I answered A through E, are they correct? I aslo need help with f, g, and h. A s

Question

I answered A through E, are they correct? I aslo need help with f, g, and h.

A student measures the potential of a cell made up with 1 M CuSO4 in one solution and 1 M AgNO3 in the other. There is a Cu electrode in the CuSo4 and an Ag electrode in the AgNO3, and the cell is set up in a "voltmeter." She finds that the potential, or voltage, of the cell, E^0cell, is 0.45 V, and that the Cu electrode is negative

a)At which electrode is oxidation occurring? Cu
b)Write the equation for the oxidation reaction. Cu ----> Cu2+ + 2e-
c)Write the equation for the reduction reaction. Ag+ + e- ---> Ag

d)If the potential of the silver, silver ion electrode, E^0Ag+, Ag is taken to be 0.000 V in oxidation or reduction, what is the value of the potential for the oxidation reaction, E^0Cu, Cu2+oxid? E^0cell=E^0oxid + E^0red. 0.45

e) If E^0Ag+, Agred equals 0.80 V, as in standard tables of electrode potentials, what is the value of the potential of oxidation reaction of copper, E^0Cu, Cu2+oxid? -0.35

f) Write the net ionic equation for the spontaneous reaction that occurs in teh cell that the student studied.

g.) The student adds 6 M NH3 to be CuSO4 solution until the Cu2+ ion is essentially all converted to Cu(NH3)4 2+ ion. The voltage of the cell, Ecell, goes up to 0.92 V and the Cu electrode is still negative. Find the residual concentration of Cu 2+ ion in the cell

_______M

h.) IN Part g, [Cu(NH3)4 2+] is about 0.05M, and [NH3] is about 3 M. Given those values and the result in Part 1g for [Cu 2+], calculate K for the reaction:

Cu(NH3)4 2+ (aq) <-> Cu 2+ (aq) + 4(NH3) (aq)

Explanation / Answer

in an electrochemical cell, electrons spontaneously flow from the negative electrode to the positive electrode (the opposite situation applies in an electrolytic cell).

a) since oxidation is a loss of electrons, it occurs at the negative electrode (Cu).

b) Cu -> Cu2+ + 2e-

c) Ag+ + e- -> Ag

d) E0cell = 0.45 V, and E0red has been defined to be zero, so E0oxid = 0.45 V

e) 0.45 = E0oxid + 0.80; rearranging E0oxid = -0.35 V

f) add answers to b) and c) such that the number of electrons on either side is the same: you can see you have to multiply c) by 2. overall you should get

Cu + 2Ag+ -> Cu2+ + 2Ag

g)
From Nernst equation: E = E

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