Consider the reaction Mg(s)+Fe2+(aq)?Mg2+(aq)+Fe(s) at 87?C , where [Fe2+]= 3.80

Question

Consider the reaction

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

at 87?C , where [Fe2+]= 3.80M and [Mg2+]= 0.210M .

Part A

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

Express your answer numerically.

Part B

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

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

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).

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

Part A

Queotient K= [Mg+2/Fe+2] = 0.21/3.8 = 0.055263158

Part B

T= 87 C = 360 K

Part C

n= No of electrons transfered = 2e-

Part D

E0 cell = E0cathode - E0anode

=?0.44 + 2.372. = 1.932 v

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