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

sodium (na), for example, has one electron in its valence shell. this is an unstable state because that valence shell needs eight electrons to be full. in order to fill its valence shell, find a way to add seven electrons to its valence shell

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