Using the voltage from the following Galvanic cell, calculate $K_{s p}$ $K_(sp)$ for $A g_{2} S O_{4}$ $Ag_2SO_4$ (s)?

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Using the voltage from the following Galvanic cell, calculate $K_{s p}$ $K_(sp)$ for $A g_{2} S O_{4}$ $Ag_2SO_4$ (s)?

$P b (s) ∣ ∣ P b^{2 +} (1.8 M) ∣ ∣ ∣ ∣ A g^{+} (satd A g_{2} S O_{4}) ∣ ∣ A g (s)$ $Pb(s)|Pb^(2+)(1.8M)||Ag^(+)("satd "Ag_2SO_4)|Ag(s)$

$E_{c e l l} = 0.83 V$ $E_(cell)=0.83" V"$

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Answer 1 · 2018-08-02T06:33:03

$K_{s p} = 1.03 \times 10^{- 5}$ $K_(sp) = 1.03 xx 10^(-5)$

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

It looks like the point of this is to:

Find $E_{c e l l}^{\circ}$ $E_(cell)^@$ using reference values you should be given.
Use $E_{c e l l}$ $E_(cell)$ to find $Q_{c}$ $Q_c$ for these concentrations, assuming that the silver sulfate precipitate is in equilibrium with its ions already.
Find $K_{s p}$ $K_(sp)$ , knowing that saturated solutions satisfy the condition given in $(2)$ $(2)$ .

You should be given:

$E_{r e d}^{\circ} ({Pb}^{2 +} \to Pb) = - 0.13 V$ $E_(red)^@("Pb"^(2+)->"Pb") = -"0.13 V"$
$E_{r e d}^{\circ} ({Ag}^{+} \to Ag) = 0.80 V$ $E_(red)^@("Ag"^(+)->"Ag") = "0.80 V"$

Thus,

$E_{c e l l}^{\circ} = 0.80 V - (- 0.13 V) = 0.93 V$ $E_(cell)^@ = "0.80 V" - (-"0.13 V") = "0.93 V"$ .

The next thing is that we can set up $Q_{c}$ $Q_c$ :

$Pb (s) + 2 {Ag}^{+} (a q) \to {Pb}^{2 +} (a q) + 2 Ag (s)$ $"Pb"(s) + 2"Ag"^(+)(aq) -> "Pb"^(2+)(aq) + 2"Ag"(s)$

$Q_{c} = \frac{[{Pb}^{2 +}]}{{[{Ag}^{+}]}^{2}}$ $Q_c = (["Pb"^(2+)])/(["Ag"^(+)]^2)$

$= \frac{1.8 M}{{[{Ag}^{+}]}^{2}}$ $= ("1.8 M")/(["Ag"^(+)]^2)$

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

$E_{c e l l} = E_{c e l l}^{\circ} - \frac{0.0592 V}{n} {log Q}_{c}$ $E_(cell) = E_(cell)^@ - ("0.0592 V")/n log Q_c$

Now we solve algebraically for an expression for $Q_{c}$ $Q_c$ .

$\frac{0.0592 V}{n} {log Q}_{c} = E_{c e l l}^{\circ} - E_{c e l l}$ $"0.0592 V"/nlog Q_c = E_(cell)^@ - E_(cell)$

For now,

$\Rightarrow {log Q}_{c} = \frac{n}{0.0592 V} (E_{c e l l}^{\circ} - E_{c e l l})$ $=> log Q_c = n/"0.0592 V"(E_(cell)^@ - E_(cell))$

$= \frac{\frac{{2 mol e}^{-}}{1 mol atoms}}{0.0592 V} \cdot (0.93 V - 0.83 V)$ $" "" "" "" "= ("2 mol e"^(-)/"1 mol atoms")/("0.0592 V") cdot ("0.93 V" - "0.83 V")$

$= 3.38$ $" "" "" "" "= 3.38$

Therefore,

$Q_{c} = 10^{3.38} = 2390$ $Q_c = 10^(3.38) = 2390$

Lastly, we can solve for $[{Ag}^{+}]$ $["Ag"^(+)]$ and therefore $K_{s p}$ $K_(sp)$ .

${2390 M}^{- 1} = \frac{1.8 M}{{[{Ag}^{+}]}^{2}}$ $"2390 M"^(-1) = ("1.8 M")/(["Ag"^(+)]^2)$

$[{Ag}^{+}] = \sqrt{\frac{1.8 M}{{2390 M}^{- 1}}} = 0.0274 M$ $["Ag"^(+)] = sqrt("1.8 M"/("2390 M"^(-1))) = "0.0274 M"$

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

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

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

$ul(["Ag"^(2+)] = 2["SO"_4^(2-)])$

From this, we find the $K_(sp)$ :

$color(blue)(K_(sp)) = ["Ag"^(+)]^2["SO"_4^(2-)]$

$= ("0.0274 M")^2("0.0274 M"/2)$

$= color(blue)(1.03 xx 10^(-5))$

This is not that far off from the actual one, yay! The actual value should have been $1.20 xx 10^(-5)$ .

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Using the voltage from the following Galvanic cell, calculate $K_{s p}$ $K_(sp)$ for $A g_{2} S O_{4}$ $Ag_2SO_4$ (s)?

$P b (s) ∣ ∣ P b^{2 +} (1.8 M) ∣ ∣ ∣ ∣ A g^{+} (satd A g_{2} S O_{4}) ∣ ∣ A g (s)$ $Pb(s)|Pb^(2+)(1.8M)||Ag^(+)("satd "Ag_2SO_4)|Ag(s)$

$E_{c e l l} = 0.83 V$ $E_(cell)=0.83" V"$

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Using the voltage from the following Galvanic cell, calculate K_(sp)KspK_(sp) for Ag_2SO_4Ag2SO4Ag_2SO_4 (s)?

Pb(s)|Pb^(2+)(1.8M)||Ag^(+)("satd "Ag_2SO_4)|Ag(s)Pb(s)∣∣Pb2+(1.8M)∣∣∣∣Ag+(satd Ag2SO4)∣∣Ag(s)Pb(s)|Pb^(2+)(1.8M)||Ag^(+)("satd "Ag_2SO_4)|Ag(s) E_(cell)=0.83" V"Ecell=0.83 VE_(cell)=0.83" V"

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Using the voltage from the following Galvanic cell, calculate $K_{s p}$ $K_(sp)$ for $A g_{2} S O_{4}$ $Ag_2SO_4$ (s)?

$P b (s) ∣ ∣ P b^{2 +} (1.8 M) ∣ ∣ ∣ ∣ A g^{+} (satd A g_{2} S O_{4}) ∣ ∣ A g (s)$ $Pb(s)|Pb^(2+)(1.8M)||Ag^(+)("satd "Ag_2SO_4)|Ag(s)$

$E_{c e l l} = 0.83 V$ $E_(cell)=0.83" V"$