Exercises — Electrochemistry

1. 2.12 marksNCERT Exercises, Chapter 2

   Arrange the following metals in the order in which they displace each other from the solution of their salts: Al, Cu, Fe, Mg and Zn.

   Hint. A metal displaces any metal whose standard electrode potential is higher than its own, so order them by increasing $E^{\circ}$.

   A metal displaces another metal from solution when it is the stronger reducing agent, which means it has the more negative standard electrode potential. Reading the relevant entries from Table 2.1:

   Couple	$E^{\circ }$/V
   Mg${}^{2+}$/Mg	$-2.36$
   Al${}^{3+}$/Al	$-1.66$
   Zn${}^{2+}$/Zn	$-0.76$
   Fe${}^{2+}$/Fe	$-0.44$
   Cu${}^{2+}$/Cu	$+0.34$

   Each metal in this list displaces every metal below it. Magnesium displaces all four others, aluminium displaces Zn, Fe and Cu, zinc displaces Fe and Cu, and iron displaces only copper. Copper displaces none of them.

   $\text{Mg}>\text{Al}>\text{Zn}>\text{Fe}>\text{Cu}$

   As a worked check, for $\text{Zn}+{\text{Cu}}^{2+}\to {\text{Zn}}^{2+}+\text{Cu}$, $E_{\text{cell}}^{\circ }=0.34-(-0.76)=+1.10$ V, which is positive and so spontaneous; but for $\text{Cu}+{\text{Zn}}^{2+}\to {\text{Cu}}^{2+}+\text{Zn}$, $E_{\text{cell}}^{\circ }=-1.10$ V, and nothing happens.

   ✦ Mg > Al > Zn > Fe > Cu, each displacing every metal that follows it

2. 2.22 marksNCERT Exercises, Chapter 2

   Given the standard electrode potentials K${}^{+}$/K $=-2.93$ V, Ag${}^{+}$/Ag $=0.80$ V, Hg${}^{2+}$/Hg $=0.79$ V, Mg${}^{2+}$/Mg $=-2.37$ V, Cr${}^{3+}$/Cr $=-0.74$ V, arrange these metals in their increasing order of reducing power.

   Hint. Reducing power runs opposite to standard electrode potential, so the most negative $E^{\circ}$ is the strongest reducing agent.

   A metal acts as a reducing agent by losing electrons, so the more negative its standard reduction potential, the more readily it is oxidised and the stronger a reducing agent it is. Increasing reducing power therefore means decreasing $E^{\circ }$.

   Sorting the five values from the highest $E^{\circ }$ to the lowest:

   Metal	$E^{\circ }$/V
   Ag	$+0.80$
   Hg	$+0.79$
   Cr	$-0.74$
   Mg	$-2.37$
   K	$-2.93$

   Reading that column downwards gives increasing reducing power:

   $\text{Ag}<\text{Hg}<\text{Cr}<\text{Mg}<\text{K}$

   Potassium is the strongest reducing agent of the five and silver the weakest. Note that the question supplies $E_{{\text{Mg}}^{2+}/\text{Mg}}^{\circ }=-2.37$ V while Table 2.1 inside the same chapter prints $-2.36$ V; the difference is a rounding inconsistency in the book and does not change the ordering.

   ✦ Ag < Hg < Cr < Mg < K

3. 2.33 marksNCERT Exercises, Chapter 2

   Depict the galvanic cell in which the reaction $\text{Zn(s)}+2{\text{Ag}}^{+}(\text{aq})\to {\text{Zn}}^{2+}(\text{aq})+2\text{Ag(s)}$ takes place. Further show: (i) which of the electrode is negatively charged, (ii) the carriers of the current in the cell, and (iii) individual reaction at each electrode.

   Hint. Anode on the left, cathode on the right, single line for a phase boundary and double line for the salt bridge.

   Zinc is oxidised and silver ion is reduced, so zinc is the anode and silver the cathode. By the convention that the anode is written on the left:

   $\text{Zn(s)}|{\text{Zn}}^{2+}(\text{aq})\Vert {\text{Ag}}^{+}(\text{aq})|\text{Ag(s)}$

   The single vertical bars mark the metal-solution phase boundaries and the double bar marks the salt bridge.

   (i) The negatively charged electrode. The zinc electrode is negative. Zinc atoms leave the electrode as Zn${}^{2+}$ and abandon their electrons on the metal, so the zinc rod accumulates negative charge and pushes electrons out into the external circuit.

   (ii) The carriers of the current. There are two distinct carriers. In the external wire, current is carried by electrons flowing from the zinc electrode to the silver electrode. Inside the cell, current is carried by ions: Zn${}^{2+}$ and Ag${}^{+}$ migrate towards the cathode and the anions of the electrolytes migrate towards the anode, with the salt bridge supplying ions to keep both solutions electrically neutral.

   (iii) The electrode reactions.

   Anode (oxidation): $\text{Zn(s)}\to {\text{Zn}}^{2+}(\text{aq})+2e^{-}$

   Cathode (reduction): $2{\text{Ag}}^{+}(\text{aq})+2e^{-}\to 2\text{Ag(s)}$

   Adding the two half-reactions returns the overall reaction, with the two electrons cancelling.

