Question

From the following information,

AgBr(s) + e → Ag(s) + Bre(aq); E° = 0.09 V Ag⊕(aq) + e → Ag(s); E° = 0.81 V If the solubility product of AgBr is 1x10⁻ˣ, then x is

$\qquad \Bigg( Given, \frac{2.303 RT}{F} = 0.06 \Bigg)$

Answer 1

12

Answer 2

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

The solubility product of AgBr is defined by the equilibrium: $AgBr(s) \rightleftharpoons Ag^+(aq) + Br^-(aq)$ $K_{sp} = [Ag^+][Br^-]$

We are given two standard electrode potentials:

1. $AgBr(s) + e^- \rightarrow Ag(s) + Br^-(aq)$; $E^\circ_1 = 0.09 V$
2. $Ag^+(aq) + e^- \rightarrow Ag(s)$; $E^\circ_2 = 0.81 V$

To obtain the reaction $AgBr(s) \rightleftharpoons Ag^+(aq) + Br^-(aq)$, we can subtract the first half-reaction from the second half-reaction: $(Ag^+(aq) + e^- \rightarrow Ag(s)) - (AgBr(s) + e^- \rightarrow Ag(s) + Br^-(aq))$ This yields: $Ag^+(aq) - AgBr(s) \rightarrow -Br^-(aq)$ Rearranging, we get the reaction: $Ag^+(aq) + Br^-(aq) \rightarrow AgBr(s)$

The standard cell potential ($E^\circ_{rxn}$) for this reaction is the difference between the standard electrode potentials of the two half-reactions: $E^\circ_{rxn} = E^\circ_2 - E^\circ_1 = 0.81 V - 0.09 V = 0.72 V$

The equilibrium constant for the reaction $Ag^+(aq) + Br^-(aq) \rightarrow AgBr(s)$ is $K = \frac{1}{[Ag^+][Br^-]} = \frac{1}{K_{sp}}$.

Now, we use the relationship between the standard cell potential and the equilibrium constant, which is derived from the Nernst equation: $E^\circ_{rxn} = \frac{2.303 RT}{nF} \log K$ Given $\frac{2.303 RT}{F} = 0.06 V$. The number of electrons transferred ($n$) for this reaction is 1 (as seen from the half-reactions).

Substitute the values into the equation: $0.72 V = \frac{0.06}{1} \log \left(\frac{1}{K_{sp}}\right)$ $0.72 = 0.06 (-\log K_{sp})$ Divide both sides by 0.06: $-\log K_{sp} = \frac{0.72}{0.06}$ $-\log K_{sp} = 12$ $\log K_{sp} = -12$

To find $K_{sp}$, take the antilog: $K_{sp} = 10^{-12}$

The problem states that the solubility product of AgBr is $1 \times 10^{-x}$. Comparing our calculated $K_{sp}$ with the given form: $1 \times 10^{-x} = 1 \times 10^{-12}$ Therefore, $x = 12$.