"Redox Reactions and Electrochemistry" — Practice

Question 1

Find the oxidation number of sulphur in $\mathrm{H_2SO_4}$, $\mathrm{S_2O_3^{2-}}$, $\mathrm{S_4O_6^{2-}}$ and $\mathrm{S_8}$.

Accepted Answer

In $\mathrm{H_2SO_4}$, hydrogen gives $+2$ and oxygen $-8$, so sulphur is $+6$. In thiosulphate, oxygen gives $-6$ against a total of $-2$, so the two sulphurs sum to $+4$ and average $+2$. In tetrathionate, oxygen gives $-12$ against $-2$, so four sulphurs sum to $+10$ and average $+2.5$. In $\mathrm{S_8}$ it is zero, being a free element. The fractional value is an average over atoms the formula cannot distinguish, not an error.

Question 2

Balance the reaction of dichromate with iodide in acidic medium by the half-reaction method.

Accepted Answer

Chromium goes from $+6$ to $+3$, so the reduction half is $\mathrm{Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O}$, with water balancing oxygen and $\mathrm{H^+}$ balancing hydrogen. The oxidation half is $\mathrm{2I^- \rightarrow I_2 + 2e^-}$, tripled to match six electrons. Adding gives $\mathrm{Cr_2O_7^{2-} + 6I^- + 14H^+ \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O}$. The charge check confirms it: left is $-2-6+14 = +6$ and right is $+6$.

Question 3

Explain why chlorine disproportionates in cold alkali but fluorine does not disproportionate at all.

Accepted Answer

Disproportionation needs an element in an intermediate oxidation state with both a higher and a lower state available. Chlorine at 0 can go up to $+1$ and down to $-1$, giving hypochlorite and chloride in cold dilute alkali. Fluorine is the most electronegative element and has no positive oxidation state whatsoever, so from 0 it can only go down to $-1$. With no upward route available, one half of the requirement is missing and disproportionation is impossible in principle rather than merely unfavourable.

Question 4

Given standard reduction potentials of $-0.76$ V for zinc and $+0.80$ V for silver, write the cell notation and find $E^\circ_{cell}$ and $\Delta G^\circ$.

Accepted Answer

Silver has the more positive reduction potential, so it is reduced and is the cathode, making zinc the anode. The notation is $\mathrm{Zn(s)\ |\ Zn^{2+}(aq)\ ||\ Ag^+(aq)\ |\ Ag(s)}$ and $E^\circ_{cell} = 0.80 - (-0.76) = 1.56$ V. The balanced reaction is $\mathrm{Zn + 2Ag^+ ightarrow Zn^{2+} + 2Ag}$, so $n = 2$ and $\Delta G^\circ = -nFE^\circ = -2 	imes 96500 	imes 1.56 = -301\ \mathrm{kJ\ mol^{-1}}$. Note that $n$ is 2 even though the silver half-reaction transfers one electron.

Question 5

For the Daniell cell with $E^\circ = 1.10$ V, find the potential when $[\mathrm{Zn^{2+}}] = 0.10$ M and $[\mathrm{Cu^{2+}}] = 0.010$ M at 298 K.

Accepted Answer

The cell reaction is $\mathrm{Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu}$, with the solids omitted, so $Q = [\mathrm{Zn^{2+}}]/[\mathrm{Cu^{2+}}] = 0.10/0.010 = 10$. Then $E = 1.10 - (0.0591/2)\log 10 = 1.10 - 0.0296 = 1.070$ V. The potential falls only slightly for a tenfold concentration change, because the logarithm compresses it and the division by $n$ halves it again.

Question 6

A cell has $E^\circ = 0.46$ V with two electrons transferred. Calculate its equilibrium constant at 298 K.

Accepted Answer

Setting $E = 0$ and $Q = K$ in the Nernst equation gives $\log K = nE^\circ/0.0591 = (2 	imes 0.46)/0.0591 = 15.6$, so $K = 4 	imes 10^{15}$. A potential under half a volt corresponds to an equilibrium constant of $10^{15}$, which is the logarithmic relation at work: every extra 0.0591 V per electron multiplies $K$ by ten.