   ✦ $\text{Zn(s)}|{\text{Zn}}^{2+}\Vert {\text{Ag}}^{+}|\text{Ag(s)}$; the zinc electrode is negative; electrons carry current in the wire and ions in the solutions and salt bridge; Zn $\to$ Zn${}^{2+}$ + 2e${}^{-}$ at the anode and 2Ag${}^{+}$ + 2e${}^{-}$ $\to$ 2Ag at the cathode

4. 2.45 marksNCERT Exercises, Chapter 2

   Calculate the standard cell potentials of galvanic cell in which the following reactions take place: (i) $2\text{Cr(s)}+3{\text{Cd}}^{2+}(\text{aq})\to 2{\text{Cr}}^{3+}(\text{aq})+3\text{Cd}$, (ii) ${\text{Fe}}^{2+}(\text{aq})+{\text{Ag}}^{+}(\text{aq})\to {\text{Fe}}^{3+}(\text{aq})+\text{Ag(s)}$. Calculate the ${\Delta }_r{G}^{\circ }$ and equilibrium constant of the reactions.

   Hint. Identify $n$ from the balanced equation, then use $\Delta_rG^{\circ}=-nFE^{\circ}_{\text{cell}}$ and $\Delta_rG^{\circ}=-RT\ln K$.

   (i) Chromium is oxidised and cadmium reduced, so Cd${}^{2+}$/Cd is the cathode. Taking $E_{{\text{Cd}}^{2+}/\text{Cd}}^{\circ }=-0.40$ V and $E_{{\text{Cr}}^{3+}/\text{Cr}}^{\circ }=-0.74$ V:

   $E_{\text{cell}}^{\circ }=-0.40-(-0.74)=+0.34\text{V}$

   Two Cr atoms each lose three electrons, so $n=6$.

   ${\Delta }_r{G}^{\circ }=-nF{E}_{\text{cell}}^{\circ }=-6\times 96487\times 0.34=-1.968\times 10^5{\text{J mol}}^{-1}=-196.8{\text{kJ mol}}^{-1}$

   $\log K=\frac{n{E}_{\text{cell}}^{\circ }}{0.0591}=\frac{6\times 0.34}{0.0591}=34.5,K=3.2\times 10^{34}$

   (ii) Silver ion is reduced and Fe${}^{2+}$ oxidised, so

   $E_{\text{cell}}^{\circ }=E_{{\text{Ag}}^{+}/\text{Ag}}^{\circ }-E_{{\text{Fe}}^{3+}/{\text{Fe}}^{2+}}^{\circ }=0.80-0.77=+0.03\text{V}$

   Only one electron is transferred, so $n=1$.

   ${\Delta }_r{G}^{\circ }=-1\times 96487\times 0.03=-2894.6{\text{J mol}}^{-1}=-2.89{\text{kJ mol}}^{-1}$

   $\log K=\frac{1\times 0.03}{0.0591}=0.507,K=3.2$

   The contrast is instructive. A cell potential of $0.34$ V with $n=6$ gives an equilibrium constant of $10^{34}$, while $0.03$ V with $n=1$ gives barely 3, so reaction (ii) hardly proceeds at all. Since $\log K$ is proportional to $n{E}_{\text{cell}}^{\circ }$, a small change in either factor moves $K$ by many powers of ten.

   One caution on the data: $E_{{\text{Cd}}^{2+}/\text{Cd}}^{\circ }$ is not listed in Table 2.1 of this chapter, so the standard value $-0.40$ V has to be brought in from outside the book.

   ✦ (i) $E_{\text{cell}}^{\circ }=+0.34$ V, ${\Delta }_r{G}^{\circ }=-196.8$ kJ mol${}^{-1}$, $K=3.2\times 10^{34}$; (ii) $E_{\text{cell}}^{\circ }=+0.03$ V, ${\Delta }_r{G}^{\circ }=-2.89$ kJ mol${}^{-1}$, $K=3.2$

5. 2.55 marksNCERT Exercises, Chapter 2

   Write the Nernst equation and emf of the following cells at 298 K: (i) $\text{Mg(s)}|{\text{Mg}}^{2+}(0.001\text{M})\Vert {\text{Cu}}^{2+}(0.0001\text{M})|\text{Cu(s)}$, (ii) $\text{Fe(s)}|{\text{Fe}}^{2+}(0.001\text{M})\Vert {\text{H}}^{+}(1\text{M})|{\text{H}}_2(g)(1\text{bar})|\text{Pt(s)}$, (iii) $\text{Sn(s)}|{\text{Sn}}^{2+}(0.050\text{M})\Vert {\text{H}}^{+}(0.020\text{M})|{\text{H}}_2(g)(1\text{bar})|\text{Pt(s)}$, (iv) $\text{Pt(s)}|{\text{Br}}^{-}(0.010\text{M})|{\text{Br}}_2(l)\Vert {\text{H}}^{+}(0.030\text{M})|{\text{H}}_2(g)(1\text{bar})|\text{Pt(s)}$.

