Redox Reactions and Electrochemistry — Practice Questions with Answers

Oxidation is the loss of electrons by a species (increase in oxidation number). Reduction is the gain of electrons by a species (decrease in oxidation number).

They must always occur together because electrons lost by the species being oxidised must be transferred to and gained by another species being reduced — electrons cannot simply disappear or be created; every oxidation reaction is necessarily coupled with a corresponding reduction reaction, together forming a "redox" reaction.

Galvanic cell: Converts chemical energy from a spontaneous redox reaction into electrical energy — the reaction occurs on its own, generating a voltage/current.

Electrolytic cell: Uses an external source of electrical energy to drive a non-spontaneous redox reaction — the reaction only occurs because electrical energy is supplied from outside the system.

Standard electrode potential measures a species' relative tendency to be reduced (gain electrons) under standard conditions — a more positive standard reduction potential indicates a greater tendency for reduction (a stronger oxidising agent).

By comparing the standard electrode potentials of two half-reactions, the overall cell potential ($E^\circ_{cell} = E^\circ_{cathode}-E^\circ_{anode}$) can be calculated — a positive $E^\circ_{cell}$ indicates the overall redox reaction is thermodynamically feasible (spontaneous) under standard conditions, while a negative value indicates it is not spontaneous as written.

Nernst equation: $E_{cell} = E^\circ_{cell} - \dfrac{RT}{nF}\ln Q$ (or, using base-10 log at 298 K: $E_{cell}=E^\circ_{cell}-\dfrac{0.0591}{n}\log Q$)

Here, $E_{cell}$ is the actual cell potential under non-standard conditions, $E^\circ_{cell}$ is the standard cell potential, $n$ is the number of electrons transferred, $F$ is Faraday's constant, and $Q$ is the reaction quotient — this equation allows calculation of actual cell potential under any given (non-standard) concentration conditions.

First law: The mass of a substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electric charge (current × time) passed through the electrolyte.

Second law: When the same quantity of electric charge is passed through different electrolytes connected in series, the masses of different substances deposited/liberated are directly proportional to their respective chemical equivalent weights.

Rusting involves iron being oxidised (losing electrons) in the presence of moisture and oxygen: $Fe \rightarrow Fe^{2+}+2e^-$ at anodic regions on the iron surface. These released electrons travel through the iron to cathodic regions, where dissolved oxygen is reduced: $O_2+2H_2O+4e^- \rightarrow 4OH^-$.

The resulting $Fe^{2+}$ and $OH^-$ ions subsequently react and undergo further oxidation, ultimately forming hydrated iron(III) oxide (rust, $Fe_2O_3\cdot xH_2O$) — this entire process, involving distinct anodic and cathodic regions with electron transfer between them, is fundamentally an electrochemical process.

An electrochemical series is a list of elements (or half-reactions) arranged in order of their standard reduction potentials, allowing direct comparison of their relative tendencies to be reduced (or oxidised).

Application: The series can be used to predict whether a given redox reaction will occur spontaneously — a species higher in reduction potential will oxidise a species lower in the series, allowing prediction of, for example, which metals can displace others from their salt solutions (reactivity series applications).

The relationship is given by: $\Delta G^\circ = -nFE^\circ_{cell}$, where $n$ is the number of moles of electrons transferred in the balanced redox reaction, and $F$ is Faraday's constant.

This equation directly links the thermodynamic spontaneity criterion ($\Delta G^\circ<0$ for spontaneous reactions) with the electrochemical measurement of cell potential — a positive $E^\circ_{cell}$ corresponds to a negative $\Delta G^\circ$ (spontaneous), while a negative $E^\circ_{cell}$ corresponds to a positive $\Delta G^\circ$ (non-spontaneous).

Oxidation number is a formal, assigned value representing the hypothetical charge an atom would have if all bonds to atoms of different elements were considered completely ionic (fully assigning shared electrons to the more electronegative atom).

Rule example: The oxidation number of an atom in its elemental (uncombined) form is always zero — for example, the oxidation number of $O$ in $O_2$ gas, or $Na$ in solid sodium metal, is 0.

Absolute electrode potentials of individual half-reactions cannot be measured directly in isolation — only the potential difference between two connected half-cells (a complete cell) can be experimentally measured.

By arbitrarily assigning the standard hydrogen electrode a reference potential of exactly 0.00 V, and then measuring the cell potential when any other half-cell is connected to this reference SHE, the "standard electrode potential" of that other half-cell can be determined relative to this common, universally agreed reference point — this consistent reference system allows electrode potentials from different sources/experiments to be meaningfully compared on the same standardised scale.

