A solution consisting of 0.0050M Cu(NO3)2 and 0.0100 M EDTA at pH 6 is placed in a cell with a Cu wire electrode. A S.C.E. reference was used. Calculate pCu and E for this cell. You may have to look up Kf of CuY2-.

A solution consisting of 0.0050M Cu(NO3)2 and 0.0100 M EDTA at pH 6 is placed in a cell with a Cu wire electrode. A S.C.E. reference was used. Calculate pCu and E for this cell. You may have to look up Kf of CuY2-.

Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Chapter17: Electrochemistry

Section: Chapter Questions

Problem 43P

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Question

(9) A solution consisting of 0.0050M Cu(NO3)2 and 0.0100 M EDTA at pH 6
is placed in a cell with a Cu wire electrode. A S.C.E. reference was used.
Calculate pCu and E for this cell. You may have to look up Kf of CuY²-.
(10) Starting with the following half-reactions:
Ca+ +2e = Cd(s)(E° = -0.402V)
Cd(C201) +2e-= Cd(s) + 2C20 (E° = -0.572)
Calculate B2 of Cd(C201)2:
Cd+ +2C,0 = Cd(C,0,)

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