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Fundamentals of Electrolysis

17-9. A lead-acid battery in a car has six cells in series, each delivering close to2.0Vfor a total of 12V when the battery is discharging. Recharging requires ~2.4V per cell, or ~14V for the entire battery. 55Explain these observations in terms of Equation 17-6.

Short Answer

Expert verified

Electrical losses decrease the magnitude of the voltage in the cell and increase the magnitude of the voltage in the opposite direction.

Step by step solution

01

Electrogravimetric Analysis

  • Electrogravimetry is a technique for separating and quantifying ions of material, most often a metal.
  • The analyte solution is electrolyzed during this procedure.
  • The analyte is deposited on the cathode as a result of the electrochemical reduction.
02

Determine the observations in terms of the equation

  • The equation17-6is:
  • E=Ecathode-Eanode-I.R-overpotentials
  • When current does not flow, the potential is E (cathode) - E (anode) = 2.2 V .
  • When there is current, the ohmic loss and overpotentials decrease the output of the cell 2.0V , and a net cell voltage is then 2.0V.
  • The cell can be recharged at a very low current when the voltage 2.2Vis used in the opposite direction.
  • If we want to charge at a significant rate, we need additional voltage to overcome ohmic loss and overpotentials.
  • The recharge is then required -2.4V.
  • Electrical losses decrease the magnitude of the voltage in the cell and increase the magnitude of the voltage in the opposite direction.

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Most popular questions from this chapter

Will(cathode) =0.19 Vreduce 0.10MSbO+at pH 2 by the reactionSbO++2H++3e-Sb(s)+H2O,E°=0.208V?

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?

How many hours are required for 0.100 mol of electrons to flow through a circuit if the current is 1.00 A?

Calculate the cathode potential (versus S.C.F.) needed to reduce cobalt(II) to 1.0μMin each of the following solutions. In each case, Co(s) is the product of the reaction. (Disregard any overpotential.)

(a) 0.10MHClO4

(b)0.10MC2O42-(Find the potential at which)[Co(C2O4)22-]=1.0μM

role="math" localid="1668354628300" Co(C2O4)22-+2c-Co(s)+2C2O42-E°=-0.474V

(c)0.10MFIJTA at pH 7.00(Find the potential at whichCo(HI)TA)2-=1.0μM.)

Consider the following electrolysis reactions.

Cathode:H2O(l)+e-12H2(g,1.0bar)+OH-(aq,0.10M)

Anode:Br-(aq,0.10M)12Br2(l)+e-

  1. Calculate the voltage needed to drive the net reaction if current is negligible.
  2. Suppose that the cell has a resistance of2.0Ω and a current of 100 mA. How much voltage is needed to overcome the cell resistance? This is the ohmic potential.
  3. Suppose that the anode reaction has an overpotential of 0.20 V and that the cathode overpotential is 0.40 V. What voltage is needed to overcome these effects combined with those of parts (a) and (b)?
  4. Suppose that concentration polarization occurs [OH-]s. at the cathode surface increases to 1.0 M and[Br-]s at the anode surface decreases to 0.010 M. What voltage is needed to overcome these effects combined with those of (b) and (c)?
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