Electrochemistry is a high-yield chapter for both CBSE and competitive exams because it connects oxidation–reduction concepts to real measurable quantities like cell emf, electrode potentials, and concentration effects. Mastery of the Nernst equation, spontaneity criteria, and electrode processes (galvanic vs electrolytic) is essential for solving both direct numerical problems and conceptual multiple-choice questions efficiently.
15
Minutes
10
Questions
1 / -0
Marking
Q1. A galvanic cell at is constructed as . Given and , calculate the cell emf at using the Nernst equation (use for ).
Q2. A concentration cell is formed by two hydrogen electrodes at : . Using the Nernst equation, compute the emf (magnitude) and identify which electrode (the side or the side) will act as the cathode.
; the side is the cathode.
; the side is the cathode.
; the side is the cathode.
; the side is the cathode.
Q3. An electrolytic deposition is attempted from an aqueous solution containing , and . Using the Nernst equation at and given , and , which species will be reduced (deposit) first on the cathode?
will deposit first.
will deposit first.
will evolve first (hydrogen reduction).
and will deposit simultaneously.
Q4. Two one-electron half-cells have standard reduction potentials for and for . A galvanic cell assembled as at exhibits an experimental emf of approximately (negative). Which of the following best explains this observation?
A large liquid junction potential at the salt bridge reversed the sign of the measured emf.
The electrodes were accidentally connected in reverse during measurement, producing a negative reading.
The given values are not comparable because the half-reactions involve different numbers of electrons.
Although under standard conditions, the very large reaction quotient makes negative at , so the cell is non-spontaneous under these concentrations.
Q5. Assertion (A): For a reversible concentration cell of the type , the emf is given by and is independent of the electrode surface area. Reason (R): The Nernst equation arises from equality of chemical potentials and depends only on activities (concentrations), not on geometric factors such as surface area; electrode area affects kinetics (current) but not the equilibrium emf.
Both A and R are true but R is NOT the correct explanation of A.
Both A and R are true and R is the correct explanation of A.
A is true but R is false.
A is false but R is true.
Q6. Calculate the emf at 298 K of the cell . Given and .
Q7. A Daniell cell at 298 K is . The cathode compartment contains of the Cu solution. By controlled electrolysis of Cu is removed from the cathode compartment (assume negligible volume change). Using and , the new emf of the cell at 298 K is closest to:
Q8. Consider the cell at 298 K. The measured emf is . Given , calculate the pH of the hydrogen electrode compartment.
Q9. Assertion (A): During electrolysis of concentrated aqueous NaCl on an inert anode, is evolved at the anode rather than , even though standard potentials under some conditions suggest oxygen evolution could be thermodynamically competitive.
Reason (R): Large overpotential for oxygen evolution on common anode materials (and the high Cl⁻ concentration shifting the chloride oxidation potential by the Nernst equation) makes chloride oxidation kinetically and electrochemically favored.
Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.
Q10. A cell is made with and half-cells at 298 K. Initial concentrations are , and . Metal electrodes are and . Given and , when the cell reaches equilibrium () the ratio will be approximately: