Question

Consider the reaction

Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 69 ∘C , where [Fe2+]= 3.00 M and [Mg2+]= 0.210 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)

Part E

For the reaction

2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g).  E∘=0.483 V

what is the cell potential at 25 ∘C if the concentrations are [Co3+]= 0.609 M , [Co2+]= 0.863 M , [Cl−]= 0.169 M , and [Cl2]= 0.300 M ?

Answer 1

Part a

.the value of reaction quotient (Q) for the cell

Q = [products] /[reactants]

= [Mg2+]/ [Fe2+]

Q = 0.210M/3.00M

Q = 0.07

Part b

the temperature (T) in Kelvin

T = 69∘C + 273

T = 342 K

Part C

the value for n = 2 mol

Because 2 moles are transferred in this reaction

Part d

The standard cell potential

Mg2+ (aq) + 2 e- ---> Mg (s) E° = -2.36 V
Fe2+ (aq) + 2 e- ---> Fe (s) E° = -0.44 V

Oxidation reaction
Mg (s) --->Mg2+ (aq) + 2 e- E° ox = 2.36 V

Reduction reaction
Fe2+ (aq) + 2 e- --->Fe (s) E° red = -0.44 V

Overall reaction
Mg (s) + Fe2+ (aq) ---> Mg2+ (aq) + Fe (s)

E° = E° ox + E° red = 2.36 - 0.44 = 1.92 V

The standard cell potential is 1.92V