A standard state galvanic cell is constructed from the half-cells described below.
Half-Cell 1: Copper rod in a solution containing 1.00 M Cu(NO3)2
Half-Cell 2: Silver rod in a solution containing 1.00 M AgNO3
a. Draw a diagram of the cell. Label the anode and cathode. Indicate the direction of electron flow
across the wire and ion flow through the salt bridge.
Write a balanced chemical equation for the cell reaction and calculate the E‹ and ΔG‹ for it.
c. The cell is allowed to run until the reaction reaches equilibrium. Calculate the concentration of
each ion at this point.
anode is the one in which oxidation occurs (left side)
cathode --> righ side, Ag will reduce
flow goes from left to right
Cu2+ + 2 e− ⇌ Cu(s) +0.337
Ag+ + e− ⇌ Ag(s) +0.7996
Copper has lower potential, so it will oxidize first
E°cell = 0.7996 - 0.337 = 0.7996 V
dG = -nF*E°cell
dG = -2*96500*0.7996
dG = -154322.8 J/mol = -154.32 kJ/mol
calcualte concnetration of each ion at this point...
Ecell is 99% gones so
[Ag+] = 0.01 M
[Cu+2] = 1.99 M
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