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

Question

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

Quizplus · Accepted Answer

The standard Gibbs free energy change (ΔG°) can be calculated using the formula ΔG° = -nFE°cell, where n is the number of moles of electrons transferred (n=2 for this reaction), F is Faraday's constant (approximately 96485 C/mol), and E°cell is the standard cell potential (2.04 V). Substituting these values gives ΔG° = -2 * 96485 * 2.04 = -3.94 × 10^5 J, which is equivalent to -3.94 × 10^2 kJ.

Consider the Reaction in the Lead-Acid Cell Pb(s)+ PbO2(s)+ 2H2SO4(aq) →\to→

Consider the Reaction in the Lead-Acid Cell
Pb(s)+ PbO₂(s)+ 2H₂SO₄(aq) $\to$