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12. The described electrochemical cell (concentration cell) was set up: Cu | CuSO4 (c1) || CuSO4 (c2) | Cu From the previous cell reaction: Determine which electrode is the positive terminal (+) of the cell and calculate the cell potential E (electromotive force, EMF) for the given concentrations of the electrolyte CuSO4 in the cell (25C): b) c1 = 1×10-4 mol dm-3, c2 = 0.01 mol dm-3 EMF=0.2808 V, left hand cathode EMF=-0.2808 V, right hand cathode EMF=-0.0592 V, left hand cathode EMF 0.0592 V, right hand cathode

12. The described electrochemical cell (concentration cell) was set up: Cu | CuSO4 (c1) || CuSO4 (c2) | Cu From the previous cell reaction: Determine which electrode is the positive terminal (+) of the cell and calculate the cell potential E (electromotive force, EMF) for the given concentrations of the electrolyte CuSO4 in the cell (25C): b) c1 = 1×10-4 mol dm-3, c2 = 0.01 mol dm-3 EMF=0.2808 V, left hand cathode EMF=-0.2808 V, right hand cathode EMF=-0.0592 V, left hand cathode EMF 0.0592 V, right hand cathode

Fundamentals Of Analytical Chemistry
Fundamentals Of Analytical Chemistry
9th Edition
ISBN:9781285640686
Author:Skoog
Publisher:Skoog
Chapter18: Introduction To Electrochemistry
Section: Chapter Questions
Problem 18.18QAP
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12. The described electrochemical cell (concentration cell) was set up:
Cu | CuSO4 (c1) || CuSO4 (c2) | Cu
From the previous cell reaction:
Determine which electrode is the positive terminal (+) of the cell and calculate the cell potential E (electromotive force, EMF) for the given concentrations of the electrolyte CuSO4 in the cell (25C):
b) c1 = 1×10-4 mol dm-3, c2 = 0.01 mol dm-3

  • EMF=0.2808 V, left hand cathode
  • EMF=-0.2808 V, right hand cathode
  • EMF=-0.0592 V, left hand cathode
  • EMF 0.0592 V, right hand cathode
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