Question

Black-and-white photographic film is coated with silver halides. Because silver is expensive, the manufacturer monitors the ${\mathbf{Ag}}^{\mathbf{+}}$content of the waste stream, $\mathbf{[}\mathbf{A}{\mathbf{g}}^{\mathbf{+}}{\mathbf{]}}_{\mathbf{waste}}$, from the plant with an ${\mathbf{Ag}}^{\mathbf{+}}$selective electrode at ${\mathbf{25}}^{\mathbf{o}}\mathbf{C}$. A stream of known ${\mathbf{Ag}}^{\mathbf{+}}$concentration, $\mathbf{[}\mathbf{A}{\mathbf{g}}^{\mathbf{+}}{\mathbf{]}}_{\mathbf{standard}}$, is passed over the electrode in turn with the waste stream and the data recorded by a computer.

(a) Write the equations relating the nonstandard cell potential to the standard cell potential and [${\mathbf{Ag}}^{\mathbf{+}}$] for each solution. (b) Combine these into a single equation to find $\mathbf{[}\mathbf{A}{\mathbf{g}}^{\mathbf{+}}{\mathbf{]}}_{\mathbf{waste}}$. (c) Rewrite the equation from part (b) to find $\mathbf{[}\mathbf{A}{\mathbf{g}}^{\mathbf{+}}{\mathbf{]}}_{\mathbf{waste}}$ in ng/L. (d) If ${\mathbf{E}}_{\mathbf{waste}}$is$0.003$V higher than , and the standard solution contains $\mathbf{1000}$ng/L, what is $\mathbf{[}\mathbf{A}{\mathbf{g}}^{\mathbf{+}}{\mathbf{]}}_{\mathbf{waste}}$?(e) Rewrite the equation in part (b) to find$\mathbf{[}\mathbf{A}{\mathbf{g}}^{\mathbf{+}}{\mathbf{]}}_{\mathbf{waste}}$for a system in which T changes and${\mathbf{T}}_{\mathbf{waste}}$androle="math" localid="1663984231693" ${\mathbf{T}}_{\mathbf{s}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{a}\mathbf{r}\mathbf{d}}$may be different.

Short answer

(a) The equations rewritten in standard cell potential and is:

$\begin{array}{l}{\mathrm{E}}_{\mathrm{waste}}={\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{\mathrm{RT}}{\mathrm{nF}}\times \ln {\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{waste}}................\left(1\right)\\ {\mathrm{E}}_{\mathrm{standard}}={\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{\mathrm{RT}}{\mathrm{nF}}\times \ln {\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{standard}}..........\left(2\right)\end{array}$

(b) The equations combined in single equation is:

${\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{waste}}={\mathrm{e}}^{\frac{-\left({\mathrm{E}}_{\mathrm{waste}}-{\mathrm{E}}_{\mathrm{standard}}\right)\times \mathrm{nF}}{\mathrm{RT}}}\times {\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{standard}}..............\left(3\right)$

(c) The equation of ${\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{waste}}$when converted in ng/L is:

${\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{waste}}={\mathrm{e}}^{\frac{-\left({\mathrm{E}}_{\mathrm{waste}}-{\mathrm{E}}_{\mathrm{standard}}\right)\times \mathrm{nF}}{\mathrm{RT}}}\times {\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{standard}}............\left(4\right)$

(d) The value ${\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{waste}}$of is obtained as: $\mathrm{A}{{\mathrm{g}}^{+}}_{\mathrm{waste}}=889.8\frac{\mathrm{ng}}{\mathrm{L}}$.

Step-by-step solution

Define chemical reaction

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

Rewrite the equation

(a) Cell potentials may be linked to ion concentrations using the Nernst equation.

$\begin{array}{c}{\mathrm{E}}_{\mathrm{waste}}{=\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{\mathrm{RT}}{\mathrm{nF}}\times \ln {\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{waste}}................\left(1\right)\\ {\mathrm{E}}_{\mathrm{standard}}{=\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{\mathrm{RT}}{\mathrm{nF}}\times \ln {\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{standard}}..........\left(2\right)\end{array}$

Therefore, the equations are:

$\begin{array}{c}{\mathrm{E}}_{\mathrm{waste}}{=\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{\mathrm{RT}}{\mathrm{nF}}\times \ln {\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{waste}}................\left(1\right)\\ {\mathrm{E}}_{\mathrm{standard}}{=\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{\mathrm{RT}}{\mathrm{nF}}\times \ln {\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{standard}}..........\left(2\right)\end{array}$

