(a) (b) Calculate the mass of copper produced by the reduction of Cu" ion at the cathode during the passage of 1.60 amperes of current through a solution of copper (ii) sulphate for 1 hour. [Cu 63.5, F 96500C mol"] Ag(s); E+0.800V and Given that, Ag te Mn +2c-- Mn(s); E=-1.18 V Use the data to calculate the voltage, Eº of the cell: Mn(s)/Mn (0.400M) // Ag' (0.150M)/ Ag(s)
(a) (b) Calculate the mass of copper produced by the reduction of Cu" ion at the cathode during the passage of 1.60 amperes of current through a solution of copper (ii) sulphate for 1 hour. [Cu 63.5, F 96500C mol"] Ag(s); E+0.800V and Given that, Ag te Mn +2c-- Mn(s); E=-1.18 V Use the data to calculate the voltage, Eº of the cell: Mn(s)/Mn (0.400M) // Ag' (0.150M)/ Ag(s)
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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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Give handwritten answer of both part otherwise I will give you downward
![(a)
(b)
Calculate the mass of copper produced by the reduction of Cu² ion at the cathode
during the passage of 1.60 amperes of current through a solution of copper (ii)
sulphate for 1 hour.
[Cu 63.5, F=96500C mol¹]
Given that,
Mn +20
Ag te
→ Ag(s);
E = +0.800V and
Mn(s);
E"=-1.18 V
Use the data to calculate the voltage, E" of the cell:
Mn(s) / Mn²
(0.400M) // Ag (0.150M)/ Ag(s)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb8c403c5-a112-493b-b322-67308d1d6c02%2F593dce49-1be2-46c4-b6e7-2ed94a0cccf8%2Fwzt7hl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(a)
(b)
Calculate the mass of copper produced by the reduction of Cu² ion at the cathode
during the passage of 1.60 amperes of current through a solution of copper (ii)
sulphate for 1 hour.
[Cu 63.5, F=96500C mol¹]
Given that,
Mn +20
Ag te
→ Ag(s);
E = +0.800V and
Mn(s);
E"=-1.18 V
Use the data to calculate the voltage, E" of the cell:
Mn(s) / Mn²
(0.400M) // Ag (0.150M)/ Ag(s)
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