Q21.144 CP - Chemistry: Molecular Nature Of Matter And Change

Question

A chemist designs an ion-specific probe for measuring [ ${\text{Ag}}^{\text{ + }}$] in an NaCl solution saturated with AgCl. One half-cell has an Ag-wire electrode immersed in the unknown AgCl saturated NaCl solution. It is connected through a salt bridge to the other half-cell, which has a calomel reference electrode [a platinum wire immersed in a paste of mercury and calomel ( ${\mathbf{\text{Hg}}}_{\mathbf{\text{2}}}{\mathbf{\text{Cl}}}_{\mathbf{\text{2}}}$)] in a saturated KCl solution. The measured ${\text{E}}_{\text{cell}}$  is  $\text{0.060}$ V. (a) Given the following standard half-reactions, calculate [${\text{Ag}}^{\text{ + }}$ ]. 

Calomel:  ${\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}{\text{(s) + 2e}}^{\text{ - }}\to {\text{2Hg + 2Cl}}^{\text{ - }}{\text{ E}}^{\text{o}}\text{ = 0.24 V}$

Silver:  ${\text{Ag}}^{+}{\text{(aq) + e}}^{\text{ - }}\to {\text{Ag(s) E}}^{\text{o}}\text{ = 0.80 V}$

(Hint: Assume [ ${\text{Cl}}^{\text{ - }}$] is so high that it is essentially constant.) (b) A mining engineer wants an ore sample analyzed with the ${\text{Ag}}^{\text{ + }}$ - selective probe. After pre-treating the ore sample, the chemist measures the cell voltage as   V. What is [ ${\text{Ag}}^{\text{ + }}$]?

Step-by-Step Solution

Answer

1. The value is obtained as:  $\left[{\text{Ag}}^{\text{ + }}\right]\text{ = 4.75 M}$
2. The value is obtained as:  $\left[{\text{Ag}}^{\text{ + }}\right]\text{ = 0.311 M}$

Step 1: Define chemical reaction

Chemical synthesis or, alternatively, chemical breakdown into two or more separate chemicals occurs when one component interacts with another to generate new material. These processes are known as chemical reactions, and they are generally irreversible until followed by other chemical reactions.

Step 2: Evaluate the value of [ Ag  +  ]

1. Silver is reduced at typical reduction potentials, while Hg and ${\text{Cl}}^{\text{ - }}$  are oxidized.

 $$\begin{gathered} {\text{Oxidisation: 2Hg}}_{\text{(l)}}\text{ + 2C}{{\text{l}}^{\text{ - }}}_{\text{(aq)}}\to {\text{2e}}^{\text{ - }}{\text{ + 2Hg}}_{\text{2}}{\text{Cl}}_{\text{2(s)}} \\ \text{Reduction: A}{{\text{g}}^{+}}_{\text{(aq)}}{\text{ + e}}^{\text{ - }}\to {\text{Ag}}_{\text{(s)}} \end{gathered}$$

$\begin{array}{rcl}& & {\text{E}}_{\text{cell}}^{\text{o}}{\text{ = E}}_{\text{reduced}}^{\text{o}}{\text{ - E}}_{\text{oxidised}}^{\text{o}}\\ & & {\text{E}}_{\text{cell}}^{\text{o}}{\text{ = E}}_{{\text{Ag}}^{\text{ + }}}^{\text{o}}{\text{ - E}}_{{\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}}^{\text{o}}\\ & & \text{ = 0.80 V - 0.24 V}\\ & & \text{ = 0.56 V}\\ & & \end{array}$

It is given the we have the value as:

 ${\text{E}}_{\text{cell}}\text{ = 0.60 V}$

The Q may be extracted from the Nernst equation.

 ${\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cell}}^{\text{o}}\text{ - }\frac{\text{RT}}{\text{nF}}\text{×InQ}$

The process involves two electrons, and the standard temperature is 298  K.

 $$\begin{gathered} {\text{E}}_{\text{cell }}{\text{ = E}}_{\text{cell }}^{\text{o}}\text{ - }\frac{{\displaystyle \text{8.314}\frac{\text{J}}{\text{mol K}}\text{×298 K}}}{n\times {\displaystyle \text{96,500 C}}}\text{×lnQ} \\ {\text{E}}_{\text{cell}}\text{ = 0.56 - }\frac{{\displaystyle \text{8.314}\frac{\text{J}}{\text{mol K}}\text{×298 K}}}{2\times {\displaystyle \text{96,500 C}}}\text{×lnQ} \\ \text{0.60 = 0.56 - }\frac{{\displaystyle \text{8.314}\frac{\text{J}}{\text{mol K}}\text{×298 K}}}{2\times {\displaystyle \text{96,500 C}}}\text{×lnQ} \\ \text{lnQ = }\frac{{\displaystyle \text{ - (0.60 - 0.56) V×2×96,500 C}}}{\text{8.314}\frac{\text{J}}{\text{mol K}}\text{×298 K}} \\ \frac{\text{1}}{{\text{[}\overline{\text{Cl}}\text{ - ]}}^{\text{2}}{\left[{\text{Ag}}^{\text{ + }}\right]}^{\text{2}}}\text{ = e}\frac{{\displaystyle \text{ - (0.60 - 0.56) V×2×96,500 C}}}{\text{8.314}\frac{\text{J}}{\text{mol K}}\text{×298 K}} \end{gathered}$$

Assume [ ${\text{Cl}}^{\text{ - }}$] is constant and [ ${\text{Cl}}^{\text{ - }}$]= 1

$$\begin{gathered} \frac{\text{1}}{{\text{[1]}}^{\text{2}}{\left[{\text{Ag}}^{\text{ + }}\right]}^{\text{2}}}\text{ = }\frac{{\displaystyle \text{ - (0.60 - 0.56) V×2×96,500 C}}}{{\displaystyle \text{8.314}\frac{\text{J}}{\text{mol K}}\text{×298K}}} \\ {\left[{\text{Ag}}^{\text{ + }}\right]}^{\text{2}}\text{ = }\frac{1}{{\displaystyle \text{e}\frac{-\; (0.60\; -\; 0.56)\mathrm{V\times 2\times 96,500}C}{\text{8.314}\frac{\text{J}}{\text{mol K}}\text{×298K}}}} \\ \left[{\text{Ag}}^{\text{ + }}\right]\text{ = }\sqrt{\frac{1}{{\displaystyle \text{e}\frac{-\; (0.60\; -\; 0.56)V\times 2\times 96,500C}{\text{8.314}\frac{\text{J}}{\text{mol K}}\text{×298K}}}}} \\ \left[{\text{Ag}}^{\text{ + }}\right]\text{ = 4.75M} \end{gathered}$$

Therefore, the value is:  $\left[{\text{Ag}}^{\text{ + }}\right]\text{ = 4.75 M}$

Step 3: Evaluate the value of [ Ag  +  ]

a. The concentration of silver ions may be calculated using the equation in (a) and  ${\text{E}}_{\text{cell}}\text{ = 0.53 V}$.

 $$\begin{gathered} \left[{\text{Ag}}^{\text{ + }}\right]\text{ = }\sqrt{\frac{1}{{\displaystyle \text{e}\frac{{\displaystyle \text{ - (0.53 - 0.56) V×2×96,500 C}}}{{\displaystyle \text{ 8.314}\frac{\text{J}}{\text{mol K}}\text{×298K}}}}}} \\ \left[{\text{Ag}}^{\text{ + }}\right]\text{ = 0.311 M} \end{gathered}$$

Therefore, the value is:  $\left[{\text{Ag}}^{\text{ + }}\right]\text{ = 0.311 M}$