Combine to form single equation

(b) Equation ($2$) should be removed from equation ($1$) in part a in order to extract ${\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{waste}}$

$\begin{array}{c}{\mathrm{E}}_{\mathrm{waste}}-{\mathrm{E}}_{\mathrm{standard}}={\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}{-\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{\mathrm{RT}}{\mathrm{nF}}\times \ln {\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{waste}}-\frac{\mathrm{RT}}{\mathrm{nF}}\times \ln {\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{standard}}\\ {\mathrm{E}}_{\mathrm{waste}}-{\mathrm{E}}_{\mathrm{standard}}=-\frac{\mathrm{RT}}{\mathrm{nF}}\times \ln \frac{{\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{waste}}}{{\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{standard}}}\\ \ln \frac{{\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{waste}}}{{\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{standard}}}=\frac{-\left({\mathrm{E}}_{\mathrm{waste}}-{\mathrm{E}}_{\mathrm{standard}}\right)\times \mathrm{nF}}{\mathrm{RT}}\\ \frac{{\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{waste}}}{{\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{standard}}}={\mathrm{e}}^{\frac{-\left({\mathrm{E}}_{\mathrm{waste}}-{\mathrm{E}}_{\mathrm{standard}}\right)\times \mathrm{nF}}{\mathrm{RT}}}\\ {\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{waste}}={\mathrm{e}}^{\frac{-\left({\mathrm{E}}_{\mathrm{waste}}-{\mathrm{E}}_{\mathrm{standard}}\right)\times \mathrm{nF}}{\mathrm{RT}}}\times {\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{standard}}\end{array}$

Therefore, the equation is:

${\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{waste}}={\mathrm{e}}^{\frac{-\left({\mathrm{E}}_{\mathrm{waste}}-{\mathrm{E}}_{\mathrm{standard}}\right)\times \mathrm{nF}}{\mathrm{RT}}}\times {\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{standard}}$

Evaluate [Ag+]waste in ng/L

The molarity of ${\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{waste}}$and ${\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{standard}}$must be multiplied by the molar mass of Ag in ng/mol to adapt the equation to use concentrations in ng/L.

The conversion factor will simply cancel out given the term $\ln \frac{{\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{waste}}}{{\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{standard}}}$. If alternative concentration units are used, the equation remains the same. It still stands:

${\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{waste}}={\mathrm{e}}^{\frac{-\left({\mathrm{E}}_{\mathrm{waste}}-{\mathrm{E}}_{\mathrm{standard}}\right)\times \mathrm{nF}}{\mathrm{RT}}}\times {\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{standard}}$

Therefore, the equation is:

${\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{waste}}={\mathrm{e}}^{\frac{-\left({\mathrm{E}}_{\mathrm{waste}}-{\mathrm{E}}_{\mathrm{standard}}\right)\times \mathrm{nF}}{\mathrm{RT}}}\times {\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{standard}}$

Evaluate the value of [Ag+]waste

Using the equation in with ${\mathrm{E}}_{\mathrm{waste}}-{\mathrm{E}}_{\mathrm{standard}}=0.003\mathrm{V}$and ${\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{standard}}=1000\mathrm{ng}/\mathrm{L}$, the concentration of ${\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{waste}}$in ng/L may be calculated (c). At 298.15 K, the reduction of silver involves only one electron:

$\begin{array}{c}\mathrm{A}{{\mathrm{g}}^{+}}_{\mathrm{waste}}={\mathrm{e}}^{\frac{-\left({\mathrm{E}}_{\mathrm{waste}}-{\mathrm{E}}_{\mathrm{standard}}\right)\times \mathrm{nF}}{\mathrm{RT}}}\times \mathrm{A}{{\mathrm{g}}^{+}}_{\mathrm{standard}}\\ \mathrm{A}{{\mathrm{g}}^{+}}_{\mathrm{waste}}\left(\frac{\mathrm{ng}}{\mathrm{L}}\right)={\mathrm{e}}^{\frac{-0.003\times 1\times 96,500}{298.15\times 8.314}}\times 1000\left(\frac{\mathrm{ng}}{\mathrm{L}}\right)\\ \mathrm{A}{{\mathrm{g}}^{+}}_{\mathrm{waste}}=889.8\frac{\mathrm{ng}}{\mathrm{L}}\end{array}$

Therefore, the value is: $\mathrm{A}{{\mathrm{g}}^{+}}_{\mathrm{waste}}=889.8\frac{\mathrm{ng}}{\mathrm{L}}$.