An electrochemical cell based on the decomposition of H2O2 can be constructed based on the half-reactions in the table below.

Half-Reaction | Standard Reduction Potential, E°
H2O2+ 2e- => 2 OH- | 0.88 V
O2+ 2 H2O + 2e- => H2O + 2 OH- | -0.15 V

(d) Calculate the value of the standard cell potential, E °, for the cell.

(e) Indicate whether ΔG° for the decomposition reaction is greater than 0, less than 0, or equal to 0. Justify your answer.

(f) The decomposition of H2O2(aq) is slow at 298 K, but a suitable catalyst greatly increases the rate of the decomposition reaction.

(i) Draw a circle around each of the quantities below that has a different value for the catalyzed reaction than for the uncatalyzed reaction.
Keq DeltaG° DeltaH° Ea

(ii) For any quantity that you circled above, indicate whether its value is greater or less for the catalyzed reaction than for the uncatalyzed reaction. Explain why.

d) The value of the standard cell potential, E°, for the cell is 0.73 V.

e) ΔG° for the decomposition reaction is less than 0, because the reaction is spontaneous.

f) (i) Keq, DeltaG°, and Ea have different values for the catalyzed reaction than for the uncatalyzed reaction.

(ii) The values of Keq, DeltaG°, and Ea are all greater for the catalyzed reaction than for the uncatalyzed reaction. This is because the catalyst increases the rate of the reaction, making it more favorable and thus increasing the value of these quantities.

(d) To calculate the standard cell potential, E°, for the cell, we need to determine the overall reaction taking place in the cell.

From the given half-reactions:
H2O2 + 2e- => 2OH- (E° = 0.88 V)
O2 + 2H2O + 2e- => H2O + 2OH- (E° = -0.15 V)

By multiplying the second half-reaction by 2 to balance the number of electrons, we can cancel out the OH- ions and obtain the overall reaction:

2H2O2 => 2H2O + O2

The standard cell potential, E°, is the sum of the reduction potentials of the half-reactions involved in the cell. Therefore, E°cell = E°(H2O2+2e- => 2OH-) + E°(O2+2H2O+2e- => H2O+2OH-).

E°cell = 0.88 V + (-0.15 V)
E°cell = 0.73 V

Therefore, the value of the standard cell potential, E °, for the cell is 0.73 V.

(e) The standard Gibbs free energy change, ΔG°, for a reaction can be determined using the equation ΔG° = -nFE°cell, where n is the number of moles of electrons transferred, and F is Faraday’s constant (96485 C/mol).

From the balanced overall reaction:
2H2O2 => 2H2O + O2, we can see that 2 moles of electrons are transferred.

ΔG° = -nFE°cell = -(2 mol)(96485 C/mol)(0.73 V)
ΔG° = -141027.7 J/mol

Since the ΔG° value is negative, ΔG° for the decomposition reaction is less than 0. This means that the reaction is spontaneous under standard conditions.

(f) (i) In the catalyzed reaction, the only quantity that may have a different value than in the uncatalyzed reaction is the activation energy (Ea). Therefore, Ea is the only quantity to be circled.

(ii) The value of Ea for the catalyzed reaction is lower than the value for the uncatalyzed reaction. A catalyst provides an alternative reaction pathway with a lower activation energy, allowing the reaction to occur faster.

To calculate the standard cell potential (E°) for the electrochemical cell, we need to use the Nernst equation, which relates the cell potential to the standard reduction potentials of the two half-reactions and the concentrations of the reactants and products involved.

The Nernst equation is:

E = E° - (RT / nF) * ln(Q)

where:
E = cell potential (in volts)
E° = standard cell potential (in volts)
R = gas constant (8.314 J/(mol K))
T = temperature in kelvin (298 K for standard conditions)
n = number of electrons transferred in the balanced half-reaction
F = Faraday's constant (96,485 C/mol)
ln = natural logarithm
Q = reaction quotient

In this case, we have two half-reactions:

1) H2O2 + 2e- => 2OH- with E° = 0.88 V
2) O2 + 2H2O + 2e- => H2O + 2OH- with E° = -0.15 V

To calculate the standard cell potential (E°), we need to find the difference in the reduction potentials of the two half-reactions.

E° = E°(cathode) - E°(anode)

In this case, the reduction potential of the first half-reaction (0.88 V) is larger than the second half-reaction (-0.15 V), so the first half-reaction will be the cathode and the second half-reaction will be the anode.

E° = 0.88 V - (-0.15 V)
E° = 1.03 V

Therefore, the standard cell potential (E°) for the cell is 1.03 V.

To determine whether ΔG° (the standard Gibbs free energy change) for the decomposition reaction is greater than 0, less than 0, or equal to 0, we can use the relationship between ΔG° and E°.

ΔG° = -n * F * E°

In this equation, n represents the number of moles of electrons transferred in the reaction (in this case, 2), F is Faraday's constant, and E° is the standard cell potential.

Since ΔG° is directly proportional to E°, if E° is positive, ΔG° will be negative (less than 0). Therefore, ΔG° for the decomposition reaction is less than 0.

Now, moving on to the catalyzed reaction and its impact on various quantities:

(i) The quantity that has a different value for the catalyzed reaction than for the uncatalyzed reaction is Ea (activation energy). Draw a circle around Ea.

(ii) For the catalyzed reaction, the value of Ea is typically lower (i.e., the activation energy is lower) when compared to the uncatalyzed reaction. This is because a catalyst provides an alternative reaction pathway with a lower activation energy, making it easier for the reactants to overcome the energy barrier and proceed to the products. Therefore, the value of Ea is greater for the uncatalyzed reaction and lower for the catalyzed reaction.