Electrochemistry

This article walks the chapter in the order NCERT builds it, each idea resting on the one before. First, galvanic and electrolytic cells and how to write and read a cell diagram correctly. Second, the standard hydrogen electrode as the universal zero of electrode potential, and the electrochemical series it produces. Third, the Nernst equation, which frees potentials from the 1 M straitjacket and, through a short derivation, hands us the link between $E^{\circ}_{\text{cell}}$, $\Delta_rG^{\circ}$ and $K_c$. Fourth, a parallel track on how well solutions conduct — resistivity, conductivity, the cell constant, molar conductivity and how each varies with concentration, capped by Kohlrausch's law. Fifth, Faraday's laws of electrolysis and how electrode potentials (with the wrinkle of overpotential) decide what actually gets deposited or liberated. Finally, three concrete applications: batteries, fuel cells, and corrosion as an electrochemical process. Four worked examples and a cheat-sheet at the end consolidate the numerical machinery for both boards and JEE/NEET.

Galvanic and electrolytic cells: the Daniell cell

The Daniell cell converts the spontaneous redox reaction

into electrical energy, delivering $1.1\ \text{V}$ when $[\text{Zn}^{2+}]$ and $[\text{Cu}^{2+}]$ are both $1\ \text{mol dm}^{-3}$. Such a device — chemical energy in, electrical work out — is a galvanic (voltaic) cell. It splits neatly into two half-cells: a reduction half-reaction, $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}(s)$, happening at the copper electrode, and an oxidation half-reaction, $\text{Zn}(s) \rightarrow \text{Zn}^{2+} + 2e^-$, happening at the zinc electrode, joined internally by a salt bridge that keeps both solutions electrically neutral as ions are consumed and produced. If an external opposing voltage smaller than $1.1\ \text{V}$ is applied, the spontaneous reaction continues, just slower. At exactly $1.1\ \text{V}$ opposing, the reaction stops: no current flows. Push the external voltage past $1.1\ \text{V}$ and the reaction reverses entirely — zinc now deposits and copper dissolves. The same physical cell has become an electrolytic cell, using electrical energy to drive a reaction that will not happen by itself. This reversibility is the conceptual heart of the chapter: a galvanic and an electrolytic cell are not different pieces of apparatus, they are the same electrode-and-electrolyte arrangement, and which one you have depends entirely on whether an external source is overpowering the cell's own chemistry. In a galvanic cell, the electrode where oxidation occurs is the anode, and because electrons are released there it carries a negative charge; the electrode where reduction occurs is the cathode, positively charged. Convention places the anode on the left and the cathode on the right when writing the cell: $\text{Zn}(s)\,|\,\text{Zn}^{2+}(aq)\,\|\,\text{Cu}^{2+}(aq)\,|\,\text{Cu}(s)$, where a single vertical line marks a phase boundary (electrode to electrolyte) and a double vertical line marks the salt bridge. Under this convention the cell EMF is always taken as

where: $E_{\text{right}}$ and $E_{\text{left}}$ are the (reduction) electrode potentials of the half-cells written on the right and left of the cell diagram respectively. Writing the cell in this standard orientation and applying this one subtraction correctly is worth disproportionate marks — get the anode and cathode on the wrong sides and every downstream sign flips.

The standard hydrogen electrode and the electrochemical series

The potential of a single, isolated electrode cannot be measured — a voltmeter always reads the difference between two electrodes. To break this impasse, IUPAC fixes one reference half-cell, the standard hydrogen electrode (SHE), $\text{Pt}(s)\,|\,\text{H}_2(g, 1\ \text{bar})\,|\,\text{H}^+(aq, 1\ \text{M})$, at exactly $E^{\circ} = 0.00\ \text{V}$ at every temperature, corresponding to $\text{H}^+(aq) + e^- \rightarrow \tfrac{1}{2}\text{H}_2(g)$. Pairing any other half-cell against the SHE (with the SHE on the left) makes the measured cell EMF equal to that half-cell's own standard electrode potential, since $E^{\circ}_{\text{left}} = 0$:

where: all species are at unit activity (approximately 1 M for solutes, 1 bar for gases, pure solids and liquids) and $T = 298\ \text{K}$. This is how the whole table of standard reduction potentials — from $\text{F}_2/\text{F}^-$ at $+2.87\ \text{V}$ down to $\text{Li}^+/\text{Li}$ at $-3.05\ \text{V}$ — is built: every entry is a cell EMF measured against the SHE. Arranged from most positive to most negative, this table is the electrochemical series. A large positive $E^{\circ}$ means the oxidised form (left side of the half-reaction) is a strong oxidising agent and gets reduced easily; a large negative $E^{\circ}$ means the reduced form (metal) is a strong reducing agent. Reading down the table, oxidising power of species on the left falls and reducing power of species on the right rises — which is exactly why fluorine gas is the strongest common oxidiser and lithium metal the strongest common reducer in aqueous solution.

