Question

Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Cr2 ] = 0.834 M and [Sn2 ] = 0.0190 M. Standard reduction potentials can be found here.

Answer 1

Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Cr2 ] = 0.834 M and [Sn2 ] = 0.0190 M. Standard reduction potentials can be found here.

Sn2+ + 2 e− ⇌ Sn(s) −0.13

Cr2+(aq) + 2e– → Cr(s) –0.91

Assume this goes in the forward direction so , Sn2+ will reduce, due to higher potential

E° = Ecathode - EAnode = -0.13 - -0.91 = 0.78 V

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

Q = [Cr+2]/Sn+2]M n = 2; F = 96500 C; T = 298K, R =8.314 ; E = 0.78 V

Ecell = E° - (RT/nF) x lnQ

Ecell = 0.78 - 8.314*298/(2*96500) * ln (0.834/0.019)

Ecell = 0.731452 V