Asked by Mark
A voltaic cell is contructed in which a copper wire is placed in a 1.0 M Cu(NO3)2 solution and a strip of gold i placed in a 1.0 M AuNO3 solution. The measured potential of the cell is found to be 1.36V and the copper electrode is negative. The Ered for the Cu2+/Cu half- cell is +0.34
a) Since the copper electrode is negative, oxidation takes place at the copper electrode. Write the oxidation half-reaction for the Cu2+/Cu half-cell and indicate its standard oxidation potential, Eox
b)Write the half-reaction for the Au2+/Au half-cell.
c) write the overall balanced oxidation-reduction and indicate the standard cell potential Ecell
d) Calculate the standard reduction potential Ered for the Au2+/Au half-cell
a) Since the copper electrode is negative, oxidation takes place at the copper electrode. Write the oxidation half-reaction for the Cu2+/Cu half-cell and indicate its standard oxidation potential, Eox
b)Write the half-reaction for the Au2+/Au half-cell.
c) write the overall balanced oxidation-reduction and indicate the standard cell potential Ecell
d) Calculate the standard reduction potential Ered for the Au2+/Au half-cell
Answers
Answered by
DrBob222
a)
Cu ==> Cu^+2 + 2e and Eo = -0.34
b)
Au^+2 + 2e ==> Au
c)
Add a and b to obtain c and Ecell = 1.36 v
d)
If (a) is -0.34 v and the total cell (part c) is 1.36 v, what must (b) be; i.e.,
-0.34 + ? = 1.36
Note: There seems to be a discrepancy between the problem statement about AuNO3 and the question in part b.
Cu ==> Cu^+2 + 2e and Eo = -0.34
b)
Au^+2 + 2e ==> Au
c)
Add a and b to obtain c and Ecell = 1.36 v
d)
If (a) is -0.34 v and the total cell (part c) is 1.36 v, what must (b) be; i.e.,
-0.34 + ? = 1.36
Note: There seems to be a discrepancy between the problem statement about AuNO3 and the question in part b.
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