Question

Part E

Please show all of the steps

Introduction to the Nernst Equation Learning Goal: To learn how to use the Nernst equation. The standard reduction potentials listed in any reference table are only valid at standard-state conditions of 25 C and 1 M. To calculate the cell potential at non-standard-state conditions, one uses the Nernst equation, E – E – 2.303 RT log.Q where E is the potential in volts, E is the standard potential in volts, R = 8.314 J/(K.mol) is the gas constant, T is the temperature in kelvins, ne is the number of moles of electrons transferred, F = 96,500 /mole is the Faraday constant and Q is the reaction quotient.. Substituting each constant into the equation the result is E = EP - 0.0592 V log10 Q

Answer 1

Answer

1.97V

Explanation

Oxidation reaction

Mg(s) ------> Mg²⁺(aq) + 2e^- E°_red = -2.372V

Reduction half reaction

Fe²⁺(aq) + 2e^- -------> Fe(s) E°_red = -0.440V

E° = E°_red,cathode - E°_red,anode

= -0.440V - ( - 2.372V)

= 1.932V

number of electron transfer, n = 2

Overall reaction

Fe²⁺(aq) + Mg(s) ------> Fe(s) + Mg²⁺(aq)

Q = [Mg²⁺] /[Fe²⁺]

Q = 0.210M/ 2.90M = 0.07241

Nernst equation is

E = E° - (2.303RT/nF)logQ

R = gas constant , 8.314J/mol K

T = Temperature, 360.15K

F = Faraday constant, 96485C/mol

E = 1.932V - ( 2.303× 8.314J/mol K × 360.15K/2 × 96485C/mol)log( 0.07241)

E = 1.932V - (0.0357V)log(0.07241)

E = 1.932V + 0.041V

E = 1.97V