Intext Questions — Electrochemistry

1. 2.13 marksNCERT Intext Questions, Chapter 2

   How would you determine the standard electrode potential of the system Mg${}^{2+}$|Mg?

   Hint. A single electrode potential cannot be measured on its own; it is always measured against the standard hydrogen electrode.

   No half-cell potential can be measured in isolation, because any voltmeter needs two terminals. The convention is to pair the unknown half-cell with the standard hydrogen electrode (SHE), whose potential is defined as exactly zero at all temperatures.

   Set up the cell

   $\text{Mg(s)}|{\text{Mg}}^{2+}(\text{aq, 1 M})\Vert {\text{H}}^{+}(\text{aq, 1 M})|{\text{H}}_2(g,1\text{bar})|\text{Pt(s)}$

   The magnesium electrode is a strip of pure Mg dipping in 1 M Mg${}^{2+}$; the SHE is platinised platinum in 1 M H${}^{+}$ with H${}_2$ bubbled at 1 bar. Both half-cells are at 298 K and are joined by a salt bridge.

   By the sign convention of the book,

   $E_{\text{cell}}^{\circ }=E_{\text{right}}^{\circ }-E_{\text{left}}^{\circ }=E_{{\text{H}}^{+}/{\text{H}}_2}^{\circ }-E_{{\text{Mg}}^{2+}/\text{Mg}}^{\circ }=0-E_{{\text{Mg}}^{2+}/\text{Mg}}^{\circ }$

   The voltmeter reads $+2.36$ V with the magnesium electrode as the negative terminal, so

   $E_{{\text{Mg}}^{2+}/\text{Mg}}^{\circ }=-2.36\text{V}$

   The negative sign is not an accident of wiring: it says magnesium loses electrons more readily than hydrogen, so magnesium is the anode and the SHE is the cathode. Reversing the leads would only reverse the reading, not the chemistry.

   ✦ Pair Mg|Mg${}^{2+}$(1 M) with the SHE at 298 K; the cell reads 2.36 V with Mg as the negative electrode, giving $E_{{\text{Mg}}^{2+}/\text{Mg}}^{\circ }=-2.36$ V

2. 2.22 marksNCERT Intext Questions, Chapter 2

   Can you store copper sulphate solutions in a zinc pot?

   Hint. Compare the two standard electrode potentials and ask whether the displacement reaction has a positive cell potential.

   No. The reaction that would occur is

   $\text{Zn(s)}+{\text{Cu}}^{2+}(\text{aq})\to {\text{Zn}}^{2+}(\text{aq})+\text{Cu(s)}$

   From Table 2.1, $E_{{\text{Cu}}^{2+}/\text{Cu}}^{\circ }=+0.34$ V and $E_{{\text{Zn}}^{2+}/\text{Zn}}^{\circ }=-0.76$ V, so

   $E_{\text{cell}}^{\circ }=0.34-(-0.76)=+1.10\text{V}$

   A positive cell potential means ${\Delta }_r{G}^{\circ }=-nF{E}_{\text{cell}}^{\circ }$ is negative, so the reaction is spontaneous. Zinc is a stronger reducing agent than copper and will displace Cu${}^{2+}$ from solution.

   In practice the pot itself is the reducing agent, so the container dissolves while copper plates out on its inner wall. The solution is destroyed and so is the vessel.

   ✦ No, because zinc displaces copper ($E_{\text{cell}}^{\circ }=+1.10$ V), so the pot dissolves and copper is deposited on it

3. 2.33 marksNCERT Intext Questions, Chapter 2

   Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.

   Hint. An oxidising agent works on Fe$^{2+}$ only if its own reduction potential is higher than that of the Fe$^{3+}$/Fe$^{2+}$ couple.

   Oxidising Fe${}^{2+}$ means driving the half-reaction

   ${\text{Fe}}^{2+}\to {\text{Fe}}^{3+}+e^{-}$

   which is the reverse of the couple listed in Table 2.1 with $E_{{\text{Fe}}^{3+}/{\text{Fe}}^{2+}}^{\circ }=+0.77$ V. Any species whose own reduction potential exceeds $+0.77$ V will therefore take the electron and give a positive $E_{\text{cell}}^{\circ }$.

