chemistry problem help!! please!!?

Question

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?
b)Write the equation for the oxidation reaction.
c)Write the equation for the reduction reaction.

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.

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?

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

Help

Answer 1

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