Consider the reaction Mg( s ) +Fe2+( a q ) Mg2+( a q ) +Fe( s ) at 85 C , where

Question

Consider the reaction

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

at 85 C , where [Fe2+]=3.50mol L1 and [Mg2+]=0.210mol L1 .

Part A

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

Express your answer numerically.

View Available Hint(s)

Submit

Part B

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

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

View Available Hint(s)

Submit

Part C

What is the value for n?

Express your answer as an integer.

View Available Hint(s)

Submit

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.

View Available Hint(s)

Submit

Explanation / Answer

Solution-

let us find which is oxidized and reduced
Oxidation: 2Fe2+--> 2Fe3+
or 2Fe2+ --> 2Fe3+ + 2e-
Reduced: Ni2+ + 2e- --> Ni

The E potential for each
Oxidized= 0.77V
Reduced= -0.23V

Now, let us subtract the smaller V from the larger. 0.77-(-0.23)= 1 V = E cell


and

Mg(s}+ Fe^2+(aq) ---->Mg^2+ (aq)+ Fe(s)

at 85C, where Fe^2+= 3.50M and Mg^2+= 0.210M .

PART A
What is the value for the reaction quotient, Q, for the cell?
Q = [prod] / [reactants]
Q = [0.210] / [3.50]
Q = 0.06


PART B
What is the value for the temperature, T, in kelvins?
85 C + 273 = 358 Kelvin


PART C
What is the value for n?
n = 2 electrons transferred from Mg to Fe+2


PART D
Calculate the standard cell potential for

Mg(s) ----> Mg^2+ and 2 e- lost .......Eo = +2.37 volts
Fe^2+(aq) and 2 e- taken --> Fe(s)......Eo = - 0.440 volts

Eo cell = 1.93 volts

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.