Question

Given the following standard electrode potentials, the $K_{sp}$ for $PbBr_2$ is

$PbBr_2(s) + 2e^- \longrightarrow Pb(s) + 2Br^-(aq); E^\circ = -0.248 V$ $Pb^{2+}(aq) + 2e^- \longrightarrow Pb(s); \quad E^\circ = -0.126 V$

• A. 7.4 x $10^{-5}$
• B. 4.9 x $10^{-14}$
• C. 5.2 x $10^{-6}$
• D. 2.3 x $10^{-13}$

Answer 1

Answer: (A)

7.4 x $10^{-5}$

Answer 2

To determine the solubility product constant ($K_{sp}$) for $PbBr_2$, we need to relate the given standard electrode potentials to the dissolution reaction of $PbBr_2$.

The dissolution reaction for $PbBr_2$ is: $PbBr_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Br^-(aq)$

We are given the following standard electrode potentials:

1. $PbBr_2(s) + 2e^- \longrightarrow Pb(s) + 2Br^-(aq); E^\circ_1 = -0.248 V$
2. $Pb^{2+}(aq) + 2e^- \longrightarrow Pb(s); \quad E^\circ_2 = -0.126 V$

To obtain the dissolution reaction, we can consider a hypothetical electrochemical cell where $PbBr_2$ dissolves. This can be achieved by subtracting the second half-reaction from the first half-reaction:

$(PbBr_2(s) + 2e^- \longrightarrow Pb(s) + 2Br^-(aq))$ $- (Pb^{2+}(aq) + 2e^- \longrightarrow Pb(s))$

$PbBr_2(s) - Pb^{2+}(aq) \longrightarrow 2Br^-(aq)$

Rearranging this gives the dissolution reaction: $PbBr_2(s) \longrightarrow Pb^{2+}(aq) + 2Br^-(aq)$

The standard cell potential ($E^\circ_{cell}$) for this reaction is the difference between the standard electrode potentials: $E^\circ_{cell} = E^\circ_1 - E^\circ_2$ $E^\circ_{cell} = (-0.248 V) - (-0.126 V)$ $E^\circ_{cell} = -0.248 V + 0.126 V$ $E^\circ_{cell} = -0.122 V$

This $E^\circ_{cell}$ is related to the equilibrium constant ($K_{sp}$ in this case) by the Nernst equation at equilibrium: $E^\circ_{cell} = \frac{RT}{nF} \ln K_{sp}$ At 298 K (standard temperature), this simplifies to: $E^\circ_{cell} = \frac{0.0591}{n} \log K_{sp}$

In the dissolution reaction $PbBr_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Br^-(aq)$, the number of electrons transferred ($n$) is 2 (as seen in the half-reactions involving $Pb^{2+}$ and $PbBr_2$).

Substitute the values into the equation: $-0.122 V = \frac{0.0591 V}{2} \log K_{sp}$ $-0.122 = 0.02955 \log K_{sp}$

Now, solve for $\log K_{sp}$: $\log K_{sp} = \frac{-0.122}{0.02955}$ $\log K_{sp} \approx -4.128595$

To find $K_{sp}$, take the antilog: $K_{sp} = 10^{-4.128595}$ $K_{sp} = 10^{(-5 + 0.871405)}$ $K_{sp} = 10^{0.871405} \times 10^{-5}$ $K_{sp} \approx 7.437 \times 10^{-5}$

Comparing this value with the given options, option A is the closest.