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

What is the cell potential for the reaction

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

at 89 C when [Fe2+]= 3.30 M and [Mg2+]= 0.310 M .

Explanation / Answer

anode reaction: oxidation takes place

Mg(s) -------------------------> Mg+2 (aq) + 2e-   ,   E0Mg+2/Mg = - 2.38 V

cathode reaction : reduction takes palce

Fe+2(aq) + 2e- -----------------------------> Fe(s) , E0Fe+2/Fe = -0.44V

--------------------------------------------------------------------------------

net reaction: Mg(s) +Fe+2(aq) -------------------------> Mg+2 (aq) + Fe(s)

E0cell= E0cathode- E0anode

E0cell= E0Fe+2/Fe - E0Mg+2/Mg

          = -0.44 - (-2.38)

          = 1.94 V

standard cell potential =E0cell = 1.94 V

cell potential from nernest equation

Ecell = E0cell -(0.0591/n)* log [Mg+2]/[Fe+2]

Ecell = 1.94 - (0.059 x1/2) *log 0.310 /3.30}

Ecell   = 1.97 V

cell potential =Ecell   = 1.97 V

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