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What is the purpose of the Nafion the membrane in Figure 17-33

Short Answer

Expert verified

The benefits of nafion membrane in given figure has to be given.

Step by step solution

01

Define Nafion membrane

Nafion membrane is made up of co-polymer nafion which consists of tetrafluoroethylene. It is used as a conductor of protons in proton exchange membrane fuel cells. It can be prepared by co-polymerizing tetrafluoroethylene and perfluoro derivatives. The Half reaction is one of oxidation or reduction in the redox reaction. The half-half reaction is given by oxidation state change in reactants or products in the redox reaction.

02

Determine the beneifts of nafion membrane

Given,

  • Nafion membrane has cation exchange membrane which is coated by carbon fiber known as nafion.
  • Cations can easily pass-through membrane because of fixed negative charges.
  • In rat brain nafion membrane is utilized to detect dopamine, a cationic neuro transmitter.
  • Nafion membrane also known as ultra-micro electrodes and its size is .

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Most popular questions from this chapter

cd2+was used as an internal standard in the analysis of Pb2+by square wave polarography. Cd2+gives a reduction wave at -0.60 V and Pb2+gives a reduction wave at โ€“0.40 V. It was first verified that the ratio of peak heights is proportional to the ratio of concentrations over the whole range employed in the experiment. Here are results for known and unknown mixtures:

The unknown mixture was prepared by mixing 25.00(ยฑ0.05)mLof unknown (containing only Pb2+) plus 10.00(ยฑ0.05)mLof 3.23(ยฑ0.01)ร—10-4MCd2+and diluting to 50.00(ยฑ0.05)mL.

(a) Disregarding uncertainties, find [Pb2+]in the undiluted unknown.

(b) Find the absolute uncertainty for the answer to part (a).

Fundamentals of Electrolysis

17 - 6.The cell in Figure 17 - 4 is:

Cu|1.0MCuSO4(aq)|KCL(aq,3M)|AgCI(s)|Ag(s)

Write half-reactions for this cell. Neglecting activity coefficients and the junction potential betweenCuSO4(aq)and KCI(aq), predict the equilibrium (zero-current) voltage expected when the Lugging capillary contacts the electrode. For this purpose, suppose that the reference electrode potential is 0.197Vvs. S.H.E. Why is the observed equilibrium potential+109mV, not the value you calculated?

How would the over potentials change if>1.000Vwere imposed by the

Potentiostat?

A solution of Sn2+is to be electrolyzed to reduce the Sn2+to Sn(s). Calculate the cathode potential (versus S.H.E.) needed to reduce[Sn2+]to1.0ร—10-8Mif no concentration polarization occurs. What would be the potential versus S.C.E. instead of S.H.E? Would the potential be more positive or more negative if concentration polarization occurred?

Explain what is done in anodic stripping voltammetry. Why is stripping the most sensitive voltammetric technique?

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?
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