± Introduction to the Nernst Equation Consider the reaction Mg(s) + Fe2+ (aq)Mg2

Question

± Introduction to the Nernst Equation Consider the reaction Mg(s) + Fe2+ (aq)Mg2+ (aq) + Fe(s) Learning Goal at 67C , where Fe2+ ] = 2.80mol L-1 and [Mg2+-0.3 10mol L-1 To learn how to use the Nernst equation The standard reduction potentials listed in any reference tables are only valid at standard state conditions (that is solutes at 1 mol L and gases at a partial pressure of 1 bar and at 25 C). To calculate the cell potential at non-standard-state conditions, one uses the Nernst equation Part A What is the value for the reaction quotient, Q, for the cell? Express your answer numerically = Eeell 0.0257 K 1 where Ecell is the potential in volts, Ecell is the standard potential in volts, R = 8.314 J K 1 mol-1 , T is the absolute temperature, n is the number of moles of electrons transferred in the balanced equation F = 96 500 C mol-1 (the Faraday's constant), and Q is the reaction quotient. Submit Hints My Answers Give Up Review Part Part B What is the value for the temperature, T, in kelvins? Express your answer to three significant figures and include the appropriate units IA TValue Units Submit Hints My Answers Give Up Review Part

Explanation / Answer

A)

consider the given reaction

Mg(s) + Fe+2 ---> Mg+2 + Fe (s)

the reaction quotient is given by

Q = [Mg+2] / [Fe+2]

we know that solids are not considered

now

given

[Mg+2] = 0.31

[Fe+2] = 2.8

so

Q = 0.31 / 2.8

Q= 0.1107

B)

given

temperature = 67 C

we know that

temperature in kelvin = temperature in celsius + 273

so

temperature = 67 + 273

temperature = 340 K

C)

we know that

oxidation takes place at anode

reduciton takes place at cathode

anode : oxidation

Mg ---> Mg+2 + 2e-

cathode : reduction

Fe+2 + 2e- ---> Fe

we can see that

two electrons are transferred

so

n = 2

D)

now

Eo cell = Eo cathode - Eo anode

Eo cell = Eo Fe+2/Fe - Eo Mg+2/Mg

using standard values

we get

Eo cell = -0.44 - ( -2.38)

Eo cell = 1.94

so

the value of standard cell potential is 1.94 V

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