The Nernst equation

The tabulated $E^{\circ}$ values assume unit concentration, but real cells rarely sit at exactly 1 M. Nernst's contribution was to show how electrode potential shifts away from $E^{\circ}$ as concentrations change, for the electrode reaction $\text{M}^{n+}(aq) + ne^- \rightarrow \text{M}(s)$:

where: $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$, $T$ is temperature in kelvin, $F = 96{,}487\ \text{C mol}^{-1}$ (often rounded to $96{,}500$), $n$ is the number of electrons in the half-reaction, and the concentration of the pure solid metal is taken as unity. Extended to a full cell reaction $aA + bB \rightleftharpoons cC + dD$ carrying $n$ electrons, the same logic (derived in full below) gives the general Nernst equation

where: $Q = \dfrac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$ is the reaction quotient, and the numeric coefficient $2.303RT/F$ evaluates to $0.0591\ \text{V}$ (rounded to $0.059\ \text{V}$) once $R$, $F$ and $T = 298\ \text{K}$ are substituted. For the Daniell cell this becomes $E_{\text{cell}} = E^{\circ}_{\text{cell}} - (0.059/2)\log([\text{Zn}^{2+}]/[\text{Cu}^{2+}])$ — raising $[\text{Cu}^{2+}]$ or lowering $[\text{Zn}^{2+}]$ pushes $E_{\text{cell}}$ up, exactly matching intuition that more "fuel" on the product-favouring side makes the reaction more spontaneous. It is essential to use the same $n$ for both half-reactions when combining them, even if that means scaling a half-reaction's coefficients.

Cell potential, Gibbs energy and the equilibrium constant

Two consequences of the Nernst equation are exam staples in their own right and are derived rigorously in the derivation blocks below. First, the electrical work a cell can deliver reversibly equals the fall in Gibbs energy of its reaction:

where: $E_{\text{cell}}$ is intensive (does not depend on how the equation is scaled) while $\Delta_rG$ is extensive and does scale with $n$ — writing the Daniell reaction with coefficient 2 throughout doubles $\Delta_rG$ but leaves $E_{\text{cell}}$ untouched. Second, once the cell reaction reaches equilibrium the voltmeter reads zero and $Q$ becomes $K_c$, giving

Equations (3) and (4), together with the general chemical-thermodynamics relation $\Delta_rG^{\circ} = -RT\ln K_c$, form a closed triangle: measure any one of $E^{\circ}_{\text{cell}}$, $\Delta_rG^{\circ}$ or $K_c$ and the other two follow by algebra alone, without ever running the reaction to equilibrium and measuring concentrations directly.

Conductivity, resistivity and the cell constant

Electrolyte solutions carry current the same way metals do, through Ohm's law, but with an added complication: you cannot clip a solution to a Wheatstone bridge like a wire. Electrical resistance obeys $R = \rho\, l/A$, where $\rho$ (resistivity) is the constant of proportionality, $l$ is the length of the conducting column and $A$ its cross-sectional area. The reciprocal of resistivity, conductivity $\kappa = 1/\rho$, measures how well a material conducts once its geometry is normalised to a unit cube; its SI unit is $\text{S m}^{-1}$, though $\text{S cm}^{-1}$ is common in the lab. Because an ionic solution is confined in a conductivity cell whose exact $l$ and $A$ are inconvenient to measure precisely, chemists define the cell constant $G^* = l/A$ and determine it once, from a standard KCl solution of known $\kappa$:

where: $R$ is the measured resistance of the solution in the cell (using AC current, since DC would electrolyse the solution and change its composition). Table 2.2 of the NCERT text shows conductivity spanning nearly 20 orders of magnitude, from good conductors like silver ($\approx 6\times10^3\ \text{S m}^{-1}$) through weak electrolytes like $0.01\ \text{M}$ acetic acid ($0.016\ \text{S m}^{-1}$) down to insulators like Teflon ($10^{-18}\ \text{S m}^{-1}$).

