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Chapter 17: Q1TY (page 398)

A monolayer (single layer of atoms) of Cuon the (100) crystal face shown in the margin has1.53×1015atoms/cm2=2.54×10-9mol/cm2. What current can deposit one layer of Cuatoms on1cm2in 1 s?

Short Answer

Expert verified

The current required to form a monolayer of copper atoms is 0.4.90mA.

Step by step solution

01

Define electric circuit

At any point in an electric circuit the charge is the product of current I and time t.

q = I.t

The unit of q is Coulombs, I is Ampere and t is seconds

q=molesofelectrons×Faraday'sconstant

02

Determine the current required to form a monolayer of copper atoms.

Number of electrons required to deposit copper is two. Then number of moles of copper is

Moles of copper=2×2.54×10-9

=5.08×10-9mol

q=molesofelectrons×FaradayconstantIt=molesofelectrons×Faraday'sconstantI=molesofelectrons×FaradayconstanttI=(5.08×10-8mol×96485C/mol)1s

I=0.490mA

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Most popular questions from this chapter

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)?

(a) which voltage,V1orV2in the diagram is constant in controlled-potential electrolysis? Which are the working, auxiliary, and reference electrodes in the diagram?

(b) Explain how the Luggin capillary in Figure 17-4 measures the electric potential at the opening of the capillary.

For a rotating disk electrode operating at sufficiently great potential, the redox reaction rate is governed by the rate at which analyte diffuses through the diffusion layer to the electrode (Figure 17-15b). The thickness of the diffusion layer is

δ=1.61D1/3V1/6ω-1/2

whereis the diffusion coefficient of reactant (m2/s),vis the kinematic viscosity of the liquid

(=viscosity/debsity=(m2/s),v and ωis the rotation rate (radians/s) of the electrode. There are 2π radians in a circle. The current densityis localid="1655441451764" (A/m2)is

localid="1655441445229" Currentdebsity=0.62nFD2/3v-1/6ω1/2C0

where nis the number of electrons in the half-reaction, Fis the Faraday constant, and localid="1655441459070" C0is the concentration of the electroactive species in bulk solution localid="1655441466748" (mol/m3,notmol/L) Consider the oxidation oflocalid="1655441474339" Fe(CN)64-in a solution of localid="1655441479067" 10.0mMK3Fe(CN)6+50.0mMK4Fe(CN)6at +0.90V(versus S.C.E.) at a rotation speed oflocalid="1655441490849" role="math" 2.00×103revolutions per minute. 27The diffusion coefficient oflocalid="1655441497131" Fe(CN)64-is2.5×10-9m2/sand the kinematic viscosity islocalid="1655441503345" 1.1×10-5m2/sCalculate the thickness of the diffusion layer and the current density. If you are careful, the current density should look like the value in Figure 17-16b.

The drug Librium gives a polarographic wave withE1/2=-0.265V(versus S.C.E.) in 0.05MH2SO4. A 50.0 - Mlsample containing Librium gave a wave height of0.37μA. When 2.00mLof 3.00mMLibrium in0.05MH2SO4were added to the sample, the wave height increased to0.80μA. Find the molarity of Librium in the unknown.

The cyclic voltammogram of the antibiotic chloramphenicol (abbreviated) is shown here. The first cathodic scan goes from 0 to -1.0 V. The first cathodic wave, , is from the reaction RNO2+4e-+4H+→RNHOH+H2O. Peak B in the reverse anodic scan could be assigned to RNHOH→RNO+2H++2e-. In the second cathodic scan from +0.9 to -0.4 V, the new peak C appears. Write a reaction for peak C and explain why peak C was not seen in the initial scan.


Cyclic voltammogram of 3.7 ×10-4 chloramphenicol in 0.1 M acetate buffer, pH 4.62. The voltage of the carbon paste working electrode was scanned at a rate of 350 mV/s. [Data from P. T. Kissinger and W. R. Heineman, “Cyclic Voltammetry,” J. Chem. Ed. 1983, 60, 702.]
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