onsider the reaction Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s) at 81 C , where [Fe2+]= 3.20 M

Question

onsider the reaction Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s) at 81 C , where [Fe2+]= 3.20 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 = SubmitHintsMy AnswersGive UpReview 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. T = SubmitHintsMy AnswersGive UpReview Part 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 = SubmitHintsMy AnswersGive UpReview Part 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.

Explanation / Answer

consider the reaction

Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s) at 81 C ,

where

[Fe2+]= 3.20 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 =

Q = [Mg 2+] / [Fe 2+]
= [0.210 / [ 3.20]
= 0.065625 ---answer

==============================================================================================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

T =
= C + 273 K = K
= 81 + 273 = 354 k

===============================================

Part 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).

As Mg loses 2 electrons while Fe gains 2 electrons in the reaction

n = 2

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=E oxidation + E reduction

E=E((2.303)(R)(T) / (nF)) (logQ)

Given that

Eo = 1.92

gas constant R= 8.314
Temperature = 354 k
n=2
Faraday = 96500
Q = 0.065625

E = 1.92 - ((2.303) (8.314) (354) / 2 * 96500 ) log 0.065625

E= 1.92 - (6778 / 193000) * -1.182

E= 1.96151 V

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