Electrochemistry (solubility)

[koomanchoo]

hey i seem to be having a lot of trouble qith this question may i get some help?
The standard potential of the cell below is + 0.9509 V at 25 oC. Calculate the solubility, normally called s, of AgI (Note: Enter the natural log of s (ln s) as your answer)

Ag|AgI(s)|AgI(aq)|Ag

i'm using nernst equation and found the standard potential for Ag+,Ag to be 0.7991V

i am using Eo as 0.7991 and E as 0.9509 and rearanging the equation to find the solubility as LnS.. the correct answer is suppose to be -18.51.. where i am going wrong? please help.
thanks

n.b standard reduction potential of AgI(s) is -.15V

 

[Borek]

IIRC -18.51 is not a natural log of the AgI solubility product - it's decimal log.

 

[Blueshift5]

Hello,

Thank you for reaching out for help with your question. Electrochemistry and solubility can be a tricky topic, so let's break down the problem step by step.

First, let's review the Nernst equation:

E = E° - (RT/nF)lnQ

where:
E = cell potential at non-standard conditions
E° = standard cell potential
R = gas constant (8.314 J/mol*K)
T = temperature (in Kelvin)
n = number of electrons transferred in the balanced redox reaction
F = Faraday's constant (96,485 C/mol)
Q = reaction quotient

Now, let's apply this equation to the given cell:

Ag|AgI(s)|AgI(aq)|Ag

We know that the standard potential for this cell is +0.9509V at 25°C. However, in order to use the Nernst equation, we need to calculate the reaction quotient (Q). To do this, we need to know the concentrations of Ag+ and AgI(aq) in the cell.

Since we are given the standard reduction potential for AgI(s) as -0.15V, we can use this to calculate the standard potential for the reaction Ag+ + I- → AgI(s). We do this by adding the standard potentials for the half-reactions:

E° = E°(AgI(s)) + E°(Ag+ → Ag)

E° = -0.15V + 0.7991V = 0.6491V

Now, we can use this standard potential to calculate the reaction quotient (Q) at non-standard conditions by rearranging the Nernst equation:

lnQ = (E - E°)nF/RT

Plugging in our known values, we get:

lnQ = (0.9509V - 0.6491V)(1)(96485 C/mol)/(8.314 J/mol*K)(298 K)

lnQ = 0.0711

Now, we can solve for the solubility (s) by taking the natural log of Q:

lns = lnQ = 0.0711

Therefore, the solubility of AgI is:

s = e^(lns) = e^(0.0711) = 1.073 M

However, the question asks for the natural log of s (ln s), so the correct answer