Learning Goal: To learn how to use the Nernst equation. The standard reduction p

Question

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=E2.303RTnFlog10Q

where E is the potential in volts, E is the standard potential in volts, R=8.314J/(Kmol) is the gas constant, T is the temperature in kelvins, n is the number of moles of electrons transferred,

F=96,500C/(mol e) is the Faraday constant, and Q is the reaction quotient.

Substituting each constant into the equation the result is

E=E0.0592 Vnlog10Q

Consider the reaction

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

at 89 C , where [Fe2+]= 3.10 M and [Mg2+]= 0.310 M .

Part A

What is the value for the reaction quotient, Q, for the cell?

Express your answer numerically.

Part B

What is the value for the temperature, T, in kelvins?

Express your answer to three significant figures and include the appropriate units.

Part C

What is the value for n?

Express your answer as an integer and include the appropriate units (i.e. enter mol for moles).

PART D

Calculate the standard cell potential for

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

Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

1.Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s), Reaction coefficient , Q= [Mg+2]/ [Fe+2] = 0.310/3.1=0.1

2. Temperature is 89 deg.c= 89+273=372K

3. The two half reactions are Mg(s)------------> Mg+2+2e-            Eo=             2.38V        (1)

and Fe+2+2e------->Fe(s)               Eo=         -0.41V              (2)

as can be seen n=2 = no of electrons exchanged

Eq.1+Eq.2 gives Mg(s)+ Fe+2(aq)------------>Mg+2(aq)+Fe(s), Eo=2.38-0.41=1.97V

4. E= Eo=(0.0591/n)*logQ, E= stadard cell potential

E= 1.97- (0.0591/2)*log(0.1)=1.99955V

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