Molar conductivity and its variation with concentration

Conductivity alone cannot compare two different electrolytes fairly, since a more concentrated solution simply has more ions per unit volume regardless of how "good" the electrolyte is. Molar conductivity corrects for this by normalising to the amount of substance:

where: if $\kappa$ is in $\text{S m}^{-1}$ and $c$ in $\text{mol m}^{-3}$, $\Lambda_m$ comes out in $\text{S m}^2\text{mol}^{-1}$; using $\kappa$ in $\text{S cm}^{-1}$ and $c$ in $\text{mol L}^{-1}$ instead gives $\Lambda_m = 1000\kappa/c$ in $\text{S cm}^2\text{mol}^{-1}$ (the factor of 1000 converts litres to cubic centimetres). Physically, $\Lambda_m$ is the conductance of the volume of solution that contains exactly one mole of electrolyte, held between electrodes one unit length apart. Conductivity always decreases on dilution — fewer ions per unit volume — but molar conductivity always increases on dilution, because the volume containing one mole grows faster than $\kappa$ shrinks. As $c \rightarrow 0$, $\Lambda_m$ approaches a ceiling, the limiting molar conductivity $\Lambda_m^{\circ}$, but strong and weak electrolytes approach it very differently. For strong electrolytes (fully dissociated at all workable concentrations), $\Lambda_m$ rises only gently with dilution and obeys the empirical Debye–Hückel–Onsager form

a straight line when $\Lambda_m$ is plotted against $\sqrt{c}$, so $\Lambda_m^{\circ}$ can simply be read off as the intercept at $\sqrt{c} = 0$. For weak electrolytes, dilution steadily pulls the dissociation equilibrium forward, so $\Lambda_m$ rises only gradually at first and then shoots up steeply as $c \rightarrow 0$ — the curve never straightens out, so $\Lambda_m^{\circ}$ cannot be found by extrapolation and must instead come from Kohlrausch's law (next section).

Kohlrausch's law of independent migration of ions

Kohlrausch noticed a striking regularity while tabulating $\Lambda_m^{\circ}$ values: the difference $\Lambda_m^{\circ}(\text{KCl}) - \Lambda_m^{\circ}(\text{NaCl})$ is very close to $\Lambda_m^{\circ}(\text{KBr}) - \Lambda_m^{\circ}(\text{NaBr})$, about $23.4\ \text{S cm}^2\text{mol}^{-1}$ in each case — a difference that depends only on which cation is present, not on the accompanying anion. This regularity generalises into Kohlrausch's law of independent migration of ions:

where: $\nu_+$ and $\nu_-$ are the number of cations and anions produced per formula unit of electrolyte on complete dissociation, and $\lambda^{\circ}_+$, $\lambda^{\circ}_-$ are the limiting ionic conductivities of those ions, tabulated once and reused for any electrolyte. At infinite dilution each ion migrates and contributes to the current entirely independently of the ion it was paired with, so the total limiting molar conductivity is simply additive. Two big applications follow immediately: $\Lambda_m^{\circ}$ for any electrolyte, including ones that are hard to prepare or unstable, can be built from a table of ionic values (for example $\Lambda_m^{\circ}(\text{CaCl}_2) = \lambda^{\circ}_{\text{Ca}^{2+}} + 2\lambda^{\circ}_{\text{Cl}^-}$); and $\Lambda_m^{\circ}$ for a weak electrolyte, which cannot be read off a graph, can be assembled indirectly by combining the $\Lambda_m^{\circ}$ values of three strong electrolytes that share its ions (illustrated in the worked-example machinery of Derivation D below).

