Question

Chemical oxygen demand by coulonetry. An electrochemical device incorporating photooxidation on a \({\rm{Ti}}{{\rm{O}}_2}\) surface could replace refluxing with \({{\rm{C}}_2}{\rm{O}}_7^{2 - }\) to measure chemical oxygen demand (Box 16-2). The diagram shows a working electrode beld at \( + 0.30\;{\rm{V}}\) versus \({\rm{Ag}}\mid {\rm{AgCl}}\) and coated with nanoparticles of 'TiO . Wltraviolet²inradiation generates electrons and holes in \({{\rm{T}}_1}{{\rm{O}}_2}\). Holes oxidize

organic matter at the surface. Electrons reduce \({{\rm{H}}_2}{\rm{O}}\) at the auxiliary electrode in a compartment connected to the working compartment by a salt bridge. The sample compartment is only 0.18 mm thick with a volume of \(13.5\mu \,{\rm{L}}\). It requires \(\~1\;\,{\rm{min}}\) for all organic matter to diffuse to the \({\rm{Ti}}{{\rm{O}}_2}\) surface and be exhaustively oxidized.

Left: Working electrode. Fight Photocument response for sample and blank Both solutions contain \(2{\rm{M}}\,{\rm{NaNO}}\). (Dst from H zhso, D. fisng. 5 . zhang K. Cutteral, and R. Jshn, "Development of a Drect Fhotselectrocherrical Method for Deterrination of Gherrical Ouygen Demand," And. Chan. 2004, 76 155.)

The blank curve in the graph shows the response when the sample compartment contains just electrolyte. Before inradiation, no current is observed. Ultraviolet radiation causes a spike in the current, followed by a decrease to a steady level near \(40\mu \). This current arises from oxidation of water at the \({\rm{Ti}}{{\rm{O}}_2}\)sufface under ultraviolet exposure. The upper curve sbows the same experiment, but with wastewater in the sample compartment. The increased current arises from oxidation of organic matter. When the organic matter is consumed, the cument decreases to the blank level. The area between the two curves tells us how many electrons flow from oxidation of organic matter in the sample.

1. Balance the oxidation half-reaction that occurs in this cell:

\({{\rm{C}}_e}{{\rm{H}}_k}{{\rm{O}}_a}\;{{\rm{N}}_s}{{\rm{X}}_x} + {\rm{A}}{{\rm{H}}_2}{\rm{O}} \to {\rm{BC}}{{\rm{O}}_2} + {\rm{CX}} + {\rm{DN}}{{\rm{H}}_3} + {\rm{E}}{{\rm{H}}^ + } + {\rm{F}}{{\rm{e}}^ - }\)

where X is any halogen. Express the stoichiometry coefficients A, B, C, D, E, and F in terms of c, h, o, n, and x.

1. How many molecules of \({{\rm{O}}_2}\)are required to balance the halfreaction in part (a) by reduction of oxygen (\({{\rm{O}}_2} + 4{{\rm{H}}^ + } + 4{{\rm{e}}^ - } \to 2{{\rm{H}}_2}{\rm{O}}\))?
2. The area between the two curves in the graph is \(\int_0^\infty {({I_{{\rm{sample }}}}} - {I_{blank}})dt = 9.43\,{\rm{mC}}{\rm{.}}\) This is the number of electrons liberated by complete oxidation of the sample. How many moles of \({{\rm{O}}_2}\) would be required for the same oxidation?
3. Chemical oxygen demand (COD) is expressed as mg of \({{\rm{O}}_2}\) required to oxidize 1 L of sample. Find the COD for this sample.
4. If the only caidizable substance in the sample were \({{\rm{C}}_9}{{\rm{H}}_6}{\rm{N}}{{\rm{O}}_2}{\rm{CIB}}{{\rm{r}}_2}\). what is its concentration in molL?

Short answer

1. A half reaction:
   \(\begin{aligned}{c}{\rm{F}} &= {\rm{E}} - {\rm{C}}\\F &= h + 4c - 2o - 3n - c\\{\rm{F}} &= {\rm{h}} - \frac{{\rm{c}}}{2} + {\rm{o}} - 3{\rm{n}}\end{aligned}\)
2. Number of molecules of oxygen that is necessary to balance the half reaction by reduction of oxygen is \(\frac{{\rm{F}}}{4}{{\rm{O}}_2}\).
3. Number of moles of oxygen that would be necessary for oxidation is \(2.223 \times {10^{ - 8}}\;{\rm{mol}}\).
4. Chemical oxygen demand of the sample is \(52.7{\rm{mg}}/{\rm{L}}\).
5. Concentration in \({\rm{mol}}/{\rm{L}}\) is \(2.26 \times {10^{ - 4}}{\rm{M}}\).

Step-by-step solution

Concept introduction

1. A half reaction is a redox reaction component that is either an oxidation or a reduction reaction. The half reaction is determined by the change in oxidation states of individual compounds in a redox reaction. Half reactions are a typical way of redox reaction balancing.
2. The quantity of oxygen molecules required to balance the half reaction by reduction of oxygen must be determined.
3. The total number of moles of oxygen required for oxidation must be estimated.
4. The sample's chemical oxygen requirement must be estimated.

