Question

Ions that react with $\mathit{A}{\mathit{g}}^{\mathbf{+}}$can be determined electrogravimetrically by deposition on a silver working anode:

$\mathit{A}\mathit{g}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}{\mathit{X}}^{\mathbf{-}}\mathbf{\to }\mathit{A}\mathit{g}\mathit{X}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}{\mathit{e}}^{\mathbf{-}}$

(a) What will be the final mass of a silver anode used to electrolyze 75.00 mL of 0.02380 M KSCN if the initial mass of the anode is 12.4638 g?

(b) At what electrolysis voltage (versus S.C.E.) will AgBr(s) be deposited from 0.10M Br? (Consider negligible current flow, so that there is no ohmic potential, concentration polarization, or overpotential.)

(c) Is it theoretically possible to separate 99.99% of0.10M Klfrom0.10MKBr by controlled-potential electrolysis?

Short answer

1. The final mass of silver anode used in electrolysis is 12.5675 g .
2. The voltage at which bromide ion deposits silver bromide 0.111 V.
3. It is impossible to separate potassium iodide from potassium bromide.

Step-by-step solution

Introduction

• Electrogravimetry is a technique for electroplating a metal onto a platinum electrode in a quantitative manner.
• The difference in the weight of the electrode before and after electroplating determines the amount of metal plated.

Calculation regarding part (a)

(a)

It is necessary to calculate the mass of silver anode used in electrolysis.

To figure out: The weight of the silver anode used in electrolysis.

AgSCN concentration is the same as SCN concentration. The concentration of SCN can be calculated from the provided concentration of KSCN. SCN is computed as

$\begin{array}{r}75.00{\mathrm{mL}}^{\text{of}}0.02380\mathrm{MKSCN}=1.785\mathrm{mmol}{\mathrm{SCN}}^{-}\\ 1.785\mathrm{mmol}{\mathrm{SCN}}^{-}=1.785\mathrm{mmolAgSCN}\\ 1.785\mathrm{mmolAgSCN}\text{contains}0.1037\mathrm{gSCN}\\ \begin{array}{rl}\text{Mass}& =12.4638+0.1037\\ & =12.5675\mathrm{g}\end{array}\end{array}$

The amount of silver anode used in electrolysis is 12.5675g.

Calculation regarding part (b)

(b)

It is necessary to compute the voltage at which silver bromide deposits from bromide ions.

The voltage of a cell is stated as when the electric current is too small$E=E\left(cathode\right)-E\left(anode\right)$

E(cathode) is the electrode potential that is connected to the current source's negative terminal.

E(anode) is the potential of an electrode that is connected to the positive terminal of a current source.

Voltage can override the activation energy of a process at an electrode, resulting in overpotential. Overpotential is the required voltage to apply.

Ohmic potential can be expressed as follows

$E_{ohmic}=IR$

Concentration Polarization is the difference in concentration of products and reactants at the electrode's surface, as opposed to the same concentration in solution.

The voltage at which silver bromide deposited from bromide ions was determined.

At anode:$AgB{r}_{(}\left(g\right))+e^{-}\rightleftarrows Ag(s)+B{r}^{-}{E}^{°}=0.071V$

$\begin{array}{r}\mathrm{E}\text{(cathode)}=\mathrm{E}\left(\text{S.C.E.}\right)=0.241\mathrm{V}\\ \mathrm{E}\text{(anode)}=0.071-0.05916\log \left[{\mathrm{Br}}^{-}\right]\\ =0.071-0.05916\log \left[0.10\right]=0.130\mathrm{V}\\ \mathrm{E}\left(\text{cell}\right)=\mathrm{E}\left(\text{cathode}\right)-\mathrm{E}\text{(anode)}\\ =0.111\mathrm{V}\end{array}$

The voltage at which bromide ion deposits silver bromide 0.111 V.

Calculation regarding part (c)

(c)

It must be established whether potassium iodide and potassium bromide can be separated.

The constant for the solubility product ($K_{sp}$In an aqueous solution, is defined as the equilibrium between a chemical and its ions.

The solubility product is the concentration of a dissolved ion multiplied by the power of coefficients.

a lonic substance$A_{3}BK{K}_{(}sp)=[A{]}^3\left[B\right]$.

Ion-based product ( $Q_c$) is defined as the product of the ion concentrations raised to the power of the ion coefficients in the solution. role="math" localid="1668358640049" $Q_c>K_{sp}$, There will be precipitation.

$Q_c$is lower than 0.10MK , In a solution with no precipitation, more solute can dissolve.

$Q_c$is equal to $K_{sp}$.

In solution, no additional solutes can dissolve or precipitate.

To determine whether potassium iodide and potassium bromide can be separated.

In equilibrium with iodide, the intensity of silver ion is

$\left[A{g}^{+}\right]=Ksp/\left[I^{-}\right]=\frac{(8.3\times {10}^{17})}{(1.0\times {10}^{-3})}=8.3\times {10}^{-12}M$

When silver ions are in equilibrium with bromide ions, their strength is

$\left[A{g}^{+}\right]=Ksp/\left[I^{-}\right]=\frac{(5.0\times {10}^{-15})}{0.10}=5.0\times {10}^{-12}$

Thus, role="math" localid="1668358876100" $8.3\times {10}^{12}N$The precipitation of silver ion begins 0.10M The bromide ion is a type of ion that exists in nature. There isn't going to be any rain.

It is impossible to separate potassium iodide from potassium bromide.