Consider the reaction Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s) at 89 C , where [Fe2+]= 3.40

Question

Consider the reaction

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

at 89 C , where [Fe2+]= 3.40 M and [Mg2+]= 0.310 M .

Part A

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

Express your answer numerically.

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Correct

Significant Figures Feedback: Your answer .0911 was either rounded differently or used a different number of significant figures than required for this part. If you need this result for any later calculation in this item, keep all the digits and round as the final step before submitting your 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.

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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.

Explanation / Answer

Part A :

Q = [Mg2+] / [Fe2+] = 0.310 / 3.40 = 9.12 x 10 -2

Part B: T = 273 + 89 = 362 K

Part C: n = 2 mol of electrons are transferred

Part D : E0Mg2+/Mg = -2.372 V; E0Fe2+/Fe = - 0.447 V

Therfore, E0cell = -0.447 - ( - 2.372) = 1.925 V

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