   Hint. Read the anode off the left of the cell diagram and the cathode off the right, write the overall reaction, then apply $E=E^{\circ}-\frac{0.0591}{n}\log Q$.

   In every case the left-hand half-cell is the anode and the right-hand one the cathode, so $E_{\text{cell}}^{\circ }=E_{\text{right}}^{\circ }-E_{\text{left}}^{\circ }$.

   (i) Reaction: $\text{Mg(s)}+{\text{Cu}}^{2+}\to {\text{Mg}}^{2+}+\text{Cu(s)}$, $n=2$.

   $E_{\text{cell}}^{\circ }=0.34-(-2.36)=2.70\text{V}$

   $E=E_{\text{cell}}^{\circ }-\frac{0.0591}{2}\log \frac{[{\text{Mg}}^{2+}]}{[{\text{Cu}}^{2+}]}=2.70-0.02955\log \frac{0.001}{0.0001}$

   $E=2.70-0.02955\times 1=2.67\text{V}$

   (ii) Reaction: $\text{Fe(s)}+2{\text{H}}^{+}\to {\text{Fe}}^{2+}+{\text{H}}_2$, $n=2$, $E_{\text{cell}}^{\circ }=0-(-0.44)=0.44$ V.

   $E=0.44-\frac{0.0591}{2}\log \frac{[{\text{Fe}}^{2+}]p_{{\text{H}}_2}}{[{\text{H}}^{+}{]}^2}=0.44-0.02955\log \frac{0.001}{1}$

   $E=0.44+0.0886=0.53\text{V}$

   (iii) Reaction: $\text{Sn(s)}+2{\text{H}}^{+}\to {\text{Sn}}^{2+}+{\text{H}}_2$, $n=2$, $E_{\text{cell}}^{\circ }=0-(-0.14)=0.14$ V.

   $Q=\frac{0.050}{(0.020{)}^2}=125$

   $E=0.14-0.02955\log 125=0.14-0.062=0.078\text{V}$

   (iv) The left electrode is the bromine electrode, so it is the anode: $2{\text{Br}}^{-}\to {\text{Br}}_2+2e^{-}$. The overall reaction is $2{\text{Br}}^{-}+2{\text{H}}^{+}\to {\text{Br}}_2+{\text{H}}_2$, $n=2$.

   $E_{\text{cell}}^{\circ }=0-1.09=-1.09\text{V}$

   $Q=\frac{1}{[{\text{Br}}^{-}{]}^2[{\text{H}}^{+}{]}^2}=\frac{1}{(0.010{)}^2(0.030{)}^2}=1.11\times 10^7$

   $E=-1.09-0.02955\log (1.11\times 10^7)=-1.09-0.208=-1.30\text{V}$

   The negative emf in (iv) is the point of that part: as written the cell is not spontaneous, since bromine is a far stronger oxidising agent than H${}^{+}$. The reaction that actually runs is the reverse one, in which Br${}_2$ oxidises H${}_2$, and the spontaneous cell is the same diagram read backwards with an emf of $+1.30$ V.

   ✦ (i) 2.67 V, (ii) 0.53 V, (iii) 0.078 V, (iv) $-1.30$ V, the last being negative because the cell as written is not spontaneous

6. 2.63 marksNCERT Exercises, Chapter 2

   In the button cells widely used in watches and other devices the following reaction takes place: $\text{Zn(s)}+{\text{Ag}}_2\text{O(s)}+{\text{H}}_2\text{O(l)}\to {\text{Zn}}^{2+}(\text{aq})+2\text{Ag(s)}+2{\text{OH}}^{-}(\text{aq})$. Determine ${\Delta }_r{G}^{\circ }$ and $E^{\circ }$ for the reaction.

   Hint. Split the overall reaction into its two half-reactions, find $E^{\circ}_{\text{cell}}$, then use $\Delta_rG^{\circ}=-nFE^{\circ}_{\text{cell}}$.

   The two half-reactions are

   Anode: $\text{Zn(s)}\to {\text{Zn}}^{2+}(\text{aq})+2e^{-}$, with $E_{{\text{Zn}}^{2+}/\text{Zn}}^{\circ }=-0.76$ V

   Cathode: ${\text{Ag}}_2\text{O(s)}+{\text{H}}_2\text{O(l)}+2e^{-}\to 2\text{Ag(s)}+2{\text{OH}}^{-}(\text{aq})$, with $E^{\circ }=+0.344$ V

   so that $n=2$ and

   $E_{\text{cell}}^{\circ }=0.344-(-0.76)=1.104\text{V}$

   ${\Delta }_r{G}^{\circ }=-nF{E}_{\text{cell}}^{\circ }=-2\times 96487\times 1.104=-2.13\times 10^5{\text{J mol}}^{-1}$

   ${\Delta }_r{G}^{\circ }=-213.0{\text{kJ mol}}^{-1}$

   The value of about 1.1 V explains why silver oxide button cells are used in watches: it is close to the mercury cell's 1.35 V, and like the mercury cell the voltage stays almost constant through the cell's life because no dissolved species changes concentration appreciably.