   Reading upward from that entry in Table 2.1, the candidates include:

   • ${\text{F}}_2(g)+2e^{-}\to 2{\text{F}}^{-}$, $E^{\circ }=+2.87$ V, so $E_{\text{cell}}^{\circ }=2.87-0.77=+2.10$ V
   • ${\text{MnO}}_4^{-}+8{\text{H}}^{+}+5e^{-}\to {\text{Mn}}^{2+}+4{\text{H}}_2\text{O}$, $E^{\circ }=+1.51$ V, so $E_{\text{cell}}^{\circ }=+0.74$ V
   • ${\text{Cr}}_2{\text{O}}_7^{2-}+14{\text{H}}^{+}+6e^{-}\to 2{\text{Cr}}^{3+}+7{\text{H}}_2\text{O}$, $E^{\circ }=+1.33$ V, so $E_{\text{cell}}^{\circ }=+0.56$ V

   The last two are exactly the reagents used in volumetric estimation of iron(II), which is why permanganate and dichromate titrations of Fe${}^{2+}$ work at all. Cl${}_2$ ($+1.36$ V), O${}_2$ in acid ($+1.23$ V) and Br${}_2$ ($+1.09$ V) would serve equally well.

   Note the phrase "under suitable conditions": the potentials for MnO${}_4^{-}$ and Cr${}_2$O${}_7^{2-}$ are quoted for acidic solution, so the medium must be acidified for these values to apply.

   ✦ Any couple above Fe${}^{3+}$/Fe${}^{2+}$ in Table 2.1, for example F${}_2$ (2.87 V), acidified MnO${}_4^{-}$ (1.51 V) and acidified Cr${}_2$O${}_7^{2-}$ (1.33 V)

4. 2.43 marksNCERT Intext Questions, Chapter 2

   Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

   Hint. Write the Nernst equation for H$^{+}$ + e$^{-}$ $\rightarrow$ $\tfrac12$H$_2$ and remember that $-\log[\text{H}^{+}]$ is the pH.

   The electrode reaction is

   ${\text{H}}^{+}(\text{aq})+e^{-}\to \frac{1}{2}{\text{H}}_2(g)$

   with $E^{\circ }=0$ by definition and $n=1$. Taking $p_{{\text{H}}_2}=1$ bar, the Nernst equation gives

   $E=E^{\circ }-\frac{0.0591}{1}\log \frac{1}{[{\text{H}}^{+}]}=0+0.0591\log [{\text{H}}^{+}]$

   Since $\log [{\text{H}}^{+}]=-\text{pH}$,

   $E=-0.0591\times \text{pH}$

   Substituting pH $=10$:

   $E=-0.0591\times 10=-0.591\text{V}$

   The result is negative because at pH 10 the hydrogen ion concentration is only $10^{-10}$ M, far below the standard 1 M, so the electrode is a much poorer oxidising agent than the SHE. This linear dependence, $-0.059$ V per pH unit at 298 K, is the entire working principle of the glass pH electrode.

   ✦ $E=-0.591$ V

5. 2.53 marksNCERT Intext Questions, Chapter 2

   Calculate the emf of the cell in which the following reaction takes place:

   $\text{Ni(s)}+2{\text{Ag}}^{+}(0.002\text{M})\to {\text{Ni}}^{2+}(0.160\text{M})+2\text{Ag(s)}$

   Given that $E_{\text{cell}}^{\circ }=1.05$ V.

   Hint. Two electrons are transferred, and the silver ion concentration enters the reaction quotient squared.

   The reaction transfers $n=2$ electrons, so the Nernst equation reads

   $E_{\text{cell}}=E_{\text{cell}}^{\circ }-\frac{0.0591}{2}\log \frac{[{\text{Ni}}^{2+}]}{[{\text{Ag}}^{+}{]}^2}$

   The silver term is squared because two Ag${}^{+}$ ions appear in the balanced equation, which is the step most often dropped.

