only correct answer please Consider the reaction Mg( s ) +Fe2+( a q ) Mg2+( a q

Question

only correct answer please

Consider the reaction

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

at 47 C , where [Fe2+]=3.80mol L1 and [Mg2+]=0.110mol L1 .

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.

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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.

Explanation / Answer

value of Q = [Mg 2+] / [Fe 2+]
= [0.110] / [ 3.80]
= 0.0289

value for T
= C + 273 K = K
= 47 + 273 = 320

value for n
= 2 mol (bcuz Mg looses 2 electrons and Fe gains 2 electrons)
= E=Eox+Ered
=Eox = 2.37 V
=Ered = -0.45 V
=2.37 + (-0.45)
= 1.92V
E=E((2.303)(R)(T) / (nF)) (logQ)
Eo = 1.92
R= 8.314
T= 320
n=2
F = 96500
Q = 0.297
E = 1.92 - ((2.303) (8.314) (320) / 2 * 96500 ) log 0.0297
E= 1.92 - (6395 / 193000) * -1.527
E = 1.92 + 0.048
E= 2.40 V

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