Question

Extra Practice on Nernst Equation.          For:       Zn²⁺ + 2e- à Zn(s)                           E^o = -0.763 V

Calculate the potential of a zinc electrode immersed in:

C.) 0.0100 M Zn(NH₃)₄²⁺ and 0.25 M NH₃ .b₄ for Zn(NH₃)₄²⁺ complex is 7.76 x 10⁸.

Answer 1

Note:
For a metal–metal half–cell equilibrium: Mⁿ⁺_(aq) + ne^– <=> M_(s)
the Nernst equation is  E_Mⁿ⁺/M = E^θ_Mⁿ⁺/M + (RT/nF) ln {[Mⁿ⁺_(aq)]/[M_(s)]}
Where,
R (the ideal gas constant).= 8.314 J mol^–1 K^–1
F (the Faraday constant). =96500 C mol^–1


Given
Zn2+ + 2e- <=>Zn(s) Eo = -0.763 V

From the note we can find answer

N:B:-Read the Note above carefully. This type of any question will answered by the given note.

• Zn²⁺_(aq) + 2e^–<=> Zn_(s) (E^θ = –0.763V)

• E_Zn2+/Zn = E^θ_Zn2+/Zn + (2.303RT/nF) lg [Zn²⁺_(aq)]

• E = –0.763 + {(2.303 x 8.314 x 298)/(2 x 96500)} lg [0.01]

• E_Zn2+/Zn = –0.763 + 0.0295(–6.6438)

• = –0.763- 0.196

• = -0.959 V