Question

A concentration cell consists of two Zn/Zn2+ half-cells. The concentration of Zn2+ in one of the half-cells is 2.0 M and the concentration in the other half-cell is 1.0×10−3M.

Indicate the half-reaction occurring at each electrode.
Express your answers as chemical equations separated by a comma. Identify all of the phases in your answer.

Answer 1

Zn2+ + Zn(s) = Zn(s) + Zn2+

then

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

E°cell = 0, since same potential, then only change in concentration is driving this

ln(Q) = [Zn2+]anode / [Zn+2]red

Ecell = 0 - (8.314*298)/(2*9600) * ln([Zn2+]anode / [Zn+2]red)

Ecell = 0 - (8.314*298)/(2*9600) * ln((10^-3) / 2))

Ecell = 0.9808 V

then

Zn2+ anode = 2 M

Zn2+ cathode = 10^-3

as time passes by,

Zn2+ (anode) decreases from 2 M to equilbirium concentration

Zn2+ cahtode increases from 10^-3 to equilbirium concentration

Zn2+(2M) + 2e- --> Zn(s)

Zn(s) --> Zn+2(10^-3.;) + 2e-