(a) Explain the difference between charging current and faradaic current.

(b) What is the purpose of waitingafter a voltage pulse before measuring current in sampled current voltammetry?

(c) Why is square wave voltammetry more sensitive than sampled

current voltammetry?

Short Answer

Expert verified

a. The charging current and faradic current and its differences are explained

b. The advantage of gap of one second before measuring current in voltammetry.

c. The sensitivity of square wave voltammetry.

Step by step solution

01

Define the charging current and faradic current and its differences.

Faradic current:

- In working electrode, reduction or oxidation of analyte leads to faradic current.

- In Faradic current, due to higher reaction rate the analyte does not diffuse fast to electrode which leads to decay of faradic current.

Charging current:

- In charging current, cations move towards the electrode in solution and anions move away from the electrode in solution.

- The movement of electrons and ions are known as charging current.

- Charging current also known as condenser current or capacitor current.At any point

02

a) To explain: The charging current and faradic current and its differences.

Faradic current:

- In working electrode, reduction or oxidation of analyte leads to faradic current.

- In Faradic current, due to higher reaction rate the analyte does not diffuse fast to electrode which leads to decay of faradic current.

Charging current:

- In charging current, cations move towards the electrode in solution and anions move away from the electrode in solution.

- The movement of electrons and ions are known as charging current.

- Charging current also known as condenser current or capacitor current.

03

b) The advantage of gap of one second before measuring current in voltammetry.

Decay of charging current is faster than faradic current. The time interval of one second after potential leads to decay of charging current to zero but still there is considerably faradic current. Now the ratio of faradic current to charging current is greater than earlier. But if the time gap is more, both current signals become very less to measure.

04

c) The sensitivity of square wave voltammetry.

- Cathodic pulse is followed by anodic pulse and the difference between them is the signal in square wave voltammetry.

- Each cathodic pulse is oxidized by anodic pulse which reloads the surface of electrode by electro active species.

- Thus, at electrode surface the analyte concentration is more in square wave voltammetry.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What is a Clark electrode, and how does it work?

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?

From the two standard additions of 50 pm Fe(III) in the figure, find the concentration of Fe(III) in the seawater. Estimate where the baseline should be drawn for each trace and measure the peak height from the baseline. Consider the volume to be constant for all three solutions.

Ti3+ is to be generated in 0.10MHClO4 for coulometric reduction of azobenzene.

TiO2++2H++eTi3++H2OE0=0.100V4Ti3++C6H5N=NC6H5+4H2O2C6H5NH2+4TiO2++4H+

At the counter electrode, water is oxidized, and \(\mathrm{O}_{2}\) is liberated at a pressure of \(0.20\) bar. Both electrodes are made of smooth Pt, and each has a total surface area of 1.00cm2. The rate of reduction of the azobenzene is 25.9nmol/s , and the resistance of the solution between the generator electrodes is 52.4Ω.

  1. Calculate the current density (A/m2)at the electrode surface. Use Table 17-1 to estimate the overpotential for O2liberation.
  2. Calculate the cathode potential (versus S.H.E.) assuming that role="math" localid="1668356673323" [TiO2+surface]=[TiO2+]bulk=0.050Mand [Ti3+]surface=0.10M.
  3. Calculate the anode potential (versus S.H.E.).
  4. What should the applied voltage be?

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)?
See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free