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A voltaic cell has one half-cell with a Cu bar in a 1.00 M Cu²⁺ salt, and the other half-cell with a Cd bar in the same volume of a 1.00 M Cd²⁺ salt.(b) As the cell operates, [Cd²⁺] increases; find E°cell and ΔG when [Cd²⁺] is 1.95 M.

Respuesta :

The value of E°cell and ∆G is 0.74V and 133.53kJ respectively.

node reaction (oxidation reaction)

Cd ------ Cd(+2) + 2e-

Cathode reaction (reduction reaction)

Cu(+2) + 2e- -------- Cu

Redox reaction

Cu(+2) + Cd ------- Cu + Cd(2+)

E°cell = E° cathode - E° anode

E°cathode = 0.34V

E°anode = -0. 40 V

E° cell = 0.34 - (-0.40)

E°cell = 0.74V

∆G° = - n FE° cell

= -2 × 96485 × 0.74

= -142798 J

= - 142.8 kJ

∆G° = -142. 8 kJ

∆G° = R T in K

= -142. 8 = - 8.314 × 10 ^ -3 × 298 × In k

Ink = 57.64

K = 1.08 × 10 ^25

b) Cd(s) + Cu+2 (aq) ------- Cd+2 (aq) + Cu (s)

E cell = E° cell - (0. 0591 /2) log (Cd+2 /Cu+2)

= 0.74V – (0.0591/2) log(1.95/0.05)

= 0.692 V.

∆G = -n F Ecell

= -2 × 96485 × 0.692

= 133.53kJ

Thus, we found that the value of E°cell and ∆G is 0.74V and 133.53kJ respectively.

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