   A data caution is needed here. Table 2.1 of this chapter contains no entry for the Ag${}_2$O/Ag couple in alkaline solution, so the value $+0.344$ V has to be supplied from a standard table outside the book. As printed, the question cannot be answered from the chapter alone.

   ✦ $E_{\text{cell}}^{\circ }=1.104$ V and ${\Delta }_r{G}^{\circ }=-213.0$ kJ mol${}^{-1}$, using $E^{\circ }({\text{Ag}}_2\text{O}/\text{Ag})=+0.344$ V, a value not printed in the chapter

7. 2.75 marksNCERT Exercises, Chapter 2

   Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

   Hint. One quantity is defined per unit volume and the other per mole; that single difference drives everything that follows.

   Conductivity. The conductivity $\kappa$ of a solution is the reciprocal of its resistivity, $\kappa =1/\rho$. Equivalently, it is the conductance of a solution held between two electrodes of unit area of cross-section separated by unit distance, so it describes one unit volume of solution. Its SI unit is S m${}^{-1}$, and S cm${}^{-1}$ is common in practice.

   Molar conductivity. The molar conductivity is defined by

   ${\Lambda }_m=\frac{\kappa }{c}$

   and is the conductance of the volume of solution that contains one mole of electrolyte, held between electrodes of unit separation. Its unit is S m${}^2$ mol${}^{-1}$, or S cm${}^2$ mol${}^{-1}$, and the two are related by $1$ S m${}^2$ mol${}^{-1}=10^4$ S cm${}^2$ mol${}^{-1}$.

   Variation of $\kappa$ with concentration. Conductivity always decreases on dilution, for strong and weak electrolytes alike. Dilution spreads the same ions through a larger volume, so each unit volume contains fewer charge carriers.

   Variation of ${\Lambda }_m$ with concentration. Molar conductivity always increases on dilution, because the volume $V$ containing one mole of electrolyte grows faster than $\kappa$ falls, and ${\Lambda }_m=\kappa V$. But strong and weak electrolytes increase in strikingly different ways.

   For a strong electrolyte the rise is gradual and follows

   ${\Lambda }_m={\Lambda }_m^{\circ }-A\sqrt{c}$

   A plot of ${\Lambda }_m$ against $\sqrt{c}$ is a straight line, and its intercept gives the limiting molar conductivity ${\Lambda }_m^{\circ }$ directly. The electrolyte is fully dissociated at all these concentrations; what changes is only the strength of interionic attraction, which weakens as the ions move further apart.

   For a weak electrolyte the rise is small at moderate concentrations and then extremely steep near infinite dilution, because here the number of ions itself is increasing as the degree of dissociation $\alpha$ rises towards 1. The curve is not linear in $\sqrt{c}$ and cannot be extrapolated, so ${\Lambda }_m^{\circ }$ for a weak electrolyte must be obtained indirectly through Kohlrausch's law of independent migration of ions.

   At any concentration $c$ the degree of dissociation of a weak electrolyte can then be read off as $\alpha ={\Lambda }_m/{\Lambda }_m^{\circ }$.

   ✦ $\kappa$ is conductance per unit volume and falls on dilution; ${\Lambda }_m=\kappa /c$ is conductance per mole and rises on dilution, gradually and linearly in $\sqrt{c}$ for strong electrolytes but steeply and non-linearly for weak ones, so only strong electrolytes allow ${\Lambda }_m^{\circ }$ by extrapolation

8. 2.82 marksNCERT Exercises, Chapter 2

   The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm${}^{-1}$. Calculate its molar conductivity.

   Hint. Use $\Lambda_m=\kappa\times1000/c$ when $\kappa$ is in S cm$^{-1}$ and $c$ in mol L$^{-1}$.

   With $\kappa$ in S cm${}^{-1}$ and the concentration in mol L${}^{-1}$, the working form of the definition is

   ${\Lambda }_m=\frac{\kappa \times 1000}{c}$

   where the factor of 1000 converts litres to cubic centimetres. Substituting:

   ${\Lambda }_m=\frac{0.0248{\text{S cm}}^{-1}\times 1000{\text{cm}}^3{\text{L}}^{-1}}{0.20{\text{mol L}}^{-1}}$

   ${\Lambda }_m=124{\text{S cm}}^2{\text{mol}}^{-1}$

   The factor of 1000 is the step most often lost. Without it the answer comes out a thousand times too small, and the units give the mistake away: $\kappa /c$ with these units is S cm${}^2$ mol${}^{-1}$ only after the litre-to-cm${}^3$ conversion is made.

   The value is consistent with Table 2.3, which gives 129.0 S cm${}^2$ mol${}^{-1}$ for 0.100 M KCl; molar conductivity falls slightly as concentration rises, so 124 at 0.20 M is exactly what is expected.

   ✦ ${\Lambda }_m=124$ S cm${}^2$ mol${}^{-1}$

9. 2.92 marksNCERT Exercises, Chapter 2

   The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 $\Omega$. What is the cell constant if conductivity of 0.001 M KCl solution at 298 K is $0.146\times 10^{-3}$ S cm${}^{-1}$?

