Question

Question:Consider the following galvanic cell:

Pb Ag

1.8 MPb²⁺ MAg⁺

? MSO₄

2-

0.83V

Ag₂SO₄ (s)

Calculate the K_sp value for Ag₂SO₄(s). Note that to obtain

silver ions in the right compartment (the cathode

compartment), excess solid Ag₂SO₄ was added and some

of the salt dissolved.

Short answer

Value of $K_{sp}=9.6\times {10}^{-6}$.

Step-by-step solution

analyzing the given data

The two half-reaction are,

$$\begin{gathered} P{b}^{2+}+2e^{-}\to Pb\left(E°=-0.13V\right) \\ A{g}^{+}+e^{-}\to Ag\left(E°=-0.80V\right) \end{gathered}$$

Flipping the first equation, we get,

$Pb\to P{b}^{2+}+2e^{-}\left(E°=0.13V\right)$

We get the final equation

$Pb+2Ag\to P{b}^{2+}+2Ag\left(E°=0.93V\right)$

calculating the value of KSP

$$\begin{gathered} E={E^0}_{CELL}--\frac{0.0591}{\text{n}}\log Q \\ {E^0}_{CELL}--\frac{0.0591}{\text{n}}\log \frac{\left[P{b}^{2+}\right]}{{\left[A{g}^{+}\right]}^2} \\ 0.83V=0.93V--\frac{{\displaystyle 0.0591}}{{\displaystyle 2}}\log \frac{{\displaystyle 1.8}}{{\displaystyle {\left[A{g}^{+}\right]}^2}} \\ \left[A{g}^{+}\right]=0.027M \\ \left[S{O}_4^{2-}\right]=0.0135M \\ {K}_{sp}={\left(0.027\right)}^2\left(0.0135\right) \\ {K}_{sp}=9.6\times {10}^{-6} \end{gathered}$$