"Redox Reactions and Electrochemistry" — Practice

Question 1

Balance the reaction of dichromate with iron(II) in acidic solution, and state the change in oxidation number of each element involved.

Accepted Answer

Reduction: $	ext{Cr}_2	ext{O}_7^{2-}+14	ext{H}^++6e^-ightarrow2	ext{Cr}^{3+}+7	ext{H}_2	ext{O}$

Oxidation: $	ext{Fe}^{2+}ightarrow	ext{Fe}^{3+}+e^-$, multiplied by six.

$	ext{Cr}_2	ext{O}_7^{2-}+6	ext{Fe}^{2+}+14	ext{H}^+ightarrow2	ext{Cr}^{3+}+6	ext{Fe}^{3+}+7	ext{H}_2	ext{O}$

Charge check: $-2+12+14=+24$ on the left and $+6+18=+24$ on the right.

Question 2

Chlorine reacts with cold dilute sodium hydroxide. Write the balanced equation and explain why this is a disproportionation.

Accepted Answer

$	ext{Cl}_2+2	ext{NaOH}ightarrow	ext{NaCl}+	ext{NaOCl}+	ext{H}_2	ext{O}$

Chlorine begins at oxidation number zero.

In sodium chloride it is $-1$, so it has been **reduced**; in sodium hypochlorite it is $+1$, so it has been **oxidised**.

The same element is both oxidised and reduced in one reaction, which is the definition of disproportionation. It is possible only because zero is an intermediate oxidation state for chlorine.

Question 3

A cell has $E^{\circ}=+1.10$ V with $n=2$. Find $\Delta G^{\circ}$ and $K$, and state $E^{\circ}$ if the cell equation is written with all coefficients doubled.

Accepted Answer

$\Delta G^{\circ}=-2\left(96500ight)\left(1.10ight)=-212$ kJ mol$^{-1}$

$\log K=\dfrac{2	imes1.10}{0.0591}=37.2\qquad\Rightarrow\qquad K=1.6	imes10^{37}$

Doubling the equation makes $n=4$, so $\Delta G^{\circ}$ doubles to $-424$ kJ and $K$ is squared.

$E^{\circ}$ is **unchanged** at $1.10$ V, because potential is intensive.

Question 4

For the cell $	ext{Zn}\,|\,	ext{Zn}^{2+}\left(0.01\ 	ext{M}ight)\,||\,	ext{Cu}^{2+}\left(0.1\ 	ext{M}ight)\,|\,	ext{Cu}$, find the potential at $298$ K given $E^{\circ}=1.10$ V.

Accepted Answer

$	ext{Zn}+	ext{Cu}^{2+}ightarrow	ext{Zn}^{2+}+	ext{Cu}$, so $Q=\dfrac{\left[	ext{Zn}^{2+}ight]}{\left[	ext{Cu}^{2+}ight]}=\dfrac{0.01}{0.1}=0.1$

$E=1.10-\dfrac{0.0591}{2}\log\left(0.1ight)=1.10+0.0296$

$E=1.13$ V

The potential is slightly **higher** than standard, because the product ion is more dilute than the reactant ion.

Question 5

Find the potential of the couple $	ext{Cr}_2	ext{O}_7^{2-}/	ext{Cr}^{3+}$ at pH $2$ with all other species at unit concentration, given $E^{\circ}=+1.33$ V.

Accepted Answer

$E=1.33-\dfrac{0.0591}{6}\log\dfrac{1}{\left[	ext{H}^+ight]^{14}}=1.33+\dfrac{0.0591	imes14}{6}\log\left[	ext{H}^+ight]$

$=1.33-\dfrac{0.0591	imes14	imes2}{6}=1.33-0.276$

$E=1.05$ V

The couple loses about $0.138$ V per pH unit, which is why dichromate oxidations are always carried out in strongly acidic solution.

Question 6

A concentration cell is built from two hydrogen electrodes at $1$ atm, dipping into solutions of pH $3$ and pH $6$. Find its potential and identify the anode.

Accepted Answer

Electrode at pH $3$: $E=-0.0591	imes3=-0.177$ V

Electrode at pH $6$: $E=-0.0591	imes6=-0.355$ V

$E_{cell}=-0.177-\left(-0.355ight)=0.178$ V

The more acidic electrode has the higher potential and is the **cathode**; the pH $6$ electrode is the anode, where hydrogen is oxidised to protons.