Faraday's laws of electrolysis and the products of electrolysis

Faraday's two laws, published in 1833–34, make electrolysis quantitative. The first law states that the amount of chemical change at an electrode is directly proportional to the quantity of electricity (charge) passed: $Q = It$, with $Q$ in coulombs when current $I$ is in amperes and time $t$ in seconds. The second law states that for a fixed quantity of charge, the amounts of different substances liberated at different electrodes are proportional to their chemical equivalent weights (atomic mass divided by the number of electrons needed per ion). Combining both laws for an electrode reaction $\text{M}^{n+} + ne^- \rightarrow \text{M}$ gives the working formula

where: $m$ is the mass deposited (or liberated as gas, converted via molar volume), $M$ is the molar mass of the substance, $n$ is the number of electrons transferred per ion, and $F = N_A \times e = 6.022\times10^{23}\ \text{mol}^{-1} \times 1.6021\times10^{-19}\ \text{C} = 96{,}487\ \text{C mol}^{-1}$ (commonly rounded to $96{,}500$) is the charge carried by one mole of electrons — one Faraday. What actually happens at each electrode is decided by competing electrode potentials: at the cathode, the reduction reaction with the higher (more positive, or least negative) $E^{\circ}$ is thermodynamically preferred; at the anode, the oxidation reaction with the lower $E^{\circ}$ (equivalently, the species that is more easily oxidised) is preferred — subject to a practical wrinkle called overpotential, an extra voltage some reactions need kinetically before they proceed at a workable rate. This is precisely why electrolysing aqueous NaCl gives $\text{H}_2$ at the cathode (preferred over depositing reactive Na metal) and, despite water's oxidation being thermodynamically favoured over chloride, $\text{Cl}_2$ at the anode in practice (chlorine evolution has a much lower overpotential than oxygen evolution on the usual electrode materials), giving the industrially important net reaction $\text{NaCl}(aq) + \text{H}_2\text{O}(l) \rightarrow \text{Na}^+(aq) + \text{OH}^-(aq) + \tfrac{1}{2}\text{H}_2(g) + \tfrac{1}{2}\text{Cl}_2(g)$.

Batteries and fuel cells

A battery is a galvanic cell (or several in series) engineered for practical use: light, compact, steady in voltage. Primary batteries discharge once and cannot be recharged. The common dry (Leclanché) cell — zinc container as anode, graphite cathode packed in $\text{MnO}_2$ and carbon — delivers about $1.5\ \text{V}$; the mercury cell (hearing aids, watches) pairs a zinc–mercury amalgam anode with an HgO/carbon cathode in KOH/ZnO paste, giving a steady $1.35\ \text{V}$ since its overall reaction, $\text{Zn(Hg)} + \text{HgO}(s) \rightarrow \text{ZnO}(s) + \text{Hg}(l)$, involves no dissolved ion whose concentration changes with use. Secondary batteries can be recharged by driving current backwards through them, chemically reversing the discharge reaction. The lead storage battery — lead anode, $\text{PbO}_2$-packed lead grid cathode, dilute sulfuric acid electrolyte — runs $\text{Pb}(s) + \text{PbO}_2(s) + 2\text{H}_2\text{SO}_4(aq) \rightarrow 2\text{PbSO}_4(s) + 2\text{H}_2\text{O}(l)$ on discharge and reverses on charging; the nickel–cadmium cell lasts longer but costs more. A fuel cell is a galvanic cell fed continuously with reactants, so it can run as long as fuel keeps arriving. The hydrogen–oxygen fuel cell, famous for powering the Apollo missions, bubbles $\text{H}_2$ and $\text{O}_2$ through porous, platinum-catalysed carbon electrodes dipped in concentrated NaOH: cathode $\text{O}_2 + 2\text{H}_2\text{O} + 4e^- \rightarrow 4\text{OH}^-$, anode $2\text{H}_2 + 4\text{OH}^- \rightarrow 4\text{H}_2\text{O} + 4e^-$, net $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$. Converting chemical energy directly to electrical work without an intervening heat engine gives efficiencies around $70\%$, against roughly $40\%$ for thermal plants — and the only product is water, anchoring the "hydrogen economy" vision of clean energy.