Change in oxidation number: Fe: $+2\rightarrow+3$ (loses 1 electron); Mn: $+7\rightarrow+2$ (gains 5 electrons)

To balance electrons lost and gained, multiply the Fe half-reaction by 5:

Balanced equation: $5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O$

(Charge balance check: LHS = $5(2)+(-1)+8(1) = 10-1+8=17$; RHS = $5(3)+2+0=15+2=17$ ✓)

$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$

Cu is the cathode (reduction, higher reduction potential); Zn is the anode (oxidation, lower reduction potential):

$E^\circ_{cell} = E^\circ_{Cu^{2+}/Cu} - E^\circ_{Zn^{2+}/Zn} = 0.34 - (-0.76)$

$E^\circ_{cell} = \mathbf{+1.10\text{ V}}$

$\Delta G^\circ = -nFE^\circ_{cell}$

$\Delta G^\circ = -2\times96500\times1.10$

$\Delta G^\circ = -212300\text{ J/mol}$

$\Delta G^\circ = \mathbf{-212.3\text{ kJ/mol}}$ (negative, confirming the reaction is spontaneous, consistent with the positive $E^\circ_{cell}$)

Charge passed: $Q = I\times t = 2\times(1\times3600) = 7200\text{ C}$

Moles of electrons: $n_e = \dfrac{Q}{F} = \dfrac{7200}{96500} \approx 0.0746\text{ mol}$

Since $Cu^{2+}+2e^- \rightarrow Cu$, moles of Cu deposited $= \dfrac{n_e}{2} = \dfrac{0.0746}{2} = 0.0373\text{ mol}$

Mass of Cu $= 0.0373\times63.5 \approx \mathbf{2.37\text{ g}}$

K is a group 1 metal, always $+1$. Oxygen is typically $-2$ (7 oxygen atoms, total $-14$).

Let oxidation number of Cr = $x$ (2 chromium atoms). Overall compound is neutral, so sum of all oxidation numbers = 0:

$2(+1) + 2(x) + 7(-2) = 0$

$2 + 2x - 14 = 0 \Rightarrow 2x = 12 \Rightarrow x = +6$

Oxidation numbers: K = +1, Cr = +6, O = −2

Nernst equation: $E_{cell} = E^\circ_{cell} - \dfrac{0.0591}{n}\log Q$, where $Q = \dfrac{[Zn^{2+}]}{[Cu^{2+}]} = \dfrac{0.01}{1} = 0.01$

$E_{cell} = 1.10 - \dfrac{0.0591}{2}\log(0.01)$

$\log(0.01) = -2$

$E_{cell} = 1.10 - \dfrac{0.0591}{2}(-2) = 1.10 + 0.0591$

$E_{cell} = \mathbf{1.159\text{ V}}$

Reduction half: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$

Oxidation half: $Fe^{2+} \rightarrow Fe^{3+} + e^-$ (multiply by 6 to balance electrons)

$6Fe^{2+} \rightarrow 6Fe^{3+} + 6e^-$

Combined, balanced equation: $Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$

Moles of electrons: $n_e = \dfrac{Q}{F} = \dfrac{9650}{96500} = 0.1\text{ mol}$

Since $Al^{3+}+3e^- \rightarrow Al$, moles of Al deposited $= \dfrac{n_e}{3} = \dfrac{0.1}{3} \approx 0.0333\text{ mol}$

Mass of Al $= 0.0333\times27 \approx \mathbf{0.9\text{ g}}$

For zinc to displace silver, zinc must be oxidised (anode) and $Ag^+$ must be reduced (cathode).

$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = E^\circ_{Ag^+/Ag} - E^\circ_{Zn^{2+}/Zn} = 0.80-(-0.76)$

$E^\circ_{cell} = \mathbf{+1.56\text{ V}}$

Since $E^\circ_{cell}$ is positive, the reaction is spontaneous — zinc metal can indeed displace silver ions, reducing them to metallic silver while zinc itself is oxidised to $Zn^{2+}$.

$\log K = \dfrac{nE^\circ_{cell}}{0.0591} = \dfrac{2\times1.10}{0.0591} = \dfrac{2.20}{0.0591}$

$\log K \approx 37.23$

$K = 10^{37.23} \approx \mathbf{1.7\times10^{37}}$

This extremely large equilibrium constant confirms the reaction proceeds essentially to completion, strongly favouring products, consistent with its large positive $E^\circ_{cell}$.