Electrochemistry Questions and Answers

Answer: C

Correct answer: falls to a minimum and then rises

Before the end point, highly mobile hydrogen ions are replaced by much less mobile sodium ions, so the conductance falls. After the end point, excess sodium hydroxide adds mobile hydroxide ions and the conductance rises again. The two straight portions are extrapolated and their intersection gives the end point - which is why no indicator is needed and why the method works in coloured solutions.

Answer: D

Correct answer: coloured or turbid

A conductometric titration locates the end point from the intersection of two straight lines drawn through measurements taken away from it, so nothing has to be seen and no indicator is needed. That makes it valuable exactly where a visual end point fails: coloured, turbid or very dilute solutions, and titrations of weak acids with weak bases where no indicator changes sharply.

Answer: A

Correct answer: sharp change in cell potential

An indicator electrode responds to the activity of the species being titrated, and at the end point that activity changes by orders of magnitude within a fraction of a millilitre, so the measured potential jumps. The end point is taken at the steepest part of the curve, most precisely from the maximum of the first derivative. Like the conductometric method it needs no visual indicator.

Answer: B

Correct answer: quantity of electricity passed

The first law states \( m \propto Q = It \), so the mass liberated depends on the current and the time and on nothing else - not on the electrode area, the applied voltage or the concentration, provided the electrolysis proceeds. It follows directly from each ion requiring a fixed number of electrons to be discharged.

Answer: C

Correct answer: equivalent masses

The second law states that the same quantity of electricity liberates masses in the ratio of the substances’ equivalent masses - atomic mass divided by the change in oxidation number. So one faraday deposits 108 g of silver, which is univalent, but only 31.75 g of copper from the divalent state, since each copper ion needs two electrons.

Answer: D

Correct answer: 96 500 coulombs

One faraday is the charge on a mole of electrons:

and it deposits one gram equivalent of any substance. It is the conversion between the electrical and the chemical accounts, and every electrolysis calculation passes through it.

Answer: A

Correct answer: continuous electrolysis begins

Below the decomposition potential only a small transient current flows while the electrodes polarise, and it then falls almost to nothing. Above it the products are discharged continuously and the current rises steadily with voltage. The value exceeds the reversible EMF of the reaction by the overvoltages at the two electrodes and by the resistance drop of the electrolyte.

Answer: B

Activation of the electrode process, and the accumulation or depletion of species at its surface, both demand extra driving force. Hydrogen's overvoltage on mercury is large, which is what makes some electrolyses possible at all.

Answer: C

Correct answer: sodium to be deposited from aqueous solution

Hydrogen has an overvoltage of nearly a volt on mercury, so its discharge is suppressed and sodium is deposited instead, dissolving into the mercury as an amalgam - which is exactly how the mercury cell in the chlor-alkali industry works. Without that overvoltage, hydrogen would be liberated preferentially and no sodium would be obtained from an aqueous solution at all.

Answer: D

Correct answer: cathode

The article is made the cathode so that metal ions in the bath are reduced onto its surface. The anode is generally the plating metal itself, dissolving to replenish the solution as it is consumed. Everything about the quality of the deposit - adhesion, thickness, uniformity - depends on current density and on the surface preparation of that cathode.