Corrosion as an electrochemical process

Rusting, tarnishing and the green coating on old copper are all forms of corrosion: a metal losing electrons to atmospheric oxygen and slowly converting to an oxide or other salt. Iron rusting is essentially a tiny galvanic cell built on the metal's own surface. At one spot the metal oxidises and behaves as an anode, $2\text{Fe}(s) \rightarrow 2\text{Fe}^{2+} + 4e^-$ ($E^{\circ} = -0.44\ \text{V}$); the released electrons travel through the metal to another spot that behaves as a cathode, reducing dissolved oxygen in the presence of $\text{H}^+$ (from dissolved atmospheric $\text{CO}_2$), $\text{O}_2(g) + 4\text{H}^+(aq) + 4e^- \rightarrow 2\text{H}_2\text{O}(l)$ ($E^{\circ} = +1.23\ \text{V}$), giving a spontaneous overall cell potential of $1.67\ \text{V}$. The $\text{Fe}^{2+}$ formed is further oxidised by atmospheric oxygen to $\text{Fe}^{3+}$, precipitating as hydrated ferric oxide, $\text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O}$ — rust. Prevention follows directly from the mechanism: paint excludes water and oxygen from the surface; coating with a less reactive metal (tin plating) or a more reactive one (galvanising with zinc) changes what corrodes preferentially; and a sacrificial anode of a more reactive metal (magnesium or zinc) deliberately corrodes itself in place of the protected object, exploiting the electrochemical series in reverse.

Board level - mass of copper deposited by a steady current

A solution of $\text{CuSO}_4$ is electrolysed with platinum electrodes for $10$ minutes using a steady current of $1.5\ \text{A}$. Calculate the mass of copper deposited at the cathode. (Use atomic mass of $\text{Cu} = 63.5\ \text{g mol}^{-1}$, $F \approx 96{,}500\ \text{C mol}^{-1}$.)

Step 1. Convert time and find the total charge passed: $t = 10 \times 60 = 600\ \text{s}$, so $Q = It = 1.5\ \text{A} \times 600\ \text{s} = 900\ \text{C}$.

Step 2. Write the cathode half-reaction and identify $n$: $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$, so $n = 2$ and depositing one mole ($63.5\ \text{g}$) of copper needs $2F = 2 \times 96{,}500 = 193{,}000\ \text{C}$.

Step 3. Scale to the actual charge passed: $m = \dfrac{63.5\ \text{g mol}^{-1} \times 900\ \text{C}}{193{,}000\ \text{C mol}^{-1}} = 0.296\ \text{g}$.

Step 4. Extend the result. The same $900\ \text{C}$ would liberate $Q/F = 900/96{,}500 = 9.33\times10^{-3}\ \text{mol}$ of electrons; if this exact charge instead passed through a second cell containing $\text{AgNO}_3$ ($\text{Ag}^+ + e^- \rightarrow \text{Ag}$, $n=1$, $M = 108\ \text{g mol}^{-1}$), it would deposit $m_{\text{Ag}} = (108 \times 900)/96{,}500 = 1.007\ \text{g}$ of silver — the same charge, different metal, different mass, exactly as Faraday's second law predicts. Answer: $0.296\ \text{g}$ of copper is deposited; the same charge would deposit $1.007\ \text{g}$ of silver in a series cell.

Board level - resistivity, conductivity and molar conductivity of a NaOH column

A column of $0.05\ \text{mol L}^{-1}$ NaOH solution, diameter $1\ \text{cm}$ and length $50\ \text{cm}$, has a measured resistance of $5.55\times10^3\ \Omega$. Calculate its resistivity, conductivity and molar conductivity.

Step 1. Find the cross-sectional area: $A = \pi r^2 = 3.14 \times (0.5\ \text{cm})^2 = 0.785\ \text{cm}^2$, and note $l = 50\ \text{cm}$.

Step 2. Resistivity from $R = \rho\,l/A$, rearranged: $\rho = \dfrac{RA}{l} = \dfrac{5.55\times10^3\ \Omega \times 0.785\ \text{cm}^2}{50\ \text{cm}} = 87.1\ \Omega\,\text{cm}$.

Step 3. Conductivity as the reciprocal of resistivity: $\kappa = 1/\rho = 1/87.1 = 0.01148\ \text{S cm}^{-1}$.

Step 4. Molar conductivity using $\Lambda_m = 1000\kappa/c$: $\Lambda_m = \dfrac{0.01148\ \text{S cm}^{-1} \times 1000\ \text{cm}^3\text{L}^{-1}}{0.05\ \text{mol L}^{-1}} = 229.6\ \text{S cm}^2\text{mol}^{-1}$. Answer: $\rho = 87.1\ \Omega\,\text{cm}$, $\kappa = 0.01148\ \text{S cm}^{-1}$ (equivalently $1.148\ \text{S m}^{-1}$), and $\Lambda_m = 229.6\ \text{S cm}^2\text{mol}^{-1}$.