   $\frac{[{\text{Ni}}^{2+}]}{[{\text{Ag}}^{+}{]}^2}=\frac{0.160}{(0.002{)}^2}=\frac{0.160}{4\times 10^{-6}}=4\times 10^4$

   $\log (4\times 10^4)=4.602$

   $E_{\text{cell}}=1.05-\frac{0.0591}{2}\times 4.602=1.05-0.136=0.914\text{V}$

   The emf falls below $E_{\text{cell}}^{\circ }$ because the product Ni${}^{2+}$ is present at a much higher concentration than the reactant Ag${}^{+}$, which pushes the reaction quotient above unity. NCERT's own answer key on the last page of the chapter gives 0.91 V.

   ✦ $E_{\text{cell}}=0.91$ V

6. 2.63 marksNCERT Intext Questions, Chapter 2

   The cell in which the following reaction occurs:

   $2{\text{Fe}}^{3+}(\text{aq})+2{\text{I}}^{-}(\text{aq})\to 2{\text{Fe}}^{2+}(\text{aq})+{\text{I}}_2(\text{s})$

   has $E_{\text{cell}}^{\circ }=0.236$ V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

   Hint. Use $\Delta_rG^{\circ}=-nFE^{\circ}_{\text{cell}}$ first, then get $K$ from $\Delta_rG^{\circ}=-RT\ln K$.

   Two electrons are transferred, so $n=2$.

   ${\Delta }_r{G}^{\circ }=-nF{E}_{\text{cell}}^{\circ }=-2\times 96487\times 0.236=-45542{\text{J mol}}^{-1}$

   ${\Delta }_r{G}^{\circ }=-45.54{\text{kJ mol}}^{-1}$

   For the equilibrium constant, use ${\Delta }_r{G}^{\circ }=-RT\ln K_c$:

   $\ln K_c=\frac{-{\Delta }_r{G}^{\circ }}{RT}=\frac{45542}{8.314\times 298}=18.38$

   $K_c=e^{18.38}=9.62\times 10^7$

   Both values match NCERT's printed answer key. A useful check on the method: the shortcut $\log K_c=n{E}_{\text{cell}}^{\circ }/0.059$ gives $\log K_c=8.00$, hence $K_c\approx 1\times 10^8$, which is close but not identical, since $0.059$ is itself a rounded value of $2.303RT/F=0.05914$. Work through ${\Delta }_r{G}^{\circ }$ when the answer is to be quoted to three significant figures.

   ✦ ${\Delta }_r{G}^{\circ }=-45.54$ kJ mol${}^{-1}$, $K_c=9.62\times 10^7$

7. 2.72 marksNCERT Intext Questions, Chapter 2

   Why does the conductivity of a solution decrease with dilution?

   Hint. Conductivity is defined per unit volume, not per mole of electrolyte.

   Conductivity $\kappa$ is the conductance of a solution held between two electrodes of unit area separated by unit distance, so it is a property of one unit volume of the solution.

   On dilution the total number of ions from a fixed amount of electrolyte does not fall, and for a weak electrolyte it actually rises as the degree of dissociation increases. What falls is the number of ions contained in each unit volume, because the same ions are now spread through more solvent.

   Since current through that unit volume is carried only by the ions inside it, fewer charge carriers per unit volume means a smaller $\kappa$. This is why conductivity decreases on dilution for both strong and weak electrolytes.

   The contrast with molar conductivity is the point of the section: ${\Lambda }_m=\kappa /c$ refers to whatever volume contains one mole of electrolyte, and that volume grows on dilution faster than $\kappa$ falls, so ${\Lambda }_m$ increases while $\kappa$ decreases.

   ✦ Conductivity is measured per unit volume, and dilution reduces the number of ions in each unit volume even though the total number of ions does not fall

8. 2.83 marksNCERT Intext Questions, Chapter 2

   Suggest a way to determine the ${\Lambda }_m^{\circ }$ value of water.

   Hint. Water is an extremely weak electrolyte, so extrapolation is impossible; use Kohlrausch's law of independent migration of ions.

   Water dissociates only to about $10^{-7}$ M in H${}^{+}$ and OH${}^{-}$, so its conductivity is far too low to measure reliably, and ${\Lambda }_m$ cannot be extrapolated to zero concentration the way it can for a strong electrolyte. Kohlrausch's law of independent migration of ions supplies the route.

