Electrochemical cells Questions in English

(C) For a spontaneous process:
$1$. The Gibbs free energy change $\Delta G$ must be negative $(\Delta G < 0)$.
$2$. The equilibrium constant $K$ must be greater than $1$ $(K > 1)$,indicating that the products are favored at equilibrium.
$3$. The standard cell potential $E^o_{cell}$ must be positive $(E^o_{cell} > 0)$ for the reaction to be spontaneous under standard conditions.

(A) In a galvanic cell,the electrical work done is equal to the decrease in Gibbs free energy of the system.
$W_{electrical} = -\Delta G = nFE_{cell}$
Thus,the work done is equal to the change in free energy.

(D) fuel cell is an electrochemical device that converts the chemical energy of a fuel (like $H_2$) and an oxidant (like $O_2$) directly into electrical energy.
$1$. Fuel cells are highly efficient compared to conventional combustion engines.
$2$. They produce water as the only byproduct,making them pollution-free.
$3$. They operate continuously as long as the supply of reactants ($H_2$ and $O_2$) is maintained.
Therefore,all the given statements are correct.

(A) salt bridge typically contains an inert electrolyte like $KCl$,$KNO_3$,or $NH_4NO_3$ in an agar-agar gel.
These electrolytes are chosen because the transport numbers of their cations and anions are approximately equal,ensuring electrical neutrality.
$CH_3COOK$ (potassium acetate) is a salt of a weak acid and a strong base,which can undergo hydrolysis to produce $OH^-$ ions,potentially interfering with the electrochemical reaction or causing precipitation.
Therefore,it is not suitable for use in a salt bridge.

(A) The cell is a concentration cell where $E_{cell} = \frac{0.0591}{n} \log \frac{C_2}{C_1}$.
For $HCl$,which is a strong acid,the concentration of $H^{+}$ ions is $C_1 = 0.1 \, M$.
For acetic acid $(CH_3COOH)$,which is a weak acid,the concentration of $H^{+}$ ions is $C_2 = 0.1 \times \alpha \, M$,where $\alpha$ is the degree of dissociation.
Since $\alpha < 1$,$C_2 \neq C_1$.
Therefore,$E_{cell} = 0.0591 \log \frac{0.1 \alpha}{0.1} = 0.0591 \log \alpha$.
Since $\alpha \neq 1$,$E_{cell} \neq 0$.

(B) In the electrochemical series,$Hg$ is a noble metal located below $Cu$.
Due to its low reactivity,its oxide,$HgO$,is thermally unstable and decomposes upon heating to form mercury and oxygen gas: $2HgO(s) \xrightarrow{\Delta} 2Hg(l) + O_2(g)$.

(B) The standard reduction potentials are given as $E^{0}_{Fe^{3+}|Fe^{2+}} = 0.77 \, V$ and $E^{0}_{Fe^{2+}|Fe} = -0.44 \, V$.
Since $E^{0}_{Fe^{3+}|Fe^{2+}} > E^{0}_{Fe^{2+}|Fe}$,the reaction $2Fe^{3+} + Fe \rightarrow 3Fe^{2+}$ is spontaneous.
In this reaction,$Fe^{3+}$ acts as an oxidizing agent and gets reduced to $Fe^{2+}$,while $Fe$ acts as a reducing agent and gets oxidized to $Fe^{2+}$.
Therefore,the concentration of $Fe^{3+}$ will decrease and the concentration of $Fe^{2+}$ will increase.

(B) dry cell (Leclanché cell) consists of a zinc container that acts as an anode and a graphite rod surrounded by powdered manganese dioxide $(MnO_2)$ and carbon as the cathode.
In the cell,$MnO_2$ acts as an oxidizing agent to depolarize the cathode.

(B) The reaction occurring is: $2Ag^{+}(aq) + Cu(s) \to Cu^{2+}(aq) + 2Ag(s)$.
Since the standard reduction potential of silver $(E^o_{Ag^{+}/Ag} = +0.80 \ V)$ is higher than that of copper $(E^o_{Cu^{2+}/Cu} = +0.34 \ V)$,copper acts as the reducing agent and undergoes oxidation to form $Cu^{2+}$ ions.
The blue colour of the solution is due to the presence of hydrated $Cu^{2+}$ ions.

(A) Anode is the electrode where oxidation (loss of electrons) occurs.
In option $(A)$,the oxidation state of $Cr$ changes from $+3$ to $+6$,which is an oxidation process.
Options $(B)$ and $(C)$ represent reduction processes (gain of electrons),which occur at the cathode.