Competitive level (JEE / NEET) - Nernst equation, equilibrium constant and Gibbs energy for one cell

The cell $\text{Mg}(s)\,|\,\text{Mg}^{2+}(0.130\ \text{M})\,\|\,\text{Ag}^+(0.0001\ \text{M})\,|\,\text{Ag}(s)$ has $E^{\circ}_{\text{cell}} = 3.17\ \text{V}$, for the reaction $\text{Mg}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Mg}^{2+}(aq) + 2\text{Ag}(s)$. Find (a) $E_{\text{cell}}$ under the stated concentrations, (b) the equilibrium constant $K_c$, and (c) $\Delta_rG^{\circ}$.

Step 1. Apply the Nernst equation with $n = 2$: $E_{\text{cell}} = E^{\circ}_{\text{cell}} - \dfrac{0.059}{2}\log\dfrac{[\text{Mg}^{2+}]}{[\text{Ag}^+]^2} = 3.17 - \dfrac{0.059}{2}\log\dfrac{0.130}{(0.0001)^2}$.

Step 2. Evaluate the log term: $(0.0001)^2 = 1\times10^{-8}$, so $0.130/1\times10^{-8} = 1.3\times10^{7}$, and $\log(1.3\times10^{7}) = 7.11$. Then $(0.059/2) \times 7.11 = 0.21\ \text{V}$, so $E_{\text{cell}} = 3.17 - 0.21 = 2.96\ \text{V}$.

Step 3. Find $K_c$ using $E^{\circ}_{\text{cell}} = (0.059/n)\log K_c$: $\log K_c = \dfrac{2 \times 3.17}{0.059} = 107.5$, so $K_c = 10^{107.5} \approx 3\times10^{107}$ — an astronomically large constant, confirming that magnesium displaces silver ions to essential completion.

Step 4. Find $\Delta_rG^{\circ}$ using $\Delta_rG^{\circ} = -nFE^{\circ}_{\text{cell}} = -2 \times 96{,}500\ \text{C mol}^{-1} \times 3.17\ \text{V} = -611{,}810\ \text{J mol}^{-1} = -611.8\ \text{kJ mol}^{-1}$. Answer: $E_{\text{cell}} = 2.96\ \text{V}$, $K_c \approx 3\times10^{107}$, and $\Delta_rG^{\circ} = -611.8\ \text{kJ mol}^{-1}$. Timing note: about three minutes. Do the Nernst-equation arithmetic first since it is the most error-prone (watch the squared term in the denominator), then reuse the same $E^{\circ}_{\text{cell}}$ twice more for $K_c$ and $\Delta_rG^{\circ}$ — do not recompute anything from $E_{\text{cell}}$, both follow-up parts use the standard potential, not the concentration-corrected one.

Competitive level (JEE / NEET) - three electrolytic cells in series

Three electrolytic cells A, B and C, containing $\text{ZnSO}_4$, $\text{AgNO}_3$ and $\text{CuSO}_4$ solutions respectively, are connected in series. A steady current of $1.5\ \text{A}$ is passed until $1.45\ \text{g}$ of silver is deposited at the cathode of cell B. Find (a) how long the current flowed, and (b) the masses of copper and zinc deposited in cells C and A. (Atomic masses: $\text{Ag} = 108$, $\text{Cu} = 63.5$, $\text{Zn} = 65.4\ \text{g mol}^{-1}$; $F \approx 96{,}500\ \text{C mol}^{-1}$.)

Step 1. Cells in series carry the same charge $Q$ through each. Find $Q$ from the silver deposited: $\text{Ag}^+ + e^- \rightarrow \text{Ag}$ needs $n=1$, so moles of Ag $= 1.45/108 = 0.01343\ \text{mol}$, and $Q = nF = 0.01343 \times 96{,}500 = 1295.6\ \text{C}$.

Step 2. Find the time from $Q = It$: $t = Q/I = 1295.6/1.5 = 863.7\ \text{s} \approx 14.4\ \text{minutes}$.

Step 3. Apply the same $Q = 1295.6\ \text{C}$ to cell C, where $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ needs $n = 2$: moles of Cu $= 1295.6/(2 \times 96{,}500) = 6.713\times10^{-3}\ \text{mol}$, so mass $= 6.713\times10^{-3} \times 63.5 = 0.426\ \text{g}$.

