A voltaic cell is constructed that uses the following reaction and operates at 298 K?

Question

Zn(s) + Ni2+(aq) -----> Zn2+(aq) + Ni(s)

a)What is the emf of this cell under standard conditions?

b)What is the emf of this cell when [Ni2+] = 3.16 M and [Zn2+] = 0.151 M?

Answer 1

To answer your question, I have to get the following Standard Reduction Potentials:

Zn(2+) + 2 e- ==> Zn(s), E° = -0.76v . . . (1)

Ni(2+) + 2 e- ==> Ni(s), E° = -0.25v . . . . (2)

(2)-(1):

Zn(s) + Ni(2+) ==> Zn(2+) + Ni(s), E° = 0.51V

a) the emf of this cell under standard conditions is 0.51V

b) the emf of this cell when [Ni2+] = 3.16M and [Zn2+] = 0.151M is:

E = E° - RT/nF Ln([Zn2+]/[Ni2+])

= 0.51-(8.314*298/(2*96485))*ln(0.151/3.16... = 0.55 (V)