   Water behaves as the weak electrolyte ${\text{H}}_2\text{O}\rightleftharpoons {\text{H}}^{+}+{\text{OH}}^{-}$, so

   ${\Lambda }_m^{\circ }({\text{H}}_2\text{O})={\lambda }_{{\text{H}}^{+}}^{\circ }+{\lambda }_{{\text{OH}}^{-}}^{\circ }$

   Using Table 2.4:

   ${\Lambda }_m^{\circ }({\text{H}}_2\text{O})=349.6+199.1=548.7{\text{S cm}}^2{\text{mol}}^{-1}$

   The same number can be assembled from three strong electrolytes whose ${\Lambda }_m^{\circ }$ values are measurable directly by extrapolation:

   ${\Lambda }_m^{\circ }({\text{H}}_2\text{O})={\Lambda }_m^{\circ }(\text{HCl})+{\Lambda }_m^{\circ }(\text{NaOH})-{\Lambda }_m^{\circ }(\text{NaCl})$

   because the Na${}^{+}$ and Cl${}^{-}$ contributions cancel, leaving exactly ${\lambda }_{{\text{H}}^{+}}^{\circ }+{\lambda }_{{\text{OH}}^{-}}^{\circ }$.

   ✦ Apply Kohlrausch's law: ${\Lambda }_m^{\circ }({\text{H}}_2\text{O})={\lambda }_{{\text{H}}^{+}}^{\circ }+{\lambda }_{{\text{OH}}^{-}}^{\circ }=349.6+199.1=548.7$ S cm${}^2$ mol${}^{-1}$, or equivalently ${\Lambda }_m^{\circ }(\text{HCl})+{\Lambda }_m^{\circ }(\text{NaOH})-{\Lambda }_m^{\circ }(\text{NaCl})$

9. 2.93 marksNCERT Intext Questions, Chapter 2

   The molar conductivity of 0.025 mol L${}^{-1}$ methanoic acid is 46.1 S cm${}^2$ mol${}^{-1}$. Calculate its degree of dissociation and dissociation constant. Given ${\lambda }^{\circ }({\text{H}}^{+})=349.6$ S cm${}^2$ mol${}^{-1}$ and ${\lambda }^{\circ }({\text{HCOO}}^{-})=54.6$ S cm${}^2$ mol${}^{-1}$.

   Hint. Build $\Lambda^{\circ}_m$ from the two ionic values, then $\alpha=\Lambda_m/\Lambda^{\circ}_m$ and $K_a=c\alpha^2/(1-\alpha)$.

   First assemble the limiting molar conductivity by Kohlrausch's law:

   ${\Lambda }_m^{\circ }(\text{HCOOH})={\lambda }_{{\text{H}}^{+}}^{\circ }+{\lambda }_{{\text{HCOO}}^{-}}^{\circ }=349.6+54.6=404.2{\text{S cm}}^2{\text{mol}}^{-1}$

   The two ionic values are supplied in the question because Table 2.4 in the chapter lists CH${}_3$COO${}^{-}$ but not HCOO${}^{-}$.

   The degree of dissociation is the ratio of the measured to the limiting value:

   $\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}=\frac{46.1}{404.2}=0.114$

   So only about 11.4% of the acid is ionised at this concentration. Then

   $K_a=\frac{c{\alpha }^2}{1-\alpha }=\frac{0.025\times (0.114{)}^2}{1-0.114}=\frac{0.025\times 0.0130}{0.886}$

   $K_a=3.67\times 10^{-4}{\text{mol L}}^{-1}$

   Both values agree with NCERT's printed answer key. For comparison, acetic acid has $K_a\approx 1.8\times 10^{-5}$, so methanoic acid is about twenty times the stronger of the two.

   ✦ $\alpha =0.114$ and $K_a=3.67\times 10^{-4}$ mol L${}^{-1}$

10. 2.103 marksNCERT Intext Questions, Chapter 2

    If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?

    Hint. Find the total charge from $Q=It$, convert to moles of electrons with the Faraday constant, then multiply by Avogadro's number.

