Question

In: Chemistry

Consider the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 69 ∘C , where [Fe2+]= 3.10 M and [Mg2+]= 0.110 M...

Consider the reaction

Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s)

at 69 ∘C , where [Fe2+]= 3.10 M and [Mg2+]= 0.110 M .

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

b) What is the value for the temperature, T, in kelvins?

c) What is the value for n?

d) Calculate the standard cell potential for Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s)

Expert Solution

a) reaction quotient( Q ) = concentrations of products / concentrations of reactants

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

= 0.110 / 3.10

= 0.0355

b) temperature in Kelvin = T + 273

= 69 + 273

= 342 K

c) value of n = 2

number of moles of electrons are trasfered = 2 from Mg to Fe+2

d) the standard cell potential = 1.94 V

explanation:

anode reaction: oxidation takes place

Mg(s) -------------------------> Mg+2 (aq) + 2e- , E⁰_Mg+2/Mg = - 2.38 V

cathode reaction : reduction takes palce

Fe+2(aq) + 2e- -----------------------------> Fe(s) , E⁰_Fe+2/Fe = -0.44V

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

net reaction: Mg(s) +Fe+2(aq) -------------------------> Mg+2 (aq) + Fe(s)

E⁰_cell= E⁰_cathode- E⁰_anode

E⁰_cell= E⁰_Fe+2/Fe - E⁰_Mg+2/Mg

= -0.44 - (-2.38)

= 1.94 V

queen_honey_blossom answered 1 hour ago

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