Question

Consider the reaction

Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s)

at 63 ∘C , where [Fe2+]= 2.90 M and [Mg2+]= 0.210 M .

Part A

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

Express your answer numerically.

Q =

7.24×10−2

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Part B

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

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

T =

336 K

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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).

n =

2 mol

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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.

E∘ =

1.92 V

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Part E

What is the cell potential for the reaction

Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s)

at 63 ∘C when [Fe2+]= 2.90 M and [Mg2+]= 0.210 M .

Answer 1

Part A

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

Express your answer numerically.

Solution :- reaction quotient Q is calculated as follows

Q= [product]/[reactant]

Q= [Mg2+]/[Fe2+]

Q= 0.210 /2.90

Q= 7.24*10^-2

Part B

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

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

Solution :-

T in Kelvin = T in Celcius + 273

                   = 63 C +273

                   = 336 K

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).

Solution :-

N is nothing but the number of electrons transferred in the reaction

Therefore for the given reaction n=2

Because 2 electrons transferred.

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.

Solution :-

Standard cell potential is calculated as

E^o Cell = E^o cathode – E^o anode

For the given reaction Mg chaning to Mg2+ is the anode reaction and Fe2+ chaning to Fe is the cathode reaction.

Using the standard reduction potential we can calculate the E^o cell

Standard reduction potential for the Mg2+ = -2.36 V and for Fe2+ = -0.44 V

E^o Cell = E^o cathode – E^o anode

             = -0.44 V - (-2.36 V)

             = 1.92 V

Part E

What is the cell potential for the reaction

Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s)

at 63 ∘C when [Fe2+]= 2.90 M and [Mg2+]= 0.210 M

Solution :-

Formula to calculate the cell potential Ecell is as follows

E cell = E^o Cell – (RT/nF)lnQ

Where R = 8.314 J per mol K

n = number of electrons transferred

F= faradays constant = 96485 C/mol

Q = reaction quotient

T= Kelvin temperature

Now lets put the values in the formula and calculate cell potential

E cell = 1.92 V – (8.314 J per mol K * 336 K/2*96485 C per mol ) ln [7.24*10^-2]

           = 1.92 V – (-0.038)

           = 1.96 V

Therefore Ecell = 1.96 V