    Convert the time to seconds first:

    $t=2\times 60\times 60=7200\text{s}$

    $Q=It=0.5\times 7200=3600\text{C}$

    One mole of electrons carries 96487 C, so

    $n(e^{-})=\frac{3600}{96487}=0.0373\text{mol}$

    Multiplying by the Avogadro constant:

    $N=0.0373\times 6.022\times 10^{23}=2.25\times 10^{22}\text{electrons}$

    The number is enormous yet the current is modest, which is a useful reminder of how small an individual electronic charge ($1.6\times 10^{-19}$ C) is. Note also that this question is about a metallic conductor, so no chemical change accompanies the flow; the same charge passed through an electrolyte would deposit or dissolve matter at the electrodes.

    ✦ $2.25\times 10^{22}$ electrons

11. 2.112 marksNCERT Intext Questions, Chapter 2

    Suggest a list of metals that are extracted electrolytically.

    Hint. These are metals so electropositive that no ordinary chemical reducing agent will reduce their cations.

    Electrolytic extraction is used for metals whose cations sit near the bottom of Table 2.1, that is, metals with strongly negative standard electrode potentials. For these, no cheap chemical reducing agent such as carbon or carbon monoxide is powerful enough, so electrons must be forced in from an external supply.

    The common examples are:

    • Sodium and magnesium, obtained by electrolysis of their fused chlorides
    • Aluminium, obtained by electrolysis of alumina dissolved in molten cryolite
    • Potassium, calcium and lithium, obtained by electrolysis of their fused salts

    The electrolyte is molten rather than aqueous in every one of these cases, because in water the metal ion would never be reduced in preference to H${}^{+}$.

    Copper appears in a related but distinct process: it is extracted chemically and then refined electrolytically, with impure copper as the anode and pure copper deposited at the cathode.

    ✦ Na, K, Mg, Ca, Al and Li, all obtained by electrolysis of their fused salts; copper is electrolytically refined rather than extracted

12. 2.123 marksNCERT Intext Questions, Chapter 2

    Consider the reaction: ${\text{Cr}}_2{\text{O}}_7^{2-}+14{\text{H}}^{+}+6e^{-}\to 2{\text{Cr}}^{3+}+7{\text{H}}_2\text{O}$. What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr${}_2$O${}_7^{2-}$?

    Hint. The balanced half-equation already states how many moles of electrons one mole of dichromate needs.

    The half-equation shows that six moles of electrons are consumed per mole of Cr${}_2$O${}_7^{2-}$, which follows from the oxidation state change: chromium goes from $+6$ in the dichromate ion to $+3$ in Cr${}^{3+}$, a drop of 3 per chromium atom, and there are two chromium atoms.

    $Q=6\times F=6\times 96487{\text{C mol}}^{-1}$

    $Q=5.79\times 10^5\text{C}$

    No other data are needed, since the stoichiometric coefficient of the electron is the whole answer. If the approximate value $F\approx 96500$ C mol${}^{-1}$ is used instead, the result is $5.79\times 10^5$ C as well.

    ✦ $Q=6F=5.79\times 10^5$ C

13. 2.133 marksNCERT Intext Questions, Chapter 2

    Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.

    Hint. On recharge every discharge reaction runs backwards, so start by writing the discharge reactions and reverse them.

    During discharge the lead storage battery runs the reactions

    Anode: $\text{Pb(s)}+{\text{SO}}_4^{2-}(\text{aq})\to {\text{PbSO}}_4(\text{s})+2e^{-}$

    Cathode: ${\text{PbO}}_2(\text{s})+{\text{SO}}_4^{2-}(\text{aq})+4{\text{H}}^{+}(\text{aq})+2e^{-}\to {\text{PbSO}}_4(\text{s})+2{\text{H}}_2\text{O(l)}$

    so the overall discharge reaction is

    $\text{Pb(s)}+{\text{PbO}}_2(\text{s})+2{\text{H}}_2{\text{SO}}_4(\text{aq})\to 2{\text{PbSO}}_4(\text{s})+2{\text{H}}_2\text{O(l)}$

    Recharging applies an external voltage that reverses this entirely:

    $2{\text{PbSO}}_4(\text{s})+2{\text{H}}_2\text{O(l)}\to \text{Pb(s)}+{\text{PbO}}_2(\text{s})+2{\text{H}}_2{\text{SO}}_4(\text{aq})$

    The materials involved are therefore lead sulphate, which is stripped from both plates; water, which is consumed; lead, which is re-formed on the electrode that was the anode during discharge; lead dioxide, which is re-formed on the other; and sulphuric acid, which is regenerated in the electrolyte.

