ElectrochemistryHard
Question
Choose the correct statement(s).
Options
A.At the anode, the species having minimum reduction potential is formed from the oxidation of corresponding oxidizable species.
B.In highly alkaline medium, the anodic process during the electrolytic process is
4OH- → O2 + 2H2O + 4e-.
4OH- → O2 + 2H2O + 4e-.
C.The standard potential of Cl- |AgCl| Ag half-cell is related to that of Ag+ | Ag through the expression
EoAg+|Ag = EoCl-|AgCl|Ag +
ln Ksp(AgCl)
EoAg+|Ag = EoCl-|AgCl|Ag +
D.Compounds of active metals (Zn, Na, Mg) are reducible by H2 whereas those of noble metals (Cu, Ag, Au) are not reducible.
Solution
In highly alkaline medium OH- ion get oxidise in preference to H2O
4OH- → O2 + 2H2O + 4e-
EoCl-/AgCl/Ag = EoAg+/Ag -
EoRed of active metal is lower than EoRed of hydrogen so not reduced by hydrogen.
4OH- → O2 + 2H2O + 4e-
EoCl-/AgCl/Ag = EoAg+/Ag -
EoRed of active metal is lower than EoRed of hydrogen so not reduced by hydrogen.
Create a free account to view solution
View Solution FreeMore Electrochemistry Questions
Given EoCr3+/Cr = - 0.72V, EoFe2+/Fe = - 0.42 V. The potential for the cell Cr | Cr3+ (0.1M)||Fe2+(0.01M)|Fe is...Four successive members of the first row transition elements are listed below with atomic numbers. Which one of them is ...The correct order of EoM2+/M values with negative sign for the four successive elements Cr, Mn, Fe and Co is...Assuming that hydrogen behaves as an ideal gas, what is the EMF of the cell at 25°C if P1 = 600 mm and P2 = 420 mm: Pt |...For the process: Cu2+ + 2e− → Cu; log[Cu2+] vs. Ered graph is shown in the figure, where OA = 0.34 V. The electrode pote...