   Hint. The cell constant is the product of measured resistance and known conductivity.

   The cell constant $G^{\ast }=l/A$ relates the measured resistance to the conductivity through $\kappa =G^{\ast }/R$, so

   $G^{\ast }=\kappa R$

   Substituting the given values:

   $G^{\ast }=0.146\times 10^{-3}{\text{S cm}}^{-1}\times 1500\Omega$

   $G^{\ast }=0.219{\text{cm}}^{-1}$

   The unit is cm${}^{-1}$ because S and $\Omega$ are reciprocals of each other and cancel, leaving only the reciprocal length. This is exactly why a standard KCl solution is used for calibration: measuring the electrode separation $l$ and area $A$ directly would be both inconvenient and unreliable, whereas one resistance reading against a solution of known conductivity fixes $G^{\ast }$ once and for all. From then on any unknown solution's conductivity follows from $\kappa =G^{\ast }/R$.

   ✦ $G^{\ast }=0.219$ cm${}^{-1}$

10. 2.105 marksNCERT Exercises, Chapter 2

    The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:

    Concentration/M	0.001	0.010	0.020	0.050	0.100
    $10^2\times \kappa$/S m${}^{-1}$	1.237	11.85	23.15	55.53	106.74

    Calculate ${\Lambda }_m$ for all concentrations and draw a plot between ${\Lambda }_m$ and $c^{1/2}$. Find the value of ${\Lambda }_m^{\circ }$.

    Hint. The table gives $10^{2}\kappa$ in S m$^{-1}$, so divide by 100 to get $\kappa$ in S m$^{-1}$ and again by 100 to reach S cm$^{-1}$.

    Read the table carefully first. The row is $10^2\times \kappa$ in S m${}^{-1}$, so for the first entry $\kappa =1.237\times 10^{-2}$ S m${}^{-1}$. Converting to S cm${}^{-1}$ divides by a further 100, giving $\kappa =1.237\times 10^{-4}$ S cm${}^{-1}$. Then ${\Lambda }_m=\kappa \times 1000/c$:

    ${\Lambda }_m=\frac{1.237\times 10^{-4}\times 1000}{0.001}=123.7{\text{S cm}}^2{\text{mol}}^{-1}$

    Repeating for every column:

    $c$/M	$\kappa$/S cm${}^{-1}$	$\sqrt{c}$/M${}^{1/2}$	${\Lambda }_m$/S cm${}^2$ mol${}^{-1}$
    0.001	$1.237\times 10^{-4}$	0.0316	123.7
    0.010	$1.185\times 10^{-3}$	0.1000	118.5
    0.020	$2.315\times 10^{-3}$	0.1414	115.75
    0.050	$5.553\times 10^{-3}$	0.2236	111.06
    0.100	$1.0674\times 10^{-2}$	0.3162	106.74

    Plotting ${\Lambda }_m$ on the $y$-axis against $\sqrt{c}$ on the $x$-axis gives a near-straight line falling from left to right, which is the behaviour ${\Lambda }_m={\Lambda }_m^{\circ }-A\sqrt{c}$ predicts for a strong electrolyte. Extending the line back to $\sqrt{c}=0$ gives the intercept.

    A least-squares line through all five points gives an intercept of about $124.7$ S cm${}^2$ mol${}^{-1}$ and a slope of $-59$. Using only the two most dilute points, the intercept rises to $126.1$.

    That drift is worth understanding. The relation ${\Lambda }_m={\Lambda }_m^{\circ }-A\sqrt{c}$ is a limiting law, valid only at low concentration, so including the 0.1 M point bends the fit and pulls the intercept down. Kohlrausch's law gives the true value independently:

    ${\Lambda }_m^{\circ }(\text{NaCl})={\lambda }_{{\text{Na}}^{+}}^{\circ }+{\lambda }_{{\text{Cl}}^{-}}^{\circ }=50.1+76.3=126.4{\text{S cm}}^2{\text{mol}}^{-1}$

    and the graphical intercept converges on exactly that figure as the fit is restricted to more dilute points.

    ✦ ${\Lambda }_m$ = 123.7, 118.5, 115.75, 111.06 and 106.74 S cm${}^2$ mol${}^{-1}$ respectively; the plot against $\sqrt{c}$ is linear and extrapolates to ${\Lambda }_m^{\circ }\approx 124$-$126$ S cm${}^2$ mol${}^{-1}$, against the Kohlrausch value of 126.4

11. 2.113 marksNCERT Exercises, Chapter 2

    Conductivity of 0.00241 M acetic acid is $7.896\times 10^{-5}$ S cm${}^{-1}$. Calculate its molar conductivity. If ${\Lambda }_m^{\circ }$ for acetic acid is 390.5 S cm${}^2$ mol${}^{-1}$, what is its dissociation constant?

    Hint. Get $\Lambda_m$ first, then $\alpha=\Lambda_m/\Lambda^{\circ}_m$, then $K_a=c\alpha^{2}/(1-\alpha)$.

