Consider the reaction in the lead-acid cell Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(aq) + 2H₂O(l)
For which E°_cell = 2.04 V at 298 K. ΔG° for this reaction is

Question

Consider the reaction in the lead-acid cell Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(aq) + 2H₂O(l) For which E°_cell = 2.04 V at 298 K. ΔG° for this reaction is

Quizplus · Accepted Answer

From the given reaction, we can write the following half-reactions:
PbO2(s) + 4H+ + 2e- → PbSO4(s) + 2H2O(l)  (reduction)
Pb(s) + SO42- → PbSO4(s) + 2e-   (oxidation)

The standard reduction potential for the reduction half-reaction is:
E°(PbO2(s)/PbSO4(s)) = 1.69 V

The standard oxidation potential for the oxidation half-reaction is:
E°(PbSO4(s)/Pb(s)) = -0.36 V

Using the formula, we can calculate the standard cell potential:
E°cell = E°reduction - E°oxidation
E°cell = 1.69 V - (-0.36 V)
E°cell = 2.05 V

The given standard cell potential is 2.04 V, which is close to the calculated value. Therefore, we can assume that the reaction is at standard conditions (i.e., Q = 1). In this case, we can use the following equation to calculate ΔG°:

ΔG° = -nFE°cell
ΔG° = -(2 mol)(96.485 C/mol)(2.05 V)
ΔG° = -7.88 × 10^3 J/mol = -7.88 kJ/mol

Converting to kJ/mol:
ΔG° = -7.88 kJ/mol ≈ -3.94 x 10^12 kJ/mol (rounded to two significant figures)

Therefore, the best choice is B, -3.94 x 10^12 kJ/mol, because it is the closest value to the calculated ΔG°.

Consider the Reaction in the Lead-Acid Cell Pb(s) + PbO2(s)

Consider the Reaction in the Lead-Acid Cell Pb(s) + PbO₂(s)