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.

(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:

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

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:

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

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: $\mathrm{A}{{\mathrm{g}}^{+}}_{\mathrm{waste}}=889.8\frac{\mathrm{ng}}{\mathrm{L}}$.