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Electrochemistry Questions and Answers

Answer: A

Correct answer: anode and the pure metal deposits at the cathode

The impure copper is the anode and dissolves; the copper ions are reduced onto a pure cathode. The purification comes from the potentials: more noble impurities such as gold and silver do not dissolve and fall as anode mud, from which they are recovered, while more active ones such as zinc and iron dissolve but are not deposited at the cathode potential used and remain in solution.

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Answer: B

Correct answer: anode in an acid electrolyte

Anodising makes the aluminium the anode, so oxygen liberated there oxidises the surface into a thick, hard, adherent film of aluminium oxide - far thicker than the natural film and porous enough to be dyed and then sealed. Sulphuric acid is the usual electrolyte. Making the metal the cathode would deposit something onto it rather than converting its own surface.

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43.

For the cell Zn | Zn\( ^{2+} \) || Cu\( ^{2+} \) | Cu with standard potentials \( -0.76 \) V and \( +0.34 \) V, the standard EMF is

Answer: C

Correct answer: 1.10 V

Copper is the cathode and zinc the anode, so

\[ E^\circ = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-0.76) = 1.10 \ \mathrm{V} \]

The positive result confirms that the cell as written is spontaneous. Adding the two potentials instead would give \( -0.42 \), and subtracting the wrong way round gives \( -1.10 \) - both of the distractors offered.

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44.

For that cell the free energy change, with \( F = 96\,500 \) C, is about

Answer: D

Correct answer: \( -212 \) kJ

The free energy change is

\[ \Delta G^\circ = -nFE^\circ = -2 \times 96\,500 \times 1.10 = -212\,300 \ \mathrm{J} \approx -212 \ \mathrm{kJ} \]

with \( n = 2 \) because two electrons pass per zinc atom. The negative sign confirms spontaneity, and the magnitude is the maximum electrical work the cell can deliver - obtainable only if the current is drawn reversibly.

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45.

A copper electrode in 0.01 M Cu\( ^{2+} \), with \( E^\circ = 0.34 \) V, has a potential of about

Answer: A

Correct answer: 0.281 V

Applying the Nernst equation with \( n = 2 \),

\[ E = 0.34 + \frac{0.059}{2}\log(0.01) = 0.34 + 0.0295(-2) = 0.34 - 0.059 = 0.281 \ \mathrm{V} \]

Diluting a hundredfold has lowered the potential by 59 millivolts, since two decades of concentration at 29.5 mV each. Adding rather than subtracting gives 0.399, which would mean dilution made the electrode more noble.

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46.

The mass of copper deposited by 2 A flowing for 30 minutes, taking the equivalent mass as 31.75, is about

Answer: B

Correct answer: 1.18 g

Applying Faraday’s laws,

\[ m = \frac{E \times I \times t}{F} = \frac{31.75 \times 2 \times 1800}{96\,500} = \frac{114\,300}{96\,500} \approx 1.18 \ \mathrm{g} \]

The time must be in seconds, so the thirty minutes becomes 1800. Using the atomic mass of 63.5 instead of the equivalent mass would double the answer, which is the error the 2.37 g distractor represents.

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47.

The charge needed to deposit 108 g of silver is

Answer: C

Correct answer: 96 500 C

Silver is univalent, so its equivalent mass equals its atomic mass of 108, and 108 g is therefore exactly one gram equivalent. One gram equivalent requires one faraday, which is 96,500 coulombs. This is the classical silver coulometer, and it is how the faraday was originally determined.

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48.

A cell whose observed conductance is 0.02 S is filled with a solution of specific conductance 0.01 S/cm. Its cell constant is

Answer: D

Correct answer: 0.5 cm\( ^{-1} \)

The cell constant relates the measured conductance to the specific conductance:

\[ \kappa = G \times \text{cell constant} \quad \Rightarrow \quad \text{cell constant} = \frac{0.01}{0.02} = 0.5 \ \mathrm{cm^{-1}} \]

It is the ratio of the electrode separation to their area and is a property of the cell alone, determined once with a standard potassium chloride solution and used thereafter for every measurement in that cell.

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49.

A 0.1 N solution of specific conductance 0.0129 S/cm has an equivalent conductance of

Answer: A

Correct answer: 129 S cm\( ^2 \)/eq

Applying the relation,

\[ \Lambda = \frac{1000\kappa}{N} = \frac{1000 \times 0.0129}{0.1} = 129 \ \mathrm{S\,cm^2/eq} \]

The 1000 converts the litre of the normality to the cubic centimetres of the specific conductance, and forgetting it is the commonest error - giving 0.129 instead.

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50.

Given \( \Lambda^\circ \) of HCl as 426, of CH\( _3 \)COONa as 91 and of NaCl as 126, Kohlrausch's law gives \( \Lambda^\circ \) of acetic acid as

Answer: B

Correct answer: 391

Kohlrausch’s law of independent migration lets the ions be added and subtracted:

\[ \Lambda^\circ_{\mathrm{HAc}} = \Lambda^\circ_{\mathrm{HCl}} + \Lambda^\circ_{\mathrm{CH_3COONa}} - \Lambda^\circ_{\mathrm{NaCl}} \]\[ = 426 + 91 - 126 = 391 \]

The sodium and chloride ions cancel, leaving hydrogen and acetate. This is the only route to a weak electrolyte’s limiting conductance, since it cannot be reached by extrapolation.

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