Question

Consider the following half reactions at 298 K

Fe²⁺ + 2 e^- → Fe    E^o = -0.441 V
Cd²⁺ + 2 e^- → Cd    E^o = -0.403 V

A galvanice cell based on these half reactions is set up under standard conditions where each solution is 1.00 L and each electrode weighs exactly 100.0 g. How much will the Cd electrode weigh when the nonstandard potential of the cell is 0.02880 V?

The answer is 139g.

Answer 1

Answer :

Given that

Fe2+ + 2 e- → Fe    Eo = -0.441 V = E⁰_anode
Cd2+ + 2 e- → Cd    Eo = -0.403 V = E⁰_cathode at 298K

So overall reaction will be:

Fe+2 + cd_(s)------------> Fe_(s) + cd2⁺

cd2+ is the cathode (reduction); Ne2+ is the anode (oxidation).

We know that Nernst equation can be written as:

E_cell = E⁰ _cell - (RT/nF) * ln(Q)

R = 8.314 J/K-mol = gas constant

n = 2 = number of electron trasfer

F = Faraday constant (96487 C mol–1) = 96500C/mol

Q = [products]^x / [reactants]^y

At standared condition E_cell = 0

therefor nernst equation will be:

E_cell = E ⁰_cell - (RT/nF) * ln(Q)

for standared condition

E ⁰_cell = (RT/nF) * ln(Q)

E⁰cell = E⁰_cathode  - E⁰_anode  

E⁰_anode = ( -0.441 V )  

E⁰_{cathode =}   -0.403 V

given E_cell = 0.0288 V

therefor E_cell = (-0.403) - (- 0.441 ) - (RT/2F)* In Q  

=> 0.0288 - 0.038 = - (8.314 * 298/2 * 96500) * In Q

=>> - 0.0092 = - 0.01283 * In Q

=> In Q = .71

=> Q = e0.71 = 2.0

= >  [products]^x / [reactants]^y = [cd+2]²/ [Fe+2]² = 2.0

=> [cd+2]/ (100g/ 1L) = 1.4

Answer = [cd+2] = 140g approximate 139g

If you have any doubt feel free to ask. Thanks!