The overall reaction and equilibrium constant value for a hydrogen-oxygen fuel cell at 298 k is given below. 2 h2(g) + o2(g) → 2 h2o(l) k = 1.34 ✕ 1083 (a) calculate ℰ° and δg° at 298 k for the fuel-cell reaction. ℰ° v δg° kj (b) predict the signs of δh° and δs° for the fuel-cell reaction. δh°: positive negative δs°: positive negative (c) as temperature increases, does the maximum amount of work obtained from the fuel-cell reaction increase, decrease, or remain the same? explain. since δs is , as t increases, δg becomes more . therefore, the maximum work obtained will as t increases.

(a) ΔG° = -474 kJ/mol; E° = 1.23 V

(b) ΔH° negative; ΔS° negative

(c) Since ΔS is negative, as T increases, ΔG becomes more positive. Therefore, the maximum work obtained will decrease as T increases.

Explanation:

Let's consider the following reaction.

2 H₂(g) + O₂(g) → 2 H₂O(l)

with an equilibrium constant K = 1.34 × 10⁸³

(a) Calculate E° and ΔG° at 298 K for the fuel-cell reaction.

We can calculate the standard Gibbs free energy (ΔG°) using the following expression:

ΔG° = - R × T × lnK

ΔG° = - 8.314 × 10⁻³ kJ . mol⁻¹.K⁻¹ × 298 K × ln 1.34 × 10⁸³ = -474 kJ/mol

To calculate the standard cell potential (E°) we need to write oxidation and reduction half-reactions.

Oxidation: 2 H₂ ⇒ 4 H⁺ + 4 e⁻

Reduction: O₂ + 4 e⁻ ⇒ 2 O²⁻

The moles of electrons (n) involved are 4.

We can calculate E° using the following expression:

(b) Predict the signs of ΔH° and ΔS° for the fuel-cell reaction. ΔH°: positive negative ΔS°: positive negative

The standard Gibbs free energy is related to the standard enthalpy (ΔH°) and standard entropy (ΔS°) through the following expression:

ΔG° = ΔH° - T.ΔS°

Usually, the major contribution to ΔG° is ΔH°. So, if ΔG° is negative (exergonic), ΔH° is expected to be negative (exothermic).

The entropy is related to the number of moles of gases. There are 3 gaseous moles in the reactants and 0 in the products, so the final state is predicted to be more ordered than the initial state, resulting in a negative ΔS°.

(c) As temperature increases, does the maximum amount of work obtained from the fuel-cell reaction increase, decrease, or remain the same?

The maximum amount of work obtained depends on the standard Gibbs free energy.

wmax = ΔG° = ΔH° - T.ΔS°

Since ΔS is negative, as T increases, ΔG becomes more positive. Therefore, the maximum work obtained will decrease as T increases.

a

explanation:

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6