Question

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.

E‹:____________________________

ΔG‹:____________________________

c. The cell is allowed to run until the reaction reaches equilibrium. Calculate the concentration of

each ion at this point.

Answer 1

Q1

anode is the one in which oxidation occurs (left side)

cathode --> righ side, Ag will reduce

flow goes from left to right

For

E°cell

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

for

dG

dG = -nF*E°cell

dG = -2*96500*0.7996

dG = -154322.8 J/mol = -154.32 kJ/mol

c)

calcualte concnetration of each ion at this point...

assume

Ecell is 99% gones so

[Ag+] = 0.01 M

[Cu+2] = 1.99 M