Question

Consider an electrochemical cell based on the following overall reaction, Fe(s) + 2Ag+(aq)  Fe2+(aq) + 2Ag(s) Fe2+(aq) + 2e– Fe(s), ℰ° = –0.44 V Ag+(aq) + e– Ag, ℰ° = 0.80 V Calculate the cell potential (in V) for this reaction at 25oC when the concentration of Ag+ ions is 0.050 M and the concentration of Fe2+ ions is 1.50 M. a. +1.32 V b. +1.50 V c. +1.20 V d. -1.32 V e. -1.20 V

Answer 1

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 = Ered - Eox = 0.80 --0.44 = 1.24 V

n = 2, Q= [Fe+2]/[Ag+]^2

Ecell = 1.24 - (8.314*298)/(2*96500) * ln(1.5 / (0.05^2))

Ecell = 1.157 V

round up to

Ecell = 1.2 V