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 : The value of and K is, -180 kJ/mol and

Explanation :

The balanced cell reaction will be,

The half-cell reactions are:

Oxidation reaction (anode) :

Reduction reaction (cathode) :

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

where,

= standard Gibbs free energy

F = Faraday constant = 96500 C

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

= standard electrode potential of the cell = 0.93 V

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

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

where,

=  standard Gibbs free energy  = -180 kJ/mol

R = gas constant =

T = temperature = 298 K

K = equilibrium constant = ?

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

Therefore, the value of and K is, -180 kJ/mol and

particle   p is an electron and particle q is a neutron

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