Question

In: Chemistry

Consider the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 89 ∘C , where [Fe2+]= 3.00 M and [Mg2+]= 0.110 M...

Consider the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 89 ∘C , where [Fe2+]= 3.00 M and [Mg2+]= 0.110 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?

Part C:

What is the value for n?

Part D:

Calculate the standard cell potential for

Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s)

Expert Solution

Mg(s) ----------------> Mg^2+ (aq) + 2e^- E0 = 2.38v

Fe^2+ (aq) + 2e^- ------> Fe(s) E0 = -0.41v

-----------------------------------------------------------------------------------

Mg(s) + Fe^2+ (aq) -----> Mg^2+ (aq) + Fe(s) E0cell = 1.97v

the standard cell potential for 1.97v

n = 2

T = 89+273 = 362K

Mg(s) + Fe^2+ (aq) -----> Mg^2+ (aq) + Fe(s)

[Fe2+]= 3.00 M and [Mg2+]= 0.110 M .

Q = [Mg^2+]/[Fe^2+]

= 0.11/3 = 0.0366

Ecell = E0cell - 0.0592/n logQ

= 1.97 -0.0592/2 log0.0366

= 1.97 -0.0296*-1.4365 = 2.0125v

part-A

Mg(s) + Fe^2+ (aq) -----> Mg^2+ (aq) + Fe(s)

[Fe2+]= 3.00 M and [Mg2+]= 0.110 M .

Q = [Mg^2+]/[Fe^2+]

= 0.11/3 = 0.0366

part-B

T = 89+273 = 362K

part-C

n = 2 electrons are involve in the reaction

part-D

Mg(s) ----------------> Mg^2+ (aq) + 2e^- E0 = 2.38v

Fe^2+ (aq) + 2e^- ------> Fe(s) E0 = -0.41v

-----------------------------------------------------------------------------------

Mg(s) + Fe^2+ (aq) -----> Mg^2+ (aq) + Fe(s) E0cell = 1.97v

the standard cell potential for 1.97v

queen_honey_blossom answered 2 years ago

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