Question

please comment if need more info. no need to show work just correct answers.

Consider the following reaction at 298K. 2 Fe3+ (aq) + 2 1 (aq) — 2 Fe2+ (aq) +12 (5) Which of the following statements are correct? Choose all that apply Eºcell <0 The reaction is reactant-favored. On= 6 mol electrons delta Gº<0 DK > 1

Answer 1

2I^-(aq) ----------> I2(s) + 2e^-     E0 = -0.535v

2Fe^3+ (aq) + 2e^- -----> 2Fe^2+ (aq)   E0 = 0.771v

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

2Fe^3+ (aq) + 2I^- (aq) -----> I2(s) + 2Fe^2+ (aq)    E0cell = 0.236v

E0cell>0

n = 2 moles electrons

$\Delta$G⁰    = -nE0cell*F

              = -2*0.236*96500

               = -45548J

$\Delta$G⁰ <0   the reaction favored products

$\Delta$G⁰    = -RTlnK

-45548   = -8.314*298lnK

lnK = -45548/(-8.314*298)

lnK   = 18.38

K    = 9.6*10^7

K>1

correct statements

E0cell>0

$\Delta$G⁰ <0

The reaction products-favored

part-B

Zn(s) ---------> Zn^2+ (aq) + 2e^-     E0 = 0.76v

I2(s) + 2e^- ------> 2I^- (aq)               E0 = 0.535v

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

Zn(s) + I2(s) -----> Zn^2+ (aq) + 2I^- (aq)    E0cell = 1.295v

E0cell>0

n = 2 moles electrons

$\Delta$G⁰    = -nE0cell*F

              = -2*1.295*96500

               = -249935J

$\Delta$G⁰ <0   the reaction favored products

$\Delta$G⁰    = -RTlnK

-249935 = -8.314*298lnK

lnK = -249935/(-8.314*298)

lnK   = 100.9

K    = 6.6*10^43

K>1

correct statements

n = 2 moles electrons

E0cell>0

The reaction products-favored

part-3

anode reaction

Sn(s) ---------------> Sn^2+ (aq) + 2e^-             E0   = 0.140v

cathode reaction

2Ag^+ (aq) + 2e^- ------> 2Ag(s)                        E0   = 0.799v

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

spontaneous cell reaction

Sn(s) + 2Ag^+ (aq) ------> Sn^2+ (aq) + 2Ag(s)     E0cell = 0.939v

cell voltage

E0cell   - 0.939v