Consider the reaction Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s) at 59 C , where [Fe2+]= 3.80

Question

Consider the reaction

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

at 59 C , where [Fe2+]= 3.80 M and [Mg2+]= 0.210 M .

Part A

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

Express your answer numerically.

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.

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).

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.

Q =

Explanation / Answer

A)

Q = [Mg2+]/[Fe2+]

= 0.210/3.80

= 0.0553

Answer: 0.0553

B)

T = 59 oC = (59 + 273) K = 332 K

C)

Number of electrons being transferred is 2

so,

n = 2

D)

from data table:

Eo(Mg2+/Mg(s)) = -2.372 V

Eo(Fe2+/Fe(s)) = -0.44 V

According to reaction Mg is getting oxidised, so it is anode

Fe is cathode

here:

cathode is (Fe2+/Fe(s))

anode is (Mg2+/Mg(s))

Eocell = Eocathode - Eoanode

= (-0.44) - (-2.372)

= 1.932 V

Answer: 1.932 V

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.