    The molar conductivity comes straight from the definition:

    ${\Lambda }_m=\frac{\kappa \times 1000}{c}=\frac{7.896\times 10^{-5}\times 1000}{0.00241}=32.76{\text{S cm}}^2{\text{mol}}^{-1}$

    Acetic acid is a weak electrolyte, so the ratio of the measured to the limiting molar conductivity gives the degree of dissociation:

    $\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}=\frac{32.76}{390.5}=0.0839$

    Only about 8.4% of the acid is ionised. The dissociation constant then follows from Ostwald's dilution law:

    $K_a=\frac{c{\alpha }^2}{1-\alpha }=\frac{0.00241\times (0.0839{)}^2}{1-0.0839}=\frac{1.696\times 10^{-5}}{0.9161}$

    $K_a=1.85\times 10^{-5}{\text{mol L}}^{-1}$

    This is the accepted value for acetic acid, and the chapter's own Example 2.9 obtains $1.78\times 10^{-5}$ from a different concentration, so the two agree to within experimental scatter. Note that $\alpha$ is small but not negligible: dropping the $(1-\alpha )$ term would give $1.70\times 10^{-5}$, about 8% low.

    ✦ ${\Lambda }_m=32.76$ S cm${}^2$ mol${}^{-1}$, $\alpha =0.0839$ and $K_a=1.85\times 10^{-5}$ mol L${}^{-1}$

12. 2.123 marksNCERT Exercises, Chapter 2

    How much charge is required for the following reductions: (i) 1 mol of Al${}^{3+}$ to Al? (ii) 1 mol of Cu${}^{2+}$ to Cu? (iii) 1 mol of MnO${}_4^{-}$ to Mn${}^{2+}$?

    Hint. Count the electrons per formula unit from the change in oxidation state, then multiply by the Faraday constant.

    Each part reduces to counting electrons and multiplying by $F=96487$ C mol${}^{-1}$.

    (i) ${\text{Al}}^{3+}+3e^{-}\to \text{Al}$, so 3 mol of electrons are needed.

    $Q=3F=3\times 96487=2.89\times 10^5\text{C}$

    (ii) ${\text{Cu}}^{2+}+2e^{-}\to \text{Cu}$, so 2 mol of electrons are needed.

    $Q=2F=2\times 96487=1.93\times 10^5\text{C}$

    (iii) In MnO${}_4^{-}$ manganese is in the $+7$ state and in Mn${}^{2+}$ it is $+2$, a change of 5, so

    ${\text{MnO}}_4^{-}+8{\text{H}}^{+}+5e^{-}\to {\text{Mn}}^{2+}+4{\text{H}}_2\text{O}$

    $Q=5F=5\times 96487=4.82\times 10^5\text{C}$

    Part (iii) is the only one that needs thought, since the electron count is not visible in the formula and has to be worked out from oxidation states. The oxygen atoms are spectators here; they leave as water without changing oxidation state.

    ✦ (i) $3F=2.89\times 10^5$ C, (ii) $2F=1.93\times 10^5$ C, (iii) $5F=4.82\times 10^5$ C

13. 2.133 marksNCERT Exercises, Chapter 2

    How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl${}_2$? (ii) 40.0 g of Al from molten Al${}_2$O${}_3$?

    Hint. Convert the mass to moles, then multiply by the number of electrons each ion needs.

    The answer is wanted in faradays, so no conversion to coulombs is needed; just count moles of electrons.

    (i) ${\text{Ca}}^{2+}+2e^{-}\to \text{Ca}$, and $M(\text{Ca})=40$ g mol${}^{-1}$.

    $n(\text{Ca})=\frac{20.0}{40}=0.5\text{mol}$

    $\text{Electricity}=0.5\times 2=1.0F$

    (ii) ${\text{Al}}^{3+}+3e^{-}\to \text{Al}$, and $M(\text{Al})=27$ g mol${}^{-1}$.

    $n(\text{Al})=\frac{40.0}{27}=1.48\text{mol}$

    $\text{Electricity}=1.48\times 3=4.44F$

    The three electrons per aluminium atom, combined with its low atomic mass, are why aluminium extraction is so power-hungry: producing a kilogram of aluminium takes over three times the charge that a kilogram of a divalent metal of similar mass would need.

    ✦ (i) 1.0 F, (ii) 4.44 F

14. 2.143 marksNCERT Exercises, Chapter 2

    How much electricity is required in coulomb for the oxidation of (i) 1 mol of H${}_2$O to O${}_2$? (ii) 1 mol of FeO to Fe${}_2$O${}_3$?

    Hint. Assign oxidation states before and after, and count how many electrons one mole of the named reactant loses.

    (i) Oxygen goes from $-2$ in water to $0$ in O${}_2$, losing 2 electrons per oxygen atom. Writing the half-reaction for one mole of water:

    ${\text{H}}_2\text{O}\to \frac{1}{2}{\text{O}}_2+2{\text{H}}^{+}+2e^{-}$

    $Q=2F=2\times 96487=1.93\times 10^5\text{C}$

    (ii) In FeO iron is $+2$ and in Fe${}_2$O${}_3$ it is $+3$, so each iron atom loses just one electron. One mole of FeO contains one mole of iron:

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

    $Q=1F=96487\text{C}$

    Part (ii) is a trap for anyone who counts from the formula Fe${}_2$O${}_3$ and doubles the answer. The question asks about one mole of FeO, not one mole of Fe${}_2$O${}_3$, and the electron count belongs to the iron atoms, of which there is exactly one mole.