Question 7

The conductivity of $0.02$ M potassium chloride is $2.77	imes10^{-3}$ S cm$^{-1}$. Find its molar conductivity.

Accepted Answer

$\Lambda_m=\dfrac{1000\kappa}{C}=\dfrac{1000	imes2.77	imes10^{-3}}{0.02}$

$\Lambda_m=138.5$ S cm$^{2}$ mol$^{-1}$

The limiting value for potassium chloride is about $149.9$, so at this concentration the molar conductivity has fallen by roughly eight per cent through interionic attraction.

Question 8

Given limiting molar conductivities of $349.6$ for $	ext{H}^+$, $40.9$ for $	ext{CH}_3	ext{COO}^-$, $50.1$ for $	ext{Na}^+$ and $76.3$ for $	ext{Cl}^-$, find $\Lambda_m^{\circ}$ for acetic acid two different ways.

Accepted Answer

Directly: $349.6+40.9=390.5$ S cm$^{2}$ mol$^{-1}$

Via salts: $\Lambda_m^{\circ}\left(	ext{HCl}ight)+\Lambda_m^{\circ}\left(	ext{CH}_3	ext{COONa}ight)-\Lambda_m^{\circ}\left(	ext{NaCl}ight)$

$=\left(349.6+76.3ight)+\left(50.1+40.9ight)-\left(50.1+76.3ight)=425.9+91.0-126.4=390.5$

Identical, because sodium and chloride cancel exactly — which is the content of Kohlrausch's law.

Question 9

The molar conductivity of $0.01$ M ammonia is $9.6$ S cm$^{2}$ mol$^{-1}$ against a limiting value of $271.4$. Find the degree of dissociation and $K_b$.

Accepted Answer

$\alpha=\dfrac{9.6}{271.4}=0.0354$

$K_b=\dfrac{C\alpha^{2}}{1-\alpha}=\dfrac{0.01	imes\left(0.0354ight)^{2}}{0.9646}$

$=\dfrac{1.253	imes10^{-5}}{0.9646}=1.30	imes10^{-5}$

Close to the accepted $1.8	imes10^{-5}$, the difference reflecting that even a weak electrolyte's conductivity is slightly reduced by interionic effects.

Question 10

A current of $1.5$ A flows for $20$ minutes through acidified water. Find the volumes of hydrogen and oxygen liberated at STP.

Accepted Answer

Charge $=1.5	imes1200=1800$ C, that is $\dfrac{1800}{96500}=0.01865$ faraday.

Hydrogen: $n=2$, so moles $=\dfrac{0.01865}{2}=9.33	imes10^{-3}$, volume $=209$ mL.

Oxygen: $n=4$, so moles $=4.66	imes10^{-3}$, volume $=104$ mL.

The two-to-one volume ratio is exactly what the formula of water requires.

Question 11

Three cells containing solutions of silver, copper and aluminium salts are connected in series. If $1.08$ g of silver is deposited, find the masses of copper and aluminium.

Accepted Answer

Silver: $n=1$, equivalents $=\dfrac{1.08}{108}=0.01$

Copper: equivalent mass $=\dfrac{63.5}{2}=31.75$, so mass $=0.01	imes31.75=0.318$ g

Aluminium: equivalent mass $=\dfrac{27}{3}=9$, so mass $=0.01	imes9=0.09$ g

The masses are in the ratio of the equivalent masses, not the molar masses.

Question 12

Explain why a hydrogen-oxygen fuel cell can exceed the efficiency of any heat engine operating between the same temperatures.

Accepted Answer

A heat engine converts chemical energy first into heat and then into work, and the second conversion is bounded by the Carnot efficiency, which depends only on the two temperatures.

A fuel cell converts chemical energy **directly** into electrical work, never passing through heat at all, so the Carnot limit simply does not apply.

Its theoretical ceiling is the ratio of the free energy change to the enthalpy change, which for hydrogen and oxygen is about $83\%$.

Practical cells reach around $60\%$, roughly double what a comparable engine achieves.

Practice — "Redox Reactions and Electrochemistry"

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