(B) In a galvanic cell representation,the left side represents the anode where oxidation occurs,and the right side represents the cathode where reduction occurs.
Given cell: $Cu_{(s)} | Cu^{2+}_{(aq)} || Hg^{2+}_{(aq)} | Hg_{(l)}$
Anode (Oxidation): $Cu_{(s)} \to Cu^{2+}_{(aq)} + 2e^-$
Cathode (Reduction): $Hg^{2+}_{(aq)} + 2e^- \to Hg_{(l)}$
Adding these two half-reactions,we get the overall cell reaction:
$Cu_{(s)} + Hg^{2+}_{(aq)} \to Cu^{2+}_{(aq)} + Hg_{(l)}$
Therefore,the correct option is $B$.

(D) The correct answer is $(D)$.
$A$ metal can displace another metal from its salt solution if it is more reactive,meaning it has a lower (more negative) standard reduction potential than the metal in the salt.
The standard reduction potentials $(E^o)$ are:
$Mg^{2+} + 2e^- \to Mg$ $(E^o = -2.37 \ V)$
$Zn^{2+} + 2e^- \to Zn$ $(E^o = -0.76 \ V)$
$Fe^{2+} + 2e^- \to Fe$ $(E^o = -0.44 \ V)$
$Cu^{2+} + 2e^- \to Cu$ $(E^o = +0.34 \ V)$
$Ag^+ + e^- \to Ag$ $(E^o = +0.80 \ V)$
Since $Mg$,$Zn$,and $Fe$ have lower reduction potentials than $Cu$,they will displace $Cu$ from $CuSO_4$ solution.
$Ag$ has a higher reduction potential $(+0.80 \ V)$ than $Cu$ $(+0.34 \ V)$,so it cannot displace $Cu$ from its solution.

(C) $fuel \ cell$ is an electrochemical device that converts the chemical energy of a fuel (like $H_2$ or $CH_4$) and an oxidizing agent (like $O_2$) directly into electricity through a redox reaction.

(C) The cell reaction is:
Anode (Oxidation): $Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^- \,;\, E^o_{ox} = -(-0.76) = 0.76 \, V$
Cathode (Reduction): $Ag_2O_{(s)} + H_2O_{(l)} + 2e^- \rightarrow 2Ag_{(s)} + 2OH^{-}_{(aq)} \,;\, E^o_{red} = 0.34 \, V$
$E^o_{\text{cell}} = E^o_{ox} + E^o_{red}$
$E^o_{\text{cell}} = 0.76 \, V + 0.34 \, V = 1.10 \, V$

(C) The relationship between Gibbs energy and cell potential is given by $\Delta G = -nFE_{cell}$.
For the reaction $\frac{2}{3} Al_2O_3 \rightarrow \frac{4}{3} Al + O_2$,the number of electrons transferred $(n)$ is calculated as follows:
$Al$ goes from $+3$ to $0$ (change of $3$ per $Al$ atom).
For $\frac{4}{3}$ moles of $Al$,$n = \frac{4}{3} \times 3 = 4$.
Given $\Delta G = +960 \ kJ \ mol^{-1} = 960000 \ J \ mol^{-1}$.
Using $\Delta G = -nFE_{cell}$:
$960000 = -4 \times 96500 \times E_{cell}$.
$E_{cell} = -\frac{960000}{386000} \approx -2.487 \ V$.
The magnitude of the potential difference required for the electrolytic reduction is approximately $2.5 \ V$.

(A) The standard Gibbs energy change is given by the formula: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$
The cell reaction is: $2 Ag^{+} + Cu \rightarrow Cu^{2+} + 2 Ag$
Here,the number of electrons transferred,$n = 2$.
Given,$E^{\circ}_{cell} = + 0.46 \ V$ and $F = 96500 \ C \ mol^{-1}$.
Substituting the values: $\Delta G^{\circ} = -2 \times 96500 \times 0.46 \ J \ mol^{-1}$
$\Delta G^{\circ} = -88780 \ J \ mol^{-1}$
Converting to $kJ \ mol^{-1}$: $\Delta G^{\circ} = -88.78 \ kJ \ mol^{-1}$
Rounding to the nearest value,we get $\Delta G^{\circ} \approx -89.0 \ kJ \ mol^{-1}$.

