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Using standard reduction potentials, calculate the standard emf, free energy, and equilibrium constant at 298K for...

Using standard reduction potentials, calculate the standard emf, free energy, and equilibrium constant at 298K for each of the following reactions:

a. Cl2(g) + 2 I-(aq) → 2 Cl-(aq) + I2(s)

b. 2 NO3-(aq) + 8 H+(aq) 3 Cu(s) → 2 NO(g) + 4 H2O(l) + 3 Cu2+(aq)

c. Fe(s) + 2 Fe3+(aq) → 3 Fe2+(aq)

Ive posted this twice already and the answers have been really inconsistant

Solutions

Expert Solution

a. Cl2(g) + 2 I-(aq) → 2 Cl-(aq) + I2(s)

2I^- (aq) -------------> I2(s)+2e^- E0 = -0.54V

Cl2(g) + 2e^- --------> 2Cl^- (aq) E0 = 1.36V

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

Cl2(g) + 2 I-(aq) → 2 Cl-(aq) + I2(s) E0 cell = 0.82V

G⁰ = -nE0cell*F

= -2*0.82*96500 = -158260J

G⁰ = -RTlnK

-159260 = -8.314*298*2.303logKeq

logKeq = -159260/-5705.8483

logKeq = 28

Keq = 10²⁸

b. 3Cu(s) -------------------------> 3Cu⁺² + 6e^- E0 = -0.34V

2NO3^- +8H⁺ + 6e^- --------> 2NO(g) + 4H2O(l) E0 = 0.96v

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

2 NO3-(aq) + 8 H+(aq)+ 3 Cu(s) → 2 NO(g) + 4 H2O(l) + 3 Cu2+(aq) E0cell = 0.62V

G⁰ = -nE0cell*F

= -6*0.62*96500 = -358980J

G⁰ = -RTlnK

-358980 =- 8.314*298*2.303logK

logK = -358980/-5705.8483

logK = 63

Keq = 10⁶³

queen_honey_blossom answered 7 months ago

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