Question

When the Ag⁺ concentration is 1.31 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 1.672V. What is the Zn²⁺ concentration?

2Ag⁺(aq) + Zn(s)---->2Ag(s) + Zn²⁺(aq)

Answer: _______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

Ag+ + e− ⇌ Ag(s) +0.7996

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 = 0.7996 - (-0.762) = 1.5616 V

E°cell = 1.5616 V

Nernst Equation must be used...

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:

From the balanced reaction

Q = [Zn2+] / [Ag+]^2

n = 2 electrons transfer, T = 298K, R = 8.314 J/molK, F = 96500 C/mol

substitute

Ecell = E° - (RT/nF) x lnQ

Ecell = 1.5616 - 8.314*298/(2*96500)*ln([Zn2+] / [Ag+]^2)

1.672 = 1.5616 - 8.314*298/(2*96500)*ln([Zn2+] / (1.31^2))

solve for [Zn+2] = ?

(1.672-1.5616)/(-8.314*298) * (2*96500) = ln([Zn2+] / (1.31^2))

exp(-8.6) = Zn+2 / (1.31^2)

[Zn+2] = (1.31^2) * exp(-8.6)

[Zn2+] = 0.00031594 M