Pb(s) -0.13 1 4 6. Cr3+(aq) + 3e- - Cr(s) -0.74 8. 6. +/- 0. 2. 7.