# Using the voltage from the following Galvanic cell, calculate K_(sp) for Ag_2SO_4 (s)?

## $P b \left(s\right) | P {b}^{2 +} \left(1.8 M\right) | | A {g}^{+} \left(\text{satd } A {g}_{2} S {O}_{4}\right) | A g \left(s\right)$ ${E}_{c e l l} = 0.83 \text{ V}$

Aug 2, 2018

${K}_{s p} = 1.03 \times {10}^{- 5}$

compared to the literature value of around $1.20 \times {10}^{- 5}$.

It looks like the point of this is to:

1. Find ${E}_{c e l l}^{\circ}$ using reference values you should be given.
2. Use ${E}_{c e l l}$ to find ${Q}_{c}$ for these concentrations, assuming that the silver sulfate precipitate is in equilibrium with its ions already.
3. Find ${K}_{s p}$, knowing that saturated solutions satisfy the condition given in $\left(2\right)$.

You should be given:

E_(red)^@("Pb"^(2+)->"Pb") = -"0.13 V"
E_(red)^@("Ag"^(+)->"Ag") = "0.80 V"

Thus,

${E}_{c e l l}^{\circ} = \text{0.80 V" - (-"0.13 V") = "0.93 V}$.

The next thing is that we can set up ${Q}_{c}$:

$\text{Pb"(s) + 2"Ag"^(+)(aq) -> "Pb"^(2+)(aq) + 2"Ag} \left(s\right)$

${Q}_{c} = \left({\left[{\text{Pb"^(2+)])/(["Ag}}^{+}\right]}^{2}\right)$

= ("1.8 M")/(["Ag"^(+)]^2)

And now let's put that off to the side, and solve for it last. At ${25}^{\circ} \text{C}$, we have the Nernst equation again:

${E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{\text{0.0592 V}}{n} \log {Q}_{c}$

Now we solve algebraically for an expression for ${Q}_{c}$.

$\frac{\text{0.0592 V}}{n} \log {Q}_{c} = {E}_{c e l l}^{\circ} - {E}_{c e l l}$

For now,

$\implies \log {Q}_{c} = \frac{n}{\text{0.0592 V}} \left({E}_{c e l l}^{\circ} - {E}_{c e l l}\right)$

" "" "" "" "= ("2 mol e"^(-)/"1 mol atoms")/("0.0592 V") cdot ("0.93 V" - "0.83 V")

$\text{ "" "" "" } = 3.38$

Therefore,

${Q}_{c} = {10}^{3.38} = 2390$

Lastly, we can solve for $\left[{\text{Ag}}^{+}\right]$ and therefore ${K}_{s p}$.

"2390 M"^(-1) = ("1.8 M")/(["Ag"^(+)]^2)

["Ag"^(+)] = sqrt("1.8 M"/("2390 M"^(-1))) = "0.0274 M"

Two silver cations and one sulfate anion are in the equilibrium:

${\text{Ag"_2"SO"_4(s) rightleftharpoons 2"Ag"^(+)(aq) + "SO}}_{4}^{2 -} \left(a q\right)$

Due to the coefficients above, it must be recognized that:

$\underline{\left[{\text{Ag"^(2+)] = 2["SO}}_{4}^{2 -}\right]}$

From this, we find the ${K}_{s p}$:

$\textcolor{b l u e}{{K}_{s p}} = \left[{\text{Ag"^(+)]^2["SO}}_{4}^{2 -}\right]$

$= \left(\frac{\text{0.0274 M")^2("0.0274 M}}{2}\right)$

$= \textcolor{b l u e}{1.03 \times {10}^{- 5}}$

This is not that far off from the actual one, yay! The actual value should have been $1.20 \times {10}^{- 5}$.