Question

Electrochemistry - Equilibrium

1. The electrochemical cell described by the balanced chemical reaction has a standard cell potential of 0.25 V. Calculate the equilibrium constant (Kc) for the reaction at 25 oC. Round your answer to 3 significant figures.
St. Red. Pot. (V) Constants Faraday's Constant
MnO2/Mn2+ +1.21
0 °C = 273.15 K 96485 C/mol
PbO2/Pb2++1.46
R = 8.314 J/(K x mol)
Mn2+(aq) + PbO2(s) → MnO2(s) + Pb2+(aq)
2. The galvanic cell described by the balanced chemical equation has an Eo of 1.23 V. Calculate the value (kJ) for the standard free energy change of the cell. Round your answer to 3 significant figures.
St. Red. Pot. (V) Faraday's Constant
H2O/H2 -0.83
F = 96485 C
O2/OH- +0.40
2H2(g) + O2(g) → 2H2O(l)
3. The electrochemical cell described by the balanced chemical reaction has a standard emf of 1.78 V. Calculate the equilibrium constant (Kc) for the reaction at 298 K. Round your answer to 3 significant figures.
St. Red. Pot. (V) Constants Faraday's Constant
H+/H2 0
0 °C = 273.15 K 96485 C/mol
H2O2/H2O+1.78
R = 8.314 J/(K x mol)
H2(g) + H2O2(aq) → 2H2O(l)

Answer 1

Step-by-step explanation:

The relation between equilibrium constant and Ecell is given below .

E⁰cell = (RT / nF ) lnK , F is faraday constant T is 273 + 25 = 298 K

E⁰cell = 1.46 - 1.21 = .25 V

n = 2

Putting the values

.25 = (8.314 x 298 lnK) / (2 x 96485 )

lnK = 19.47

K = 2.85 x 10⁸

2 )

Change in free energy Δ G

Δ G ⁰ = nE⁰ F

n = 4

E⁰ = .4 + .83 = 1.23 V

Δ G ⁰= 4 x 1.23 x 96485

= 474706 J / mol

3 )

E⁰cell = (RT / nF ) lnK

n = 2

1.78 = 8.314 x 298 lnK / 2 x 96485

lnK = 138.638

K = 1.62 x 10⁶⁰