Question

A voltaic cell is constructed that uses the following reaction and operates at 298 K:
Zn(s)+Ni2+(aq)→Zn2+(aq)+Ni(s).

a. What is the emf of this cell when [Ni2+]= 2.10 M and [Zn2+]= 0.160 M ?

b. What is the emf of the cell when [Ni2+]= 0.100 M and [Zn2+]= 0.860 M ?

Thanks!

Answer 1

First, get E°cell

Ni2+ + 2 e− ⇌ Ni(s) −0.25

Zn2+ + 2 e− ⇌ Zn(s) −0.7618

E°cell = Ered - Eox = -0.25 - -0.7618

E°cell = 0.5118 V

Now...

When the cell is NOT under standard conditions, i.e. 1M of each reactants at T = 25°C and P = 1 atm; then we must use Nernst Equation.

The equation relates E°cell, number of electrons transferred, charge of 1 mol of electron to Faraday and finally, the Quotient retio between products/reactants

The Nernst Equation:

Ecell = E0cell - (RT/nF) x lnQ

In which:

Ecell = non-standard value

E° or E0cell or E°cell or EMF = Standard EMF: standard cell potential
R is the gas constant (8.3145 J/mol-K)
T is the absolute temperature = 298 K
n is the number of moles of electrons transferred by the cell's reaction
F is Faraday's constant = 96485.337 C/mol or typically 96500 C/mol
Q is the reaction quotient, where

Q = [C]^c * [D]^d / [A]^a*[B]^b

pure solids and pure liquids are not included. Also note that if we use partial pressure (for gases)

Q = P-A^a / (P-B)^b

substitute in Nernst Equation:

Ecell = E° - (RT/nF) x lnQ

a. What is the emf of this cell when [Ni2+]= 2.10 M and [Zn2+]= 0.160 M ?

Q = [Zn+2]/[Ni2+]

n = 2 e-, F = 96500 C/mol

Ecell = E° - (RT/nF) x lnQ

Ecell = 0.5118 - (8.314*298)/(2*96500) * ln(0.16/2.10)

Ecell = 0.544849 V

b. What is the emf of the cell when [Ni2+]= 0.100 M and [Zn2+]= 0.860 M ?

Q = [Zn+2]/[Ni2+]

n = 2 e-, F = 96500 C/mol

Ecell = E° - (RT/nF) x lnQ

Ecell = 0.5118 - (8.314*298)/(2*96500) * ln(0.86/0.10)

Ecell = 0.48417 V