    The practical consequence is worth noting: because H${}_2$SO${}_4$ is consumed during discharge and regenerated on charging, the specific gravity of the electrolyte is a direct measure of the battery's state of charge, which is exactly how a garage tests a car battery with a hydrometer.

    ✦ Recharging reverses the discharge reaction: $2{\text{PbSO}}_4(\text{s})+2{\text{H}}_2\text{O(l)}\to \text{Pb(s)}+{\text{PbO}}_2(\text{s})+2{\text{H}}_2{\text{SO}}_4(\text{aq})$, consuming PbSO${}_4$ and water while regenerating Pb, PbO${}_2$ and sulphuric acid

14. 2.142 marksNCERT Intext Questions, Chapter 2

    Suggest two materials other than hydrogen that can be used as fuels in fuel cells.

    Hint. The chapter names the fuels directly when it defines a fuel cell.

    A fuel cell is a galvanic cell in which the reactants are fed in continuously and the products removed continuously, so any substance whose combustion is strongly exothermic and which can be oxidised at an electrode is a candidate.

    The chapter names methane (CH${}_4$) and methanol (CH${}_3$OH) alongside hydrogen when it defines fuel cells. Both are oxidised at the anode while oxygen is reduced at the cathode.

    Methanol is the more practical of the two for portable devices, since it is a liquid at room temperature and so avoids the storage and compression problems that dog hydrogen. Methane has the advantage of an existing distribution network in the form of piped natural gas.

    ✦ Methane (CH${}_4$) and methanol (CH${}_3$OH)

15. 2.153 marksNCERT Intext Questions, Chapter 2

    Explain how rusting of iron is envisaged as setting up of an electrochemical cell.

    Hint. Two different spots on the same piece of iron act as the two electrodes, and the water film is the electrolyte.

    Rusting is not a simple direct reaction of iron with oxygen. It requires both water and air, and the chapter treats it as a miniature galvanic cell set up on the metal surface itself.

    At one spot on the iron, oxidation occurs and that spot behaves as the anode:

    $2\text{Fe(s)}\to 2{\text{Fe}}^{2+}(\text{aq})+4e^{-},E_{{\text{Fe}}^{2+}/\text{Fe}}^{\circ }=-0.44\text{V}$

    The electrons travel through the metal itself to another spot, which behaves as the cathode, where dissolved oxygen is reduced in the presence of H${}^{+}$ supplied mainly by carbonic acid formed from atmospheric CO${}_2$:

    ${\text{O}}_2(g)+4{\text{H}}^{+}(\text{aq})+4e^{-}\to 2{\text{H}}_2\text{O(l)},E^{\circ }=+1.23\text{V}$

    The overall cell reaction is

    $2\text{Fe(s)}+{\text{O}}_2(g)+4{\text{H}}^{+}(\text{aq})\to 2{\text{Fe}}^{2+}(\text{aq})+2{\text{H}}_2\text{O(l)},E_{\text{cell}}^{\circ }=1.67\text{V}$

    The thin film of moisture on the surface is the electrolyte and the metal is its own external circuit, so every requirement of a galvanic cell is met. The Fe${}^{2+}$ formed is then oxidised further by atmospheric oxygen to hydrated ferric oxide, Fe${}_2$O${}_3\cdot x$H${}_2$O, which is the flaky brown solid we call rust.

    This picture also explains the cure: attaching a block of a more easily oxidised metal such as magnesium or zinc makes that metal the anode instead, so the iron is forced to act as cathode and cannot dissolve. That is sacrificial protection.

    ✦ Iron acts as its own galvanic cell, with an anodic spot oxidising Fe to Fe${}^{2+}$, a cathodic spot reducing O${}_2$ in the presence of H${}^{+}$, the moisture film as electrolyte and the metal as external circuit, giving $E_{\text{cell}}^{\circ }=1.67$ V

Solutions written by the tuition.in editorial team and checked against lech102.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.