Question

Determine the cell potential and the equilibrium concentrations of the following concentration cell. Assume both cells have the same volume. Please see Concentration Cell for assistance. Cu | Cu2+ (0.013 M) || Cu2+ (2.3 M) | Cu

E V

anode M

cathode M

Answer 1

Remember that each species will have a specific reduction potential. Remember that this is, as the name implies, a potential to reduce. We use it to compare it (numerical) with other species.

Note that the basis if 2H+ + 2e- -> H2(g) reduction. Therefore E° = 0 V

All other samples are based on this reference.

Find the Reduction Potential of each reaction (Tables)

Zn2+ + 2 e− <-> Zn(s) Ered = −0.762 V

F2(g) + 2 e− <-> 2 F−   Ered = +2.87 V

The most positive has more potential to reduce, it will be reduced

The most negative will be oxidized, since it will donate it selectrons

For total E°cell potential:

E°cell = Ered – Eox

Eox = -Ered of the one being oxidized

E°cell = +0.34 - 0.34 = 0 V

this makes sense since it is the SAME material

the only thing that makes this possibl eis the chang ein concentration...

Nernst Equation

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

Ecell = 0 - 8.314*298/(2*96500) * ln(0.013/2.3)

Ecell = 0.013 V

Equilibrium concentrations:

anode --> 0.013 + x

cathode --> 2.3-x

0.013 + x = 2.3-x

2x = 2.3-0.013

x = (2.3-0.013 )/2

x = 1.1435

in equilibrium

anode --> 0.013 + 1.1435 = 1.1565 M

cathode --> 2.3-1.1435 = 1.1565 M