A voltaic (galvanic) cell is sketched below. (a) Write the balanced equations fo

Question

A voltaic (galvanic) cell is sketched below.

(a) Write the balanced equations for the oxidation and reduction half-reactions taking place in the cell. (Include states-of-matter under SATP conditions in your answer.)

oxidation half-reaction:

reduction half-reaction:

(b) Write the balanced equation for the overall cell reaction. (Include states-of-matter under SATP conditions in your answer.)

THEN

Calculate the standard free energy change (G°) for each of the following reactions (after first calculating the cell potential).

(a)    3 Pb(s) + 2 Au3+(aq) 3 Pb2+(aq) + 2 Au(s)
___ J

(b)    3 Be(s) + 2 Cr3+(aq) 3 Be2+(aq) + 2 Cr(s)
___ J

Explanation / Answer

a)

Mg2+ + 2 e Mg(s) 2.372

Cd2+ + 2 e Cd(s) 0.40

The anode = electrode in which the OXIDATION takes place, therefore, the reducing agent will get oxidized (lose electrons). The electrons flow from the solid electrode to the voltmeter flow. The solid species will then turn ionic, and go into solution. Since the solid converts to ion in solution, then this electrode will lose mass as time passes by.

The cathode = electrode in which REDUCTION takes place, therefore, the oxidizing agent will get reduced (gain electrons). The electrons are received from the voltmeter ( which come from the anode ). The ions in solution will then form solid over the cathode. Since there is solid formation, there will be an increase in the mass of this electrode.

a) oxidation half cell

Mg(s) Mg2+ + 2 e

b) reduction half cell

Cd2+ + 2 e Cd(s)

B)

balanced reaction

Cd2+(aq) + Mg(s) <-> Mg2+(aq) + Cd(s)

PART 2

(a)    3 Pb(s) + 2 Au3+(aq) 3 Pb2+(aq) + 2 Au(s)

dG = -n*F*E°cell

n = 3*2 = 6 electrons, F = 96500 C/mol

Pb2+ + 2 e Pb(s) 0.126

Au3+ + 3 e Au(s) +1.52

E° = 1.52 - -0.126 = 1.646

dG = -6*96500*1.646

dG = -953034 J

(b)    3 Be(s) + 2 Cr3+(aq) 3 Be2+(aq) + 2 Cr(s)

Cr3+ + 3 e Cr(s) 0.74

Be2+ + 2 e Be 1.847

E° = Ered - Eox = -0.74 - -1.847 = 1.107 V

dG = -nF*E°cell

dG = -2*3*96500*1.107

dG = -640953 J

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