Step 4. Apply the same $Q$ to cell A, where $\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}$ also needs $n=2$: moles of Zn are identical to moles of Cu found in Step 3 (same $n$, same $Q$), $6.713\times10^{-3}\ \text{mol}$, so mass $= 6.713\times10^{-3} \times 65.4 = 0.439\ \text{g}$. Answer: the current flowed for about $864\ \text{s}$ ($\approx 14.4\ \text{min}$); cell C deposits $0.426\ \text{g}$ of copper and cell A deposits $0.439\ \text{g}$ of zinc. Timing note: about three to four minutes. The trap is forgetting that "same charge, same series circuit" does not mean "same moles deposited" — only cells with the same $n$ (here, Cu and Zn, both $n=2$) deposit equal moles; Ag, with $n=1$, deposits twice as many moles as either for the same charge. Always find $Q$ or the moles of electrons first, then branch out per electrode using each electrode's own $n$.

• Anode polarity flips between cell types. Anode is negative in a galvanic cell (oxidation releases electrons there) but positive in an electrolytic cell (the external source pulls electrons out of it). This single fact, misapplied, silently ruins electrolysis-product questions.
• Mixing up $E_{\text{right}} - E_{\text{left}}$ with the wrong sign. The convention is always cathode-minus-anode with the cell diagram's anode on the left; writing the diagram backwards and forgetting to flip the subtraction is a common one-mark loss.
• Using different $n$ for the two half-reactions. Before combining half-reactions or applying the Nernst equation, balance electrons so both halves use the same $n$ — otherwise $Q$ is built incorrectly.
• Forgetting the factor of 1000 when converting $\kappa$ in $\text{S cm}^{-1}$ to $\Lambda_m$ in $\text{S cm}^2\text{mol}^{-1}$ (litres to cubic centimetres), or mixing $\text{S m}^{-1}$ and $\text{S cm}^{-1}$ within one calculation.
• Extrapolating a weak electrolyte's $\Lambda_m$ vs $\sqrt{c}$ graph to find $\Lambda_m^{\circ}$. This only works for strong electrolytes; weak electrolytes need Kohlrausch's law and three strong-electrolyte values instead.
• Believing conductivity and molar conductivity move the same way on dilution. Conductivity always falls; molar conductivity always rises. They are not the same story told twice.
• Predicting electrolysis products from $E^{\circ}$ alone, ignoring overpotential. Aqueous NaCl electrolysis gives $\text{Cl}_2$ at the anode in practice even though water's oxidation has the lower (more favourable) $E^{\circ}$, because oxygen evolution suffers a large overpotential.
• Treating the $0.059\ \text{V}$ constant as universal. It comes from $2.303RT/F$ at $T = 298\ \text{K}$ specifically; at any other temperature it must be recalculated, not quoted from memory.
• Confusing corrosion with electroplating. Corrosion is spontaneous and galvanic, needing no external source; electroplating is non-spontaneous and electrolytic, needing one — both look like "metal chemistry at an electrode" on the page but are opposite phenomena.

• Distinguish galvanic and electrolytic cells and correctly assign anode/cathode identity and charge sign in each.
• Write a galvanic cell in standard notation (anode left, cathode right, single and double vertical lines) and compute $E_{\text{cell}} = E_{\text{right}} - E_{\text{left}}$.
• Use the standard hydrogen electrode convention to interpret and build an electrochemical series from measured cell EMFs.
• Apply the Nernst equation to find $E_{\text{cell}}$, or a single electrode potential, at non-standard concentrations.
• Derive and use $E^{\circ}_{\text{cell}} = (0.059/n)\log K_c$ and $\Delta_rG^{\circ} = -nFE^{\circ}_{\text{cell}}$ to move freely between electrochemistry, thermodynamics and equilibrium.
• Define and correctly interconvert resistivity, conductivity, cell constant and molar conductivity, with correct units in each system.
• Explain, with a reason, why conductivity falls but molar conductivity rises on dilution, for both strong and weak electrolytes.
• State and apply Kohlrausch's law of independent migration of ions to find $\Lambda_m^{\circ}$ for a weak electrolyte, and hence its degree of dissociation and $K_a$.
• Apply Faraday's laws of electrolysis to find charge, time, or mass deposited or liberated, including problems with multiple cells in series.
• Describe how batteries, fuel cells and corrosion are all applications, useful or unwanted, of the same galvanic-cell electrochemistry.

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