Question

Consider the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 79 ∘C , where [Fe2+]= 3.30 M and [Mg2+]= 0.310 M...

Consider the reaction

Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s)

at 79 ∘C , where [Fe2+]= 3.30 M and [Mg2+]= 0.310 M .

Part A

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

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.

Homework Answers

Answer #1

PART A:

Mg(s) + Fe2+(aq) → Mg2+(aq) + Fe(s)

[Fe2+] = 3.30 M and [Mg2+] = 0.310 M .

The reaction quotient for the cell,

Q = [products] / [reactants]

Q = 0.310 / 3.30

Q = 9.4 x 10-2

PART B:

Temperature, T, kelvins = 79 + 273 = 352K

PART C:

Mg(s) + Fe2+(aq) → Mg2+(aq) + Fe(s)

n =2

Because Mg looses 2 electrons and Fe gains 2 electrons.

PART D:

Mg(s) Mg^2+ + 2 e- Eo = +2.37 volts

Fe2+(aq) + 2 e- Fe(s) Eo = - 0.44 volts

Eo cell = +2.37 volts - 0.44 volts

= 1.93 volts

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