Question 7

Explain how a concentration cell produces a potential when both electrodes are made of the same metal, and compute the potential for $\mathrm{Cu\ |\ Cu^{2+}(0.0010\ M)\ ||\ Cu^{2+}(0.10\ M)\ |\ Cu}$.

Accepted Answer

Both electrodes are identical, so $E^\circ = 0$ exactly and the entire potential comes from the logarithmic term. The cell moves copper ions from the concentrated compartment to the dilute one, so $Q = 0.0010/0.10 = 0.010$ and $E = -(0.0591/2)\log(0.010) = +0.0591$ V. Copper dissolves on the dilute side and plates on the concentrated side until the two concentrations meet, at which point $Q = 1$, the log vanishes and the cell is dead. It is diffusion doing electrical work.

Question 8

A current of 1.5 A is passed through molten aluminium oxide for 2 hours. Calculate the mass of aluminium deposited, taking its molar mass as 27.

Accepted Answer

The charge is $Q = It = 1.5 	imes 7200 = 10800$ C, so the electrons amount to $10800/96500 = 0.1119$ mol. Aluminium is deposited as $\mathrm{Al^{3+}}$, needing three electrons per atom, so the aluminium is $0.1119/3 = 0.0373$ mol and the mass is $0.0373 	imes 27 = 1.01$ g. The division by 3 is the step most often skipped, and omitting it triples the answer.

Question 9

Predict the products of electrolysing aqueous copper sulphate with platinum electrodes, and then with copper electrodes, and explain the difference.

Accepted Answer

With platinum, copper deposits at the cathode because $\mathrm{Cu^{2+}}$ is easier to reduce than water, and oxygen is released at the anode from the oxidation of water, since sulphate is extremely resistant to oxidation. With a copper anode the cathode is unchanged, but the anode dissolves as $\mathrm{Cu \rightarrow Cu^{2+} + 2e^-}$ instead of releasing oxygen, because oxidising the electrode itself needs a lower potential than oxidising water. Copper transfers from anode to cathode, and that is exactly how copper is refined to 99.99 per cent purity.

Question 10

Explain why conductivity decreases on dilution while molar conductivity increases.

Accepted Answer

Conductivity $\kappa$ is the conductance of a unit cube of solution, so it counts the ions present in a fixed volume, and dilution reduces that number. Molar conductivity is $\Lambda_m = 1000\kappa/c$, which asks instead how well each mole conducts, and dilution increases it because the ions are further apart and interfere with one another less, and for a weak electrolyte because more of it has ionised. One question is about the beaker and the other about the mole, so there is no contradiction.

Question 11

Given limiting molar conductivities of 91.0 for sodium acetate, 426.2 for hydrochloric acid and 126.5 for sodium chloride, all in S cm$^2$ mol$^{-1}$, find that of acetic acid, then find $K_a$ if a 0.00103 M solution has $\Lambda_m = 49.1$.

Accepted Answer

By Kohlrausch's law of independent migration, adding sodium acetate to hydrochloric acid and subtracting sodium chloride cancels the sodium and chloride contributions exactly, leaving acetic acid: $\Lambda_m^\circ = 91.0 + 426.2 - 126.5 = 390.7$ S cm$^2$ mol$^{-1}$. Then $\alpha = 49.1/390.7 = 0.126$, and Ostwald's law gives $K_a = c\alpha^2/(1-\alpha) = (0.00103 	imes 0.0158)/0.874 = 1.86 	imes 10^{-5}$. This agrees with the value from pH measurement, which is an independent confirmation that the conductivity route is sound.

Question 12

Why does galvanising protect iron even where the zinc coating has been scratched through?

Accepted Answer

Because zinc protects electrochemically rather than by acting as a barrier. Zinc has a more negative reduction potential than iron, $-0.76$ V against $-0.44$ V, so wherever both metals are exposed to the same electrolyte the zinc is oxidised preferentially and the iron becomes the cathode, where reduction rather than corrosion occurs. This is cathodic protection, and it continues to work across a scratch because current can still flow through the moisture film. Paint, which protects only by exclusion, fails the moment it is breached.

Practice — "Redox Reactions and Electrochemistry"

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