5. The number of moles per litre of the solution is called as Molarity. Molarity can be defined as the moles of solute (in grams) to the volume of the solution (in litres). The molarity of a solution can be given by the formula,\({\rm{Molarity }}({\rm{M}}) = \frac{{{\rm{ Moles of solute(ing) }}}}{{{\rm{ Volume ofsolution(in L) }}}}\)

Find the oxidation half-reaction

Balance of Carbon: \({\rm{B}} = {\rm{C}}\)

Balance of Halogen: \(C = x\)

Balance of Nitrogen: \(D = n\)

Balance of Oxygen:

\(\begin{aligned}{c}{\rm{o}} + {\rm{A}} &= 2\;{\rm{B}}\\0 + A &= 2c\\{\rm{A}} &= 2{\rm{c}} - {\rm{o}}\end{aligned}\)

Balance of Hydrogen:

\(\begin{aligned}{l}{\rm{h}} + 2\;{\rm{A}} - 3{\rm{D}} + {\rm{E}}\\h + 2(2c - 0) &= 3n + E\\{\rm{E}} &= {\rm{h}} + 4{\rm{c}} - 20 - 3{\rm{n}}\end{aligned}\)

Charge balance,

\(\begin{aligned}{c}{\rm{F}} &= {\rm{E}} - {\rm{C}}\\{\rm{F}} &= {\rm{h}} + 4{\rm{c}} - 20 - 3{\rm{n}} - {\rm{c}}\\{\rm{F}} &= {\rm{h}} - \frac{{\rm{c}}}{2} + {\rm{o}} - 3{\rm{n}}\end{aligned}\)

Find the number of mocules of oxygen

To consume \({\rm{F}}{{\rm{e}}^ - }\), the necessary molecules of oxygen are \(\frac{{\rm{F}}}{4}{{\rm{O}}_2}\) therefore, each molecule of oxygen consumes \(2{{\rm{e}}^ - }\).

Find the area

\(\begin{aligned}{c}{\rm{F}} &= \left( {9.43 \times {{10}^{ - 3}}{\rm{C}}} \right)\left( {9.6485 \times {{10}^4}{\rm{C}}/{\rm{mol}}} \right)\\ &= 9.774 \times {10^{ - 8}}\;{\rm{mol}}{{\rm{e}}^ - }\\{\rm{mol}}{{\rm{O}}_{\rm{2}}} &= \frac{F}{4} &= 2.223 \times {10^{ - 8}}\end{aligned}\)

Find the COD for this sample

The number of moles of Oxygen \( = 2.223 \times {10^{ - 8}}\;{\rm{mol}}\)

Mass of Oxygen \( = \left( {2.223 \times {{10}^{ - 8}}\;{\rm{mol}}} \right)(32.00\;{\rm{g}}/{\rm{mol}})\)

Mass of Oxygen \( = 7.114 \times {10^{ - 7}}\;{\rm{g}}\)

\(7.114 \times {10^{ - 7}}\;{\rm{g}}\)of Oxygen is required to react with \(13.5\mu {\rm{L}}\) of the sample. The mass of Oxygen that is needed to react with \(1\;{\rm{L}}\) of the sample is

\(\begin{aligned}{l} &= \left( {7.114 \times {{10}^{ - 7}}\;{\rm{g}}} \right)\left( {13.5 \times {{10}^{ - 6}}\;{\rm{L}}} \right)\\ &= 0.0527\;{\rm{g}}/{\rm{L}}\\ &= 52.7{\rm{mg}}/{\rm{L}}\end{aligned}\)

So, chemical oxygen demand of the sample \( = 52.7{\rm{mg}}/{\rm{L}}\).

Find the concentration in molL

The balanced oxidation half reaction can be given as,

\({{\rm{C}}_9}{{\rm{H}}_6}{\rm{N}}{{\rm{O}}_2}{\rm{ClB}}{{\rm{r}}_2} + 16{{\rm{H}}_2}{\rm{O}} \to 9{\rm{C}}{{\rm{O}}_2} + 3{{\rm{X}}^ - } + {\rm{N}}{{\rm{H}}_3} + 35{{\rm{H}}^ + } + 32{{\rm{e}}^ - }\)

Number of electrons observed in the reaction is \(9.774 \times {10^{ - 8}}\;{\rm{mol}}{{\rm{e}}^ - }\), therefore, there must be \(\frac{{\left( {9.774 \times {{10}^{ - 8}}\;{\rm{mol}}{{\rm{e}}^ - }} \right)}}{{\left( {32\;{\rm{mole}} - /{\rm{mol}}{{\rm{C}}_9}{{\rm{H}}_6}{\rm{N}}{{\rm{O}}_2}{\rm{ClB}}{{\rm{r}}_2}} \right)}} = 3.054 \times {10^{ - 9}}\;{\rm{mol}}\,{{\rm{C}}_9}{{\rm{H}}_6}{\rm{N}}{{\rm{O}}_2}{\rm{ClB}}{{\rm{r}}_2}\) in \(13.5\mu {\rm{L}}\).

The molarity of \({{\rm{C}}_9}{{\rm{H}}_6}{\rm{N}}{{\rm{O}}_2}{\rm{ClB}}{{\rm{r}}_2} = \frac{{\left( {3.054 \times {{10}^{ - 9}}\;{\rm{mol}}} \right)}}{{\left( {13.5 \times {{10}^{ - 6}}\;{\rm{L}}} \right)}}\)

The molarity of \({{\rm{C}}_9}{{\rm{H}}_6}{\rm{N}}{{\rm{O}}_2}{\rm{ClB}}{{\rm{r}}_2} = 2.26 \times {10^{ - 4}}{\rm{M}}\)