Question

Halide ions can be deposited at a silver anode, the reaction being

$\mathrm{Ag}\left(\mathrm{s}\right)+{\mathrm{X}}^{-}\to \mathrm{AgX}\left(\mathrm{s}\right)+{\mathrm{e}}^{-}$

Suppose that a cell was formed by immersing a silver anode in an analyte solution that was 0.0250 M Cl^-, Br^-_, and I^- ions and connecting the half-cell to a saturated calomel cathode via a salt bridge.

(a) Which halide would form first and at what potential? Is the cell galvanic or electrolytic?

(b) Could I^- and Br^- be separated quantitatively? (Take 1.00 x 10^-5 M as the criterion for quantitative removal of an ion.) If a separation is feasible, what range of cell potential could be used?

(c) Repeat part (b) for I^- and Cl^-.

(d) Repeat part (b) for Br^- and Cl^-.

Short answer

(a) The halide which will form first is AgI at 0.3 V and the type of cell is galvanic.

(b) There is a possibility of I^- and Br^-being separate quantities and the range of cell potential for this possibility is above 0.076 V.

(c) There is a possibility of I^- and Cl^-being separate quantities and the range of cell potential for this possibility is above 0.073 V.

(d) There is no possibility of Cl^- and Br^- being separate quantities.

Step-by-step solution

Part (a) Step 1: Given Information

The halide which will form first and the potential at which it will form along with the type of the cell is to be stated.

Part (a) Step 2: Explanation

The concentration of Br^-, I^-, and Cl^- is 0.0250 M.

The expression the Nernst equation for half cell reduction potential at room temperature is:

$E=E_c-\left(E^0-\frac{0.05916\mathrm{V}}{n}{\log }_{10}\frac{a_{red}}{a_{ox}}\right)$……. (I)

Here, the half cell reduction potential is E, the cathode potential is E_c, the standard half cell reduction potential is E⁰, the cell potential is E_cell, chemical activity of reductant is a_red, chemical activity of oxidant is a_ox, and the number of moles of electrons is n.

The reaction of halides with silver is shown below.

data-custom-editor="chemistry" $\mathrm{AgI}\left(\mathrm{s}\right)+{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Ag}\left(\mathrm{s}\right)+{\mathrm{I}}^{-}$…… (II)

The value of electrode potential for the Equation (II) as -0.151 V.

data-custom-editor="chemistry" $\mathrm{AgBr}\left(\mathrm{s}\right)+{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Ag}\left(\mathrm{s}\right)+{\mathrm{Br}}^{-}$…… (III)

The value of electrode potential for the Equation (III) as 0.073 V.

data-custom-editor="chemistry" $\mathrm{AgCl}\left(\mathrm{s}\right)+{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Ag}\left(\mathrm{s}\right)+{\mathrm{Cl}}^{-}$ …… (IV)

The value of electrode potential for the Equation (IV) as 0.222 V.

Substitute E_I for E, 0.244 V for E_c, -0.151 V for E₀, 1 for n, and 0.0250 for $\frac{a_{red}}{a_{ox}}$in Equation (I).

role="math" localid="1649308981048" $$\begin{gathered} {\mathrm{E}}_{\mathrm{I}}=0.244\mathrm{V}-\left(-0.151\mathrm{V}-\frac{0.05916\mathrm{V}}{1}{\log }_{10}0.0250\right) \\ =0.244\mathrm{V}-\left(-0.151\mathrm{V}+0.0948\mathrm{V}\right) \\ =0.3\mathrm{V} \end{gathered}$$

Substitute E_Br for E, 0.244 V for E_c, 0.073 V for E₀, 1 for n, and 0.0250 for $\frac{a_{red}}{a_{ox}}$in Equation (I).

$$\begin{gathered} {\mathrm{E}}_{Br}=0.244\mathrm{V}-\left(0.073\mathrm{V}-\frac{0.05916\mathrm{V}}{1}{\log }_{10}0.0250\right) \\ =0.244\mathrm{V}-\left(0.073\mathrm{V}+0.0948\mathrm{V}\right) \\ =0.076\mathrm{V} \end{gathered}$$

Substitute E_Cl for E, 0.244 V for E_c, 0.222 V for E₀, 1 for n, and 0.0250 for $\frac{a_{red}}{a_{ox}}$in Equation (I).

$$\begin{gathered} {\mathrm{E}}_{Cl}=0.244\mathrm{V}-\left(0.222\mathrm{V}-\frac{0.05916\mathrm{V}}{1}{\log }_{10}0.0250\right) \\ =0.244\mathrm{V}-\left(0.222\mathrm{V}+0.0948\mathrm{V}\right) \\ =-0.073\mathrm{V} \end{gathered}$$

The potential of iodine is the largest out of the three halides.

A salt bridge is used to connect the electrochemical cell to carry out the process. Hence, this cell is a type of galvanic cell.

Therefore, the halide which will form first is AgI at 0.3 V and the type of cell is galvanic.

Part (b) Step 1: Given Information 

The possibility of I^- and Br^- being separate quantities and if possible, the range of cell potential that can be used is to be stated.

Part (b) Step 2: Explanation

The ratio of a_red to a_ox is 1.00 x 10^-5 .

Substitute E_I for E, 0.244 V for E_C, -0.151 V for E⁰, 1 for n, and 1.00 x 10^-5 for $\frac{a_{red}}{a_{ox}}$in Equation (I).

$$\begin{gathered} {\mathrm{E}}_{\mathrm{I}}=0.244\mathrm{V}-\left(-0.151\mathrm{V}-\frac{0.05916\mathrm{V}}{1}{\log }_{10}\left(1.00\times {10}^{-5}\right)\right) \\ =0.244\mathrm{V}-\left(-0.151\mathrm{V}+0.2592\mathrm{V}\right) \\ =0.01\mathrm{V} \end{gathered}$$

The potential needed for the formation of AgBr is 0.076 V. Hence, the separation of I^- is only possible above the potential of 0.076 V.

Therefore, there is a possibility of I^- and Br^- being separate quantities and the range of cell potential for this possibility is above 0.076 V.

Part (c) Step 1: Given Information

The possibility of I^- and Cl^-being separate quantities and if possible, the range of cell potential that can be used is to be stated.

Part (c) Step 2: Explanation

The potential needed for the formation of AgCl is 0.073 V. Hence, the separation of I^- is only possible above the potential of 0.073 V .

Therefore, there is a possibility of I^- and Cl^- being separate quantities and the range of cell potential for this possibility is above 0.073 V .

Part (d) Step 1: Given Information

The possibility of Cl^- and Br^-being separate quantities and if possible, the range of cell potential that can be used is to be stated.

Part (d) Step 2: Explanation

Substitute E_I for E, 0.244 V for E_c, -0.151 V for E⁰, 1 for n, and 1.00 x 10^-5 for $\frac{a_{red}}{a_{ox}}$in Equation (I).

$$\begin{gathered} {\mathrm{E}}_{\mathrm{I}}=0.244\mathrm{V}-\left(0.073-\frac{0.05916\mathrm{V}}{1}{\log }_{10}\left(1.00\times {10}^{-5}\right)\right) \\ =0.244\mathrm{V}-\left(0.073\mathrm{V}+0.2592\mathrm{V}\right) \\ =-0.124\mathrm{V} \end{gathered}$$

The potential needed for the formation of AgCl is 0.073 V. Hence, the separation of I is not possible

Therefore, there is no possibility of I^- and Br^- being separate quantities.