Question

Given the half-cells:

Ni/Ni⁺² and Fe/Fe⁺²

Chose the anode and cathode. Calculate E° and ΔG° for this reaction. Calculate what the new cell potential (E) would be if the concentrations were [Ni⁺²] = 0.40 M and [Fe⁺²] = 0.30 M. Calculate how long you would have to run the battery at 6 amps to make the E drop by 0.03 volts, and calculate how many grams of anode you lose by that time.

Using the equations:

ΔG° = -nFE°

ln K = (nE°) / 0.0257

E = E° - (0.0257 / n) ln Q

Charge (coulombs) = Amps x time (seconds)

Answer 1

The half cell reduction potentials are :

Ni/Ni⁺² = -0.257 Volts

Fe/Fe⁺² = -0.447 Volts

Answer 1:

According to the above data: Anode is Fe/Fe²⁺ half cell and Cathode is Ni/Ni²⁺ half cell. Oxidation takes place on anode and reduction takes place in cathode.

reaction is represented as=

Fe + Ni²⁺ ---------------> Fe²⁺ + Ni

Answer 2:

Now, E^o_cell = E_cathode - E_anode

E⁰_cell= (-0.257)-(-0.447) = 0.19 Volts

Answer 3:

ΔG°_reaction = -nFE⁰_cell ,

where n (number of electron change in reaction) = 2 mol e-, F (Faraday's constant)= 96485 coulomb/mol e^-

ΔG°_reaction = -2*96485*(0.19) = 36664.3 coulomb.volt = 36664.3 Joule = 36.664 kJ

Answer 4:

E_cell = E⁰_cell - (0.0591/n)*log(Q)

where Q= [Fe²⁺]/[Ni²⁺]

We have to find the E value when the concentrations are ([Ni²⁺]=0.40M and [Fe²⁺]=0.30M

E_cell = 0.19 - (0.0591/2)*log(0.40/0.30) = 0.19 - (0.02955)*.124938 = 0.18630 Volt

E_cell = 0.18630 Volt