BEGINNING
Reasons for choosing this topic
Coulometry is a quantitative electrochemical analysis technique that involves exhaustive electrolysis of the analyte, ensuring complete oxidation or reduction at the working electrode There are two main types of coulometry: controlled-potential coulometry, which maintains a constant potential, and controlled-current coulometry, which applies a constant current through the electrochemical cell This method is highly valuable in modern analytical chemistry for its accuracy and reliability.
Therefore, I have carried out the topic: " EXERCISES ABOUT COULOMETRIC METHODS” with the desire to provide a useful reference to teachers and students.
This topic serves as a valuable resource for students seeking to enhance their chemistry knowledge through reference exercises and foreign textbooks It aims to provide useful insights for readers interested in studying chemistry effectively.
Research Purposes
- Promoting my ability to explore, learn myself From that, building good exercises for students.
- Synthesizing exercise-system to improve knowledge and skill solving it for students at the university.
- Learn more deeply about analytical exercises, especially electrolytic analysis methods.
- Approaching the method of solving electrochemical analysis exercises of some countries in the world.
Research tasks
Search and synthesized chemically some exercises electrochemical analysis by electrolysis of our country and other countries.
Objects of study
Exercises about electrochemical methods of analysis :
Electrolysis (electrogravimetric and coulometric methods)
Methods for study
Based on the knowledge learned, conduct searches, research, analyze, compare and synthesize sources: textbooks of our country and other countries, the examination questions, etc.
To analyze a brass alloy, a 0.442-g sample is dissolved in acid and diluted to a volume of 500 mL Electrolysis of a 10.00-mL aliquot at –0.3 V versus a saturated calomel electrode (SCE) successfully reduces Cu²⁺ to Cu, necessitating a total charge of 16.11 C When the potential is adjusted to –0.6 V versus SCE, an additional charge of 0.442 C is required to complete the reduction process.
Pb 2+ to Pb Report the %w/w Cu and Pb in the alloy.
The reduction of Cu 2+ to Cu requires two electrons per mole of Cu (n = 2) We calculate the moles and the grams of Cu in the portion of sample being analyzed.
63.55g Cu 8.348 10 mol Cu 5.301 10 g Cu mol Cu
This is the Cu from a 10.00 mL portion of a 500.0 mL sample; thus, the %w/w copper in the original sample of brass is
� For lead, we follow the same process, thus
207.2g Pb 2.19 10 mol Pb 4.53 10 g Pb mol Pb 500.0 mL 4.53 10 g Pb
The purity of a Na2S2O3 sample was assessed through coulometric redox titration, utilizing I- as a mediator and I3- as the titrant A precise 0.1342 g sample was dissolved in a 100-mL volumetric flask with distilled water A 10.00-mL aliquot was placed in an electrochemical cell, combined with 25 mL of 1M KI, 75 mL of a pH 7.0 phosphate buffer, and starch indicator solution The electrolysis process, conducted at a constant current of 36.45 mA, took 221.8 seconds to reach the endpoint indicated by the starch The resulting data will be used to determine the sample's purity.
The coulometric titration of S O 2 3 2 with I 3 is :
Oxidizing S O 2 3 2 to S O 4 6 2 requires one electron per S O 2 3 2 ( n=1)
Solving for the grams of Na2S2O3 gives
The analysis indicates that a 10.00-mL portion of the 100-mL sample contains 0.1325 g of Na2S2O3 Consequently, the purity of the original sample can be determined based on this measurement.
Note that the calculation is worked as if is S O 2 3 2 oxidized directly at the working electrode instead of in solution.
Suppose we wish to electrolyze I to I 3 in a 0.10 M KI solution containing 3.0 ×10 -5 M I 3 at pH 10.00 with fixed at 1.00 bar
(a) Find the cell voltage if no current is flowing
Electrolysis increases the concentration of I3- to 3.0 × 10^-4 M while leaving other concentrations unchanged With a cell resistance of 2.0 Ω and a current of 63 mA, the cathode overpotential is measured at 0.382 V, and the anode overpotential is 0.025 V.
V What voltage is needed to drive the reaction?
(a) The open-circuit voltage is E(cathode)- E(anode):
We would have to apply 1.081 V to force the reaction to occur.
(b) Now E(cathode) is unchanged but E(anode) changes because [I ] 3 s is different from [I ] 3 in bulk solution.
Instead of 1.081 V, we need to apply 1.644 V to drive the reaction.
To dissolve 0.125 g of an alloy sample completely, mix it with 100 ml of H2SO4, which facilitates the transfer of iron into the solution as Fe 2+ Once dissolved, transfer the entire solution into the electrolyte flask.
Add 50,00 ml of Ce (III) 0,0400M solution Electrolysis with current intensity I 0.0500A using Pt electrode The titration stop is achieved after 511.2 seconds. Calculate the% Fe content in the alloy.
When electrolysis occurs oxidation of Fe 2+ at anot:
At the end of electrolysis, Ce 3+ oxidation will occur :
And when Fe 2+ is completely finished, a small amount of Ce 4+ electrolyte will soar and we can end electrolysis.
To analyze the antimony content in Sb2O3 from the sample, we start by weighing 15 grams of the sample and acidifying it with hydrochloric acid (HCl) to create a solution Next, we prepare an excess of sodium bromide (NaBr) solution and proceed to quantify the antimony through titration, applying a constant current of 2.4125 mA for 12 minutes during the electrolysis process The reaction that occurs involves the reduction of antimony ions, and to calculate the Sb2O3 content in parts per million (ppm), we use the molecular weight of Sb2O3, which is 291.5 g/mol.
The reaction occurs when sample dissolution:
The reaction occurs when titration of electric quantity:
Applying Faraday's law, we have:
� From (1) and (2) we have the mass of Sb2O3 in the sample:
The electrolysis of Co²⁺ (0.100 M) and Cd²⁺ (0.0500 M) solutions involves determining the concentration of Co²⁺ when Cd²⁺ begins electrolysis The reaction for Cd²⁺ electrolysis is represented as Cd²⁺ + 2e⁻ → Cd Additionally, calculations are required to find the cathode potential needed to reduce Co²⁺ ion concentration to 1.0 × 10⁻⁶ M.
Solution: a Calculate Co concentration so that Cd 2+ starts electrolysis.
The reactions in the cathode:
Let Cd start electrolysis, then the cathode
So we can not completely separate Co from Cd b Calculation of the cathode to reduce Co 2+ concentration to 1.0 �10 -6 M
So when the cathode potential is -0.457V, the concentration of Co 2+ is only 1.0 �
Co 2+ is considered completely separated from the electrolyte solution.
To calculate the cathode potential required for the complete separation of copper (Cu) from a mixture of 0.010M CuSO4 and 0.010M ZnSO4 using two platinum electrodes in 1M H2SO4, one must consider the electrochemical properties of both ions During the electrolysis process, as zinc ions (Zn²⁺) begin to deposit, the concentration of remaining copper ions (Cu²⁺) decreases It is essential to determine the concentration of Cu²⁺ left in the solution at the onset of zinc electrolysis to understand the efficiency of the separation process.
Cu Cu Cu Cu Cu V
Zn Zn Zn Zn Zn V
So, at the cathode Cu 2+ electrolysis first
To separate Cu 2+ from solution, the concentration of Cu 2+ remains less than 10 -6 M
Cu Cu Cu Cu Cu V
To Cu separated completely, we have : E Zn 2 Zn E