if you answer all of these questions, i will give the most points chegg will let
ID: 798411 • Letter: I
Question
if you answer all of these questions, i will give the most points chegg will let me,
Calculate the standard free-energy change at 25 degree C for the following reaction: Mg(s)+Fe2+)aq)rightarrowMg2+(aq)+Fe(s) Calculate the standard cell potential at 25 degree C for the reaction X(s)+2Y+(aq) rightarrow X2+(aq)+2Y(s) Where Delta H degree = -795kJ and Delta S degree = -213J/K. Calculate Delta G degreer xn and E degree cell for a redox reaction with n = 2 that has an equilibrium constant of K = 5.8times 10-2. Which of the following redox reactions do you expect to occur spontaneously in the forward direction?Explanation / Answer
1) Fe2+(aq) + Mg(s) ---> Fe(s) + Mg+2 (aq)
E0 cell = E0cathode - E0 anode
= Eo Fe+2/Fe - Eo Mg+2/Mg
= -0.44 - ( -2.37 )
= 1.93
Eo cell is 1.93 V
dG0 rxn = -nFEo
here n = 2 as 2 moles of electrons are transferred .
so dGo rxn = - 2 x 96500 x 1.93
= -372490
dGo rxn = - 372.49 kJ/mol
2) Given X + 2Y+ ----> X+2 + 2Y
dHo = -795kJ ,dS0 = -213 J/K
temp =25 C =298 K
we know that dGo = dHo - TdSo
dGo = -795 x 103 - (298 x -213)
dGo = -731526
dGo = -731.526 kJ
so dGo is -731.526 kJ
we know that dGo = -nFEo
here n =2 as 2 moles of electrons are transferred.
-731.526 x 103 = - 2 x 96500 x Eo
Eo =3.79 V
So E0 is 3.79 V
Item 3)
Given n = 2 ,K =5.8 x 10-2 and temp T=298
we know that dGo = -RTlnK
dGo = -8.314 x 298 x ln 5.8 x 10-2
dGo = 7054.42
dGo is 7.054 kJ
we know that dGo = -nFEo
7054.42 = - 2 x 96500 x Eo
Eo = 0.03655
Eo is 0.03655
Item 2)
Given 2Al(s) + 3Ag+(aq) ----> 2Al3+(aq) + 3 Ag (s)
E0 cell = E0cathode - E0 anode
= Eo Ag+/Ag - Eo Al+3/Al
= 0.80 - ( -1.66 )
=
Eo cell is 2.46 V
as the E0 cell is positive , the reaction is spontaneous in the forward reaction .
Mn+2 + Pb ----> Mn + Pb+2
E0 cell = E0cathode - E0 anode
= Eo Mn+2 /Mn - Eo Pb+2/Pb
= -1.185 - ( -0.13 )
= -1.055
Eo cell is -1.055 V
as the E0 cell is negative , the reaction is non - spontaneous in the forward reaction .
Given Ni + Zn+2 -----> Ni+2 + Zn
E0 cell = E0cathode - E0 anode
= Eo Zn+2/Zn - Eo Ni+2/Ni
= -0.76 - ( -0.25)
= -0.51
Eo cell is -0.51 V
as the E0 cell is negative , the reaction is non spontaneous in the forward reaction .
Given Fe 2+(aq) + Zn(s) ----------> Fe(s) + Zn2+(aq)
E0 cell = E0cathode - E0 anode
= Eo Fe+2/Fe - Eo Zn+2/Zn
= -0.44 - ( -0.76)
= 0.32
Eo cell is 0.32
as the E0 cell is positive , the reaction is spontaneous in the forward reaction .
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