Compute thermodynamic potential:

Problem:

Nickel is to be deposited from a solution that is 0.l0 M in Ni²⁺ and buffered to pH 2.0. The Oxygen is evolved at a partial pressure of l atm at a platinum anode. A cell has resistance of 3.l5 ? and temperature is 25^oC. Compute

a) the thermodynamic potential needed to initiate the deposition of nickel

b) the IR drop for a current of 1.00A

c) the initial applied potential, given in which oxygen voltage is 0.85 V

d) the applied potential, needed when [Ni²⁺] is l × l0^-4 M  assuming that all other variables remain unchanged.

Ni²⁺ + 2e                ?              Ni(s) E^o = - 0.250V

O₂ (g) + 4H⁺ + 4e   ?            2H₂O E^o = + 1.23V

Answer:

a)         Depositon of Ni²⁺

E_cell    = E_cathode - E_anode

E_cathode   = E^oNi²⁺ - 0.0592/2 log 1/ [Ni²⁺]

= -0.25 - 0.0592/2 log l/0.l

= -0.25 -0.0296 = -0.2796V approximately - 0.28V

E_anode = 1.23 - 0.0592/4 log l/(l.00)(l.00 × l0^-2 )4

pH      =  2.00  or l0^-2 M

= 1.23 - 0.0592/4 log 1/l0^-8

=  l.23 - 0.118  =  1.112V

E_cell     = E_cathode   - E_anode

= -0.28 - (l.ll2) = -1.39 V

b)         IR        3.l5 × l.0 = - 3.l5V

c)         Initial applied potential

E_appl     =   E_cathode -     E_anode - IR - overvoltage

=          - 0.28 - l.ll - 3.l5 - 0.85   =             - 5.39 V

d)        E_cathode   =        E^oNi²⁺ - 0.0592/2 log 1/l0^-4=        - 0.25 - 0.118

E_cathode =          - 0.368V

E_applied potential = E_c - E_a - IR - overvoltage

= - 0.368 - 1.112 - 3.l5 - 0.85   = -5.48V

Applied potential needed when Ni²⁺ is l × l0^-4 M is   = - 5.48 V.