Use the tabulated half-cell potentials to calculate ΔG° for the following balanced redox reaction. Pb2+(aq) + Cu(s) → Pb(s) + Cu2+(aq)

The standard Gibbs free energy for the reaction

Pb²⁺(aq) + Cu(s) → Pb(s) + Cu²⁺(aq) is 89.3 kJ/mol.

Let's consider the following redox reaction.

Pb²⁺(aq) + Cu(s) → Pb(s) + Cu²⁺(aq)

We can identify both half-reactions.

Reduction (Cathode): Pb²⁺(aq) + 2 e⁻ → Pb(s)    E°red = -0.126 V

Oxidation (Anode): Cu(s) → Cu²⁺(aq) + 2 e⁻       E°red = +0.337 V

We can calculate the standard potential of the cell (E°cell) using the following expression.

E°cell = E°red,cathode - E°red,anode = -0.126 V - 0.337 V = -0.463 V

The standard Gibbs free energy (ΔG°) is a way to measure the spontaneity of a reaction. We can calculate it using the following expression.

ΔG° = − n × F × E°cell =

ΔG° = − 2 mol × (96,485 J/V.mol) × (-0.463 V) × (1 kJ/1000 J) = 89.3 kJ/mol

where,

Since ΔG° > 0, the reaction is not spontaneous.

The standard Gibbs free energy for the reaction

Pb²⁺(aq) + Cu(s) → Pb(s) + Cu²⁺(aq) is 89.3 kJ/mol.

Learn more about Gibbs free energy: link

90.71 KJ

Explanation:

From the reaction equation, lead is the cathode while copper was the anode.

Hence;

E°anode = +0.34V

E°cathode = -0.13 V

E°cell =E°cathode - E°anode

E°cell = -0.13 V - 0.34V

E°cell = -0.47 V

But;

ΔG° = -nFE°cell

n= number of electrons transferred = 2

F = Faraday's constant = 96500 C

E°cell = -0.47 V

ΔG° = -(2 * 96500 * (-0.47))

ΔG° = 90,710 J or 90.71 KJ

first one is the correct answer