Question

 1)  What is the solubility of PbI₂ in g/l, if the solubility product is 7.9x10^-9

2) Cu(s)/ Cu+2 (0.02M) //Zn2+ (0.02M)/Zn(s); the standard reduction potential for Cu and Zn is 0.34 and -0.76 respectively. Calculate the standard cell potential, change in free energy and cell potential

Answer 1

1.) Let the solubility of $PbI_2$ in 0.01M NaI solution be s M.

Then in this case,

$[Pb^{2+}] = sM$

And,

$[I^{-}] = (2s+0.01)M$

So, $K_{sp} = [Pb^{2+}][I^{-}]^2$

$1.4 \times 10^{-8} = s \times (4s^2+4s(0.01)+(0.01)^2)$

Neglecting $s^2$ and $s^3$ terms, we get:

We get,

$s = 1.4 \times 10^{-4}M$

2.) In this part we can use the Nernst Equation,

$E_{cell} = E_0 - \dfrac{0.0591}{n}log\dfrac{[Reduced]}{[Oxidised]}$