Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant for AgCl at 298 K is 1.77 × 10 − 10 and the standard reduction potential of the half‑reaction Ag + ( aq ) + e − − ⇀ ↽ − Ag ( s ) is + 0.799 V .

The value of for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

Its solubility product will be as follows.

Cell reaction for this equation is as follows.

Reduction half-reaction: ,  

Oxidation half-reaction: ,   = ?

Cell reaction:

So, for this cell reaction the number of moles of electrons transferred are n = 1.

    Solubility product,

                                               =

Therefore, according to the Nernst equation

At equilibrium, = 0.00 V

Putting the given values into the above formula as follows.

                  =

                  = 0.577 V

Hence, we will calculate the standard cell potential as follows.

        = 0.222 V

Thus, we can conclude that value of for the half-cell reaction is 0.222 V.

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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