chemistry problem help!! please!!?

Question

A student measures the potential of a galvanic cell composed of 1 M CuSO4 in one solution and 1 M AgNO3 in the other. The cell contains a copper electrode in the CuSO4 solution and a silver electrode in the AgNO3 solution, and it is connected to a voltmeter. She finds that the cell potential, E^0_cell, is 0.45 V, with the copper electrode acting as the negative electrode. 
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 standard potential of the silver/silver ion electrode, E^0_Ag+/Ag, is taken to be 0.000 V for both oxidation and reduction, what is the value of the potential for the oxidation reaction of copper, E^0_Cu/Cu2+_oxid? (Note: E^0_cell = E^0_oxid + E^0_red.)
e) If E^0_Ag+/Ag_red equals 0.80 V, as listed in standard tables of electrode potentials, what is the value of the potential for the oxidation reaction of copper, E^0_Cu/Cu2+_oxid?
f) Write the net ionic equation for the spontaneous reaction that occurs in the cell studied by the student.

Anonymous · Accepted 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

Alexa Lakin

chemistry problem help!! please!!?

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