Consider the reaction Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s) at 79 C , where [Fe2+]= 3.00

Question

Consider the reaction

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

at 79 C , where [Fe2+]= 3.00 M and [Mg2+]= 0.210 M .

Part A

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

Express your answer numerically.

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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.

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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).

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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.

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Explanation / Answer

A)
Q = [Mg2+]/[Fe2+]
= 0.210/3.00
= 0.070
Answer: 0.070

B)
T = 79 oC
= (79+273) K
= 352 K

C)
n = 2
Since there are 2 electrons being transferred

D)
Lets find Eo 1st
from data table:
Eo(Mg2+/Mg(s)) = -2.372 V
Eo(Fe2+/Fe(s)) = -0.44 V

As per given reaction/cell notation,
cathode is (Fe2+/Fe(s))
anode is (Mg2+/Mg(s))

Eocell = Eocathode - Eoanode
= (-0.44) - (-2.372)
= 1.932 V
Answer: 1.93 V

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