Electrochemistry Questions and Answers

Answer: C

Correct answer: 10 per cent

The degree of dissociation is the ratio of the two conductances:

so nine tenths of the acetic acid is present as undissociated molecules - which is what makes it a weak acid, and what the conductance measurement establishes quantitatively.

Answer: D

Dividing 0.177 by 0.0591 gives three pH units, and a more positive potential means a higher hydrogen ion activity, so the pH is three units below seven. A reading below the buffer value would indicate an alkaline solution.

Answer: A

Correct answer: 0.059 V

For a concentration cell the standard potentials cancel and

Two decades of concentration difference, halved by the divalent charge. Forgetting the \( n \) of 2 gives 0.118 V, which would be the answer for a univalent ion.

Answer: B

The hydrogen electrode defines the scale but is awkward to use, calomel replaces it in practice, and the glass electrode is the indicator electrode for hydrogen ion activity. All three appear together in a pH measurement's history.

Answer: C

Correct answer: 1-Q, 2-P, 3-R

The three behave differently on dilution because they are normalised differently. Specific conductance, being per unit volume, falls as the ions are spread through more solvent. Equivalent conductance, referring to a fixed quantity of solute, rises as dissociation increases and interionic attraction weakens. And the degree of dissociation itself rises towards unity, complete dissociation being the limit at infinite dilution.

Answer: D

Correct answer: 4, 2, 1, 3

Calibration comes first and needs two points: a pH 7 buffer to set the offset, and a second buffer bracketing the expected sample to set the slope. Only then are the electrodes rinsed and immersed in the sample and the reading taken. Calibrating after measuring, or with one buffer only, leaves either the offset or the slope unverified.

Answer: A

Correct answer: 1 and 3 only are correct

Statements 1 and 3 are correct: oxidation is at the anode by definition in both cell types, and one faraday deposits one gram equivalent of anything. Statement 2 reverses the behaviour - specific conductance falls on dilution, since it is measured per unit volume and dilution reduces the ions in each cubic centimetre. It is the equivalent conductance that rises.

Answer: B

Correct answer: increasing the surface area of the electrode

Electrode potential is an intensive property fixed by the activities of the species and the temperature, so it does not depend on how much electrode there is - a large sheet and a small wire of the same metal in the same solution give the same potential. Area affects the current a cell can deliver, not its EMF. The other three all change the ion activity or the Nernst term and so do shift the potential.

Answer: C

Correct answer: a Daniell cell delivering current to a load

A Daniell cell is galvanic: its reaction is spontaneous and it supplies electrical energy rather than consuming it. The other three are electrolytic, forcing a non-spontaneous reaction with an external supply - anodising to build an oxide film, electroplating to deposit nickel, and electrorefining to dissolve impure copper at one electrode and deposit pure copper at the other.

Answer: D

Correct answer: polarisation and the internal resistance drop

Drawing current makes the cell irreversible: products accumulate at the electrodes and oppose the reaction, and the current through the internal resistance produces a drop that comes off the terminal voltage. The terminal voltage is therefore \( E - Ir \) less the polarisation, and it approaches the true EMF only as the current approaches zero. This is exactly why a potentiometer is used for accurate EMF measurement.