Consider the titration of 100.0 mL of 0.010 0 M Ce4+ in 1 M HClO4 with 0.040 0 M Cu+ to give Ce3+ and Cu2+. Calculate the potential of the indicator electrode after adding 24.5 mL of Cu+.
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- The following cell was found to have a potential of —0.492 V: Ag|AgCl(sat’d)||HA(0.200 M),NaA(0.300 M)|H2(1.00 atm),Pt Calculate the dissociation constant of HA, neglecting the junction potential.Calculate the cell potential when 150.00 mL of 0.100 M Ag+ solution is titrated with 25.00 mL of 0.150 M Cl- solution. Silver wire indicator electrode and Ag-AgCl electrode are used as reference electrodes. E0 (Ag/AgCl) = 0.197 V , E0 Ag+/Ag = 0.799 V, Kçç (AgCl) = 1.8 x10-8By how many volts will the potential of an ideal Mg2+ ion selective electrode (ISE) change if the electrode is removed from a 1.0 x 10-4 M solution of MgCl2 and placed in 1.0 x 10-3 M MgCl2? 28 mL of 0.1M Ce4+ is mixed with 50 mL of 0.05M Fe2+. The solution potential is monitored with a Pt indicator electrode and an Ag/AgCl reference electrode. Calculate the potential of the Pt electrode and the cell potential. An ion selective electrode for NO2- has the following values of Log k for the listed interferents. Interferent Logk F- -3.1 SO42- -4.1 I- -1.2 NO3- -3.3 Which ion…
- The selectivity coefficient of the fluoride ion-selective electrode is kF-,OH- = 0,15. How does the electrode potential of 1.4 x 10-5 M F- ions at pH 5.5 change to pH 11.0 ? ( aF- = [F-] dilute solution is considered).Calculate the cell potential when 50.00 mL of 0.100 M Ag+ solution is titrated with 50.00 mL of 0.150 M Cl- solution. Silver wire indicator electrode and Ag-AgCl electrode are used as reference electrodes. E0 (Ag/AgCl) = 0.197 V , E0 Ag+/Ag = 0.799 V, Kçç (AgCl) = 1.8 x10-8 A. 0.704v B. 0.507V C.0.238v D. 0.435VA 0.2g sample of toothpaste containing fluoride was treated with 50 cm3 of a suitable buffer solution and diluted to 100 cm3. Using a fluoride ion-selective electrode, a 25.00 cm3 aliquot of this solution gave cell potentials of –155.3 mV before and –176.2 mV after spiking with 0.1 cm3 of a 0.5 mg/cm3 fluoride standard. Calculate the pF– corresponding to each cell potential and the percentage by weight of fluoride in the toothpaste.
- Briefly discuss the following: Different types of membrane indicator electrode How a pH glass electrode works Errors encountered in the use of pH electrode Contribution of other variables in the glass indicator potentialA 40.00-mL aliquot of 0.05000 M HNO2 is diluted to 75.00 mL and titrated with 0.0800 M Ce4+ . The pH of the solution is maintained at 1.00 throughout the titration; the formal potential of the cerium system is 1.44 V. Calculate the potential of the indicator electrode with respect to a saturated calomel reference electrode after the addition of 5.00 mL of cerium (IV). (Use a MW value in 4 decimal places)A titration of 50.0 mL of 0.10 M Sn2+ with 0.2 M Fe3+ in 1 M HCl to give Fe2+ and Sn4+ using Pt and calomel electrodes. Assuming the standard potential for Sn2+/Sn4+ = 0.139V ; calomel, 0.241 V; Fe2+/Fe3+ = 0.732 V. This my cathode reaction: Fe3+ (aq) + 1 electron <-> Fe2+(s) and anode reaction: Sn2+(s) <-> Sn4+(aq) + 2 electrons. How will I write my Nernst equation for the cell voltage for the oxidation reaction?