    ✦ (i) $2F=1.93\times 10^5$ C, (ii) $1F=96487$ C

15. 2.153 marksNCERT Exercises, Chapter 2

    A solution of Ni(NO${}_3$)${}_2$ is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

    Hint. Find $Q=It$ in coulombs, convert to moles of electrons, then halve because Ni$^{2+}$ needs two electrons.

    Convert the time to seconds and find the charge:

    $t=20\times 60=1200\text{s}$

    $Q=It=5\times 1200=6000\text{C}$

    Moles of electrons passed:

    $n(e^{-})=\frac{6000}{96487}=0.0622\text{mol}$

    The cathode reaction is ${\text{Ni}}^{2+}+2e^{-}\to \text{Ni}$, so two electrons deposit one nickel atom:

    $n(\text{Ni})=\frac{0.0622}{2}=0.0311\text{mol}$

    With $M(\text{Ni})=58.7$ g mol${}^{-1}$:

    $m=0.0311\times 58.7=1.83\text{g}$

    The nitrate ion plays no part at the cathode; with platinum electrodes it is oxygen from water that is released at the anode, since NO${}_3^{-}$ is not oxidised under these conditions.

    ✦ 1.83 g of nickel

16. 2.165 marksNCERT Exercises, Chapter 2

    Three electrolytic cells A, B, C containing solutions of ZnSO${}_4$, AgNO${}_3$ and CuSO${}_4$ respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

    Hint. Cells in series pass the same charge, so solve cell B first and carry that charge into the other two.

    Cells in series carry the same current for the same time, so exactly the same charge passes through all three. Cell B is the one with data, so start there.

    With $M(\text{Ag})=108$ g mol${}^{-1}$ and ${\text{Ag}}^{+}+e^{-}\to \text{Ag}$ needing one electron:

    $n(\text{Ag})=\frac{1.45}{108}=0.01343\text{mol}=n(e^{-})$

    $Q=0.01343\times 96487=1295\text{C}$

    $t=\frac{Q}{I}=\frac{1295}{1.5}=863.6\text{s}=14.4\text{minutes}$

    The same 0.01343 mol of electrons flows through cells A and C, but both Cu${}^{2+}$ and Zn${}^{2+}$ are divalent, so each metal takes two electrons per atom:

    $n(\text{Cu})=n(\text{Zn})=\frac{0.01343}{2}=0.006713\text{mol}$

    With $M(\text{Cu})=63.5$ and $M(\text{Zn})=65.4$ g mol${}^{-1}$:

    $m(\text{Cu})=0.006713\times 63.5=0.426\text{g}$

    $m(\text{Zn})=0.006713\times 65.4=0.439\text{g}$

    This arrangement is Faraday's second law made visible: the same quantity of electricity liberates masses in the ratio of the chemical equivalents, which is why the silver figure alone is enough to fix the other two.

    ✦ $t=863.6$ s, about 14.4 minutes; 0.426 g of copper and 0.439 g of zinc are deposited

17. 2.175 marksNCERT Exercises, Chapter 2

    Using the standard electrode potentials given in Table 2.1, predict if the reaction between the following is feasible: (i) Fe${}^{3+}$(aq) and I${}^{-}$(aq), (ii) Ag${}^{+}$(aq) and Cu(s), (iii) Fe${}^{3+}$(aq) and Br${}^{-}$(aq), (iv) Ag(s) and Fe${}^{3+}$(aq), (v) Br${}_2$(aq) and Fe${}^{2+}$(aq).

    Hint. In each pair decide which species is being reduced, then check whether $E^{\circ}_{\text{cathode}}-E^{\circ}_{\text{anode}}$ is positive.

    A reaction is feasible when $E_{\text{cell}}^{\circ }>0$, since ${\Delta }_r{G}^{\circ }=-nF{E}_{\text{cell}}^{\circ }$ is then negative. The values needed from Table 2.1 are $E_{{\text{Fe}}^{3+}/{\text{Fe}}^{2+}}^{\circ }=0.77$ V, $E_{{\text{I}}_2/{\text{I}}^{-}}^{\circ }=0.54$ V, $E_{{\text{Br}}_2/{\text{Br}}^{-}}^{\circ }=1.09$ V, $E_{{\text{Ag}}^{+}/\text{Ag}}^{\circ }=0.80$ V and $E_{{\text{Cu}}^{2+}/\text{Cu}}^{\circ }=0.34$ V.

    (i) Fe${}^{3+}$ oxidises I${}^{-}$: $E_{\text{cell}}^{\circ }=0.77-0.54=+0.23$ V. Feasible. The reaction is $2{\text{Fe}}^{3+}+2{\text{I}}^{-}\to 2{\text{Fe}}^{2+}+{\text{I}}_2$, and it is the basis of the iodometric estimation of iron(III).

