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)
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.
a) Since the copper electrode is negative, oxidation takes place at the copper electrode. The oxidation half-reaction for the Cu2+/Cu half-cell can be written as:
Cu(s) -> Cu2+(aq) + 2e-
The standard oxidation potential, Eox, for this half-reaction is +0.34 V.
b) The half-reaction for the Au2+/Au half-cell can be written as:
Au(NO3)2 + 2e- -> Au(s)
c) To write the overall balanced oxidation-reduction reaction, we need to multiply the half-reactions by appropriate coefficients so that the number of electrons in both reactions is equal. In this case, we need to multiply the Au2+/Au half-reaction by 2 to balance the electrons:
2Au(NO3)2 + 4e- -> 2Au(s)
Now, we can combine the half-reactions to get the overall balanced reaction:
Cu(s) + 2Au(NO3)2 + 4e- -> Cu2+(aq) + 2Au(s) + 4NO3-
The standard cell potential, Ecell, can be determined by summing the standard reduction potential (Ered) for the reduction half-reaction and the standard oxidation potential (Eox) for the oxidation half-reaction:
Ecell = Ered + Eox
Ecell = Ered - (-0.34 V)
Ecell = Ered + 0.34 V
d) Given that the Ecell is measured to be 1.36 V, and we know that Ecell = Ered + 0.34 V, we can rearrange the equation to solve for Ered:
Ered = Ecell - Eox
Ered = 1.36 V - 0.34 V
Ered = 1.02 V.
Therefore, the standard reduction potential Ered for the Au2+/Au half-cell is +1.02 V.
a) To determine the oxidation half-reaction for the Cu2+/Cu half-cell, we need to know that oxidation involves the loss of electrons. Since the copper electrode is negative, that means it is the site of oxidation. The Cu2+ ions in the solution are reduced to copper metal, Cu. The half-reaction can be written as follows:
Cu2+ (aq) + 2e- -> Cu (s)
The standard oxidation potential, Eox, can be found using the given Ered value for the Cu2+/Cu half-cell. Since the standard reduction potential, Ered, is +0.34V, the standard oxidation potential, Eox, is the negative of Ered:
Eox = -0.34V
b) The half-reaction for the Au2+/Au half-cell can be determined using the information given. Since the copper electrode is negative, the gold electrode must be positive, which means reduction takes place at the gold electrode. The Au2+ ions in the solution are reduced to gold metal, Au. The half-reaction can be written as follows:
Au2+ (aq) + 3e- -> Au (s)
c) To write the overall balanced oxidation-reduction reaction, we need to ensure that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction. We will multiply each half-reaction by the appropriate coefficients to balance the electrons.
Cu2+ (aq) + 2e- -> Cu (s) (Oxidation - multiplied by 3)
3Au2+ (aq) + 9e- -> 3Au (s) (Reduction)
Adding these two equations together yields the overall balanced oxidation-reduction reaction:
3Cu2+ (aq) + 6Au2+ (aq) -> 3Cu (s) + 3Au (s)
The standard cell potential, Ecell, can be calculated by summing the standard reduction potential, Ered, and the standard oxidation potential, Eox:
Ecell = Ered + Eox
Ecell = (+0.34V) + (-0.34V)
Ecell = 0V
d) Since the standard cell potential, Ecell, is 1.36V and the standard oxidation potential, Eox, is -0.34V, we can find the standard reduction potential, Ered, using the equation:
Ered = Ecell - Eox
Ered = 1.36V - (-0.34V)
Ered = 1.7V
Therefore, the standard reduction potential, Ered, for the Au2+/Au half-cell is 1.7V.