Question

Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution: Pb(s) + 2Ag+ (aq) LaTeX: \longrightarrow⟶ Pb2+(aq) + 2Ag(s) As a consequence, silver is a valuable byproduct in the industrial extraction of lead from its ores. Calculate K and LaTeX: \DeltaΔG° at 298K for this reaction. E°cell = .93 K: Enter as e notation to 1 decimal place (eg 1.2e3) LaTeX: \DeltaΔG°: Enter in kJ/mol to 0 decimal places. Use 96.5 kJ/Vmol e- for F (Faraday's constant). Do not use e notation.

Answer 1

Answer : The value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

Explanation :

The balanced cell reaction will be,

$Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)$

The half-cell reactions are:

Oxidation reaction (anode) : $Pb(s)\rightarrow Pb^{2+}(aq)+2e^-$

Reduction reaction (cathode) : $2Ag^+(aq)+2e^-\rightarrow 2Ag(g)$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

where,

$\Delta G^o$ = standard Gibbs free energy

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential of the cell = 0.93 V

Now put all the given values in the above formula, we get:

$\Delta G^o=-2\times 96500\times 0.93$

$\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol$

Now we have to calculate the value of 'K'.

$\Delta G^o=-RT\ln K$

where,

$\Delta G_^o$ =  standard Gibbs free energy  = -180 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

$-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K$

$K=3.6\times 10^{31}$

Therefore, the value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

Answer 2

The change in free energy is -1.79 * 10^4 J while the rate constant is 2.6 * 10^31.

What is ΔG°?

The symbol ΔG° stands for the standard free energy change of the reaction. The equation of the reaction is; Pb(s) + 2Ag+ (aq) ⟶ Pb2+(aq) + 2Ag(s) . We can see that two moles of electrons were transferred in the process.

E°cell = 0.0592/n log K

When  E°cell = 0.93 V

0.93 = 0.0592/2 log K

31.42 = log K

K = Antilog(31.42)

K = 2.6 * 10^31

Now;

ΔG° = -nFE°cell

ΔG° = -(2 * 96500 * 0.93)

ΔG° = -1.79 * 10^4 J

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