    (ii) Ag${}^{+}$ oxidises Cu: $E_{\text{cell}}^{\circ }=0.80-0.34=+0.46$ V. Feasible, giving $\text{Cu}+2{\text{Ag}}^{+}\to {\text{Cu}}^{2+}+2\text{Ag}$.

    (iii) Fe${}^{3+}$ would have to oxidise Br${}^{-}$: $E_{\text{cell}}^{\circ }=0.77-1.09=-0.32$ V. Not feasible. Bromine is the stronger oxidising agent, so bromide holds on to its electron.

    (iv) Fe${}^{3+}$ would have to oxidise Ag: $E_{\text{cell}}^{\circ }=0.77-0.80=-0.03$ V. Not feasible, though only just. The margin is so small that under non-standard conditions, with a large excess of Fe${}^{3+}$ or something to remove Ag${}^{+}$ from solution, the Nernst equation could tip it the other way.

    (v) Br${}_2$ oxidises Fe${}^{2+}$: $E_{\text{cell}}^{\circ }=1.09-0.77=+0.32$ V. Feasible, giving ${\text{Br}}_2+2{\text{Fe}}^{2+}\to 2{\text{Br}}^{-}+2{\text{Fe}}^{3+}$. This is (iii) run backwards, as it must be.

    A note on the printed question: it refers to "Table 3.1", but the table of standard electrode potentials in this chapter is Table 2.1. The reference is a leftover from an earlier edition in which Electrochemistry was Unit 3.

    ✦ Feasible for (i), (ii) and (v); not feasible for (iii) and (iv), the latter by only $-0.03$ V

18. 2.185 marksNCERT Exercises, Chapter 2

    Predict the products of electrolysis in each of the following: (i) an aqueous solution of AgNO${}_3$ with silver electrodes, (ii) an aqueous solution of AgNO${}_3$ with platinum electrodes, (iii) a dilute solution of H${}_2$SO${}_4$ with platinum electrodes, (iv) an aqueous solution of CuCl${}_2$ with platinum electrodes.

    Hint. At the cathode the species with the highest reduction potential wins; at the anode the species with the lowest oxidation potential wins, unless overpotential intervenes.

    In every aqueous case there are extra competitors, H${}^{+}$ and OH${}^{-}$ from water, alongside the ions of the salt. The species reduced at the cathode is the one with the highest $E^{\circ }$, and the species oxidised at the anode is the one that is oxidised most easily.

    (i) AgNO${}_3$ with silver electrodes. The anode is reactive here, so it dissolves in preference to anything in solution.

    Cathode: ${\text{Ag}}^{+}+e^{-}\to \text{Ag}$   Anode: $\text{Ag}\to {\text{Ag}}^{+}+e^{-}$

    Silver dissolves from the anode and plates onto the cathode; the concentration of the solution is unchanged. This is exactly the arrangement used for electroplating.

    (ii) AgNO${}_3$ with platinum electrodes. Platinum is inert, so it cannot dissolve.

    Cathode: ${\text{Ag}}^{+}+e^{-}\to \text{Ag}$, since $E^{\circ }=0.80$ V beats the reduction of water

    Anode: $2{\text{H}}_2\text{O}\to {\text{O}}_2+4{\text{H}}^{+}+4e^{-}$, since NO${}_3^{-}$ is not oxidised

    Silver is deposited and oxygen is evolved, and the solution becomes acidic.

    (iii) Dilute H${}_2$SO${}_4$ with platinum electrodes.

    Cathode: $2{\text{H}}^{+}+2e^{-}\to {\text{H}}_2$   Anode: $2{\text{H}}_2\text{O}\to {\text{O}}_2+4{\text{H}}^{+}+4e^{-}$

    Hydrogen and oxygen are evolved in the volume ratio 2:1, which is simply the electrolysis of water. The chapter notes that at high H${}_2$SO${}_4$ concentration the anode reaction switches to $2{\text{SO}}_4^{2-}\to {\text{S}}_2{\text{O}}_8^{2-}+2e^{-}$, but the question specifies a dilute solution.

    (iv) CuCl${}_2$ with platinum electrodes.

    Cathode: ${\text{Cu}}^{2+}+2e^{-}\to \text{Cu}$, since $E^{\circ }=0.34$ V beats the reduction of water

    Anode: $2{\text{Cl}}^{-}\to {\text{Cl}}_2+2e^{-}$

    Copper is deposited and chlorine is liberated. On $E^{\circ }$ values alone water ($1.23$ V) should be oxidised in preference to chloride ($1.36$ V), but the large overpotential for oxygen evolution on platinum makes chlorine the product in practice. The same effect is what allows chlorine, rather than oxygen, to be obtained from brine.

    ✦ (i) Ag dissolves at the anode and deposits at the cathode; (ii) Ag at the cathode and O${}_2$ at the anode; (iii) H${}_2$ at the cathode and O${}_2$ at the anode; (iv) Cu at the cathode and Cl${}_2$ at the anode, chlorine winning because of oxygen overpotential

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.