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 for the reaction can be calculated using the equation 
ΔG° = -nFE°cell 
where n is the number of electrons transferred in the reaction, F is the Faraday constant (96,485 C/mol), and E°cell is the standard cell potential. 
In this case, n = 2 (because two electrons are transferred in the reaction), and E°cell = 2.04 V. 
Therefore, 
ΔG° = -2 × 96,485 C/mol × 2.04 V 
ΔG° = -3.94 × 10^12 J/mol 
Converting to kJ/mol, we get -3.94 × 10^9 kJ/mol or -3.94 × 10^12 kJ. 
So, the answer is B, -3.94 × 10^12 kJ.

Consider the Reaction in the Lead-Acid Cell Pb(s) →\to→ 2PbSO4(aq) + 2H2O(l) for Which E°cell = 2

Consider the Reaction in the Lead-Acid Cell
Pb(s) $\to$ 2PbSO₄(aq) + 2H₂O(l)
for Which E°_cell = 2