(A) The standard Gibbs free energy of reaction is calculated as: $\Delta G^{\circ}_{rxn} = \Sigma \Delta G^{\circ}_f (\text{products}) - \Sigma \Delta G^{\circ}_f (\text{reactants})$.
For the combustion of pentane: $C_5H_{12(g)} + 8O_{2(g)} \rightarrow 5CO_{2(g)} + 6H_2O_{(l)}$.
$\Delta G^{\circ}_{rxn} = [5 \times \Delta G^{\circ}_f(CO_2) + 6 \times \Delta G^{\circ}_f(H_2O)] - [\Delta G^{\circ}_f(C_5H_{12}) + 8 \times \Delta G^{\circ}_f(O_2)]$.
Given $\Delta G^{\circ}_f(O_2) = 0 \ kJ/mol$,$\Delta G^{\circ}_f(CO_2) = -394.4 \ kJ/mol$,$\Delta G^{\circ}_f(H_2O) = -237.2 \ kJ/mol$,and $\Delta G^{\circ}_f(C_5H_{12}) = -8.2 \ kJ/mol$.
$\Delta G^{\circ}_{rxn} = [5(-394.4) + 6(-237.2)] - [-8.2 + 0] = -1972 - 1423.2 + 8.2 = -3387 \ kJ/mol = -3387000 \ J/mol$.
In the fuel cell reaction,the number of electrons transferred $(n)$ is $32$.
Using $\Delta G^{\circ} = -nFE^{\circ}_{cell}$,where $F = 96500 \ C/mol$:
$E^{\circ}_{cell} = \frac{-\Delta G^{\circ}}{nF} = \frac{3387000}{32 \times 96500} = 1.0968 \ V$.

(B) The efficiency of a fuel cell $(\phi)$ is defined as the ratio of the useful work done (Gibbs free energy change,$\Delta G$) to the total heat energy available from the combustion of the fuel (Enthalpy change,$\Delta H$).
Efficiency $(\phi) = \frac{\Delta G}{\Delta H} \times 100$.
Ideally,fuel cells are expected to have an efficiency of $100 \%$.

(A) In an electrochemical cell representation,the left side represents the anode (oxidation) and the right side represents the cathode (reduction).
Given cell: $A \,|\, A^{+} \,(x \ M) \, ||\, B^{+} \,(y \ M) \, |\, B$
At the anode (left): $A \rightarrow A^{+} + e^-$
At the cathode (right): $B^{+} + e^- \rightarrow B$
Adding these two half-reactions,the overall cell reaction is: $A + B^{+} \rightarrow A^{+} + B$
Since the measured $emf$ is positive $(+ 0.20 \ V)$,the reaction is spontaneous in the direction written.

(B) The cell reactions are:
$(I) \ Ag \rightarrow Ag^{+} + e^-$,$E^o = -0.800\, V$
$(II) \ Ag + I^{-} \rightarrow AgI + e^-$,$E^o = 0.152\, V$
Subtracting $(I)$ from $(II)$ gives the reaction for the solubility product:
$AgI \rightarrow Ag^{+} + I^{-}$
$E^o_{cell} = E^o_{(II)} - E^o_{(I)} = 0.152 - (-0.800) = -0.952\, V$
Using the relation $E^o_{cell} = \frac{0.059}{n} \log K_{sp}$ at $25\, ^oC$:
$-0.952 = \frac{0.059}{1} \log K_{sp}$
$\log K_{sp} = -\frac{0.952}{0.059} = -16.135 \approx -16.13$

(A) The efficiency of a cell is given by $\eta = \frac{\Delta G^o}{\Delta H^o}$.
Given $\eta = 0.70$,$\Delta H^o = -300 \times 10^3 \ J$,and $n = 1$ (from the reaction $A + B^+ \to A^+ + B$),
$\Delta G^o = -nFE^o = -1 \times 96500 \times E^o$.
Substituting these into the efficiency formula:
$0.70 = \frac{-nFE^o}{\Delta H^o} = \frac{-1 \times 96500 \times E^o}{-300000}$.
$0.70 = \frac{96500 \times E^o}{300000}$.
$E^o = \frac{0.70 \times 300000}{96500} = \frac{210000}{96500} \approx 2.176 \ V$.

(A) The chemical reactions occurring during the discharge of a lead storage battery are as follows:
At the anode: $Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2e^-$
At the cathode: $PbO_2(s) + SO_4^{2-}(aq) + 4H^+(aq) + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l)$
Overall reaction: $Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)$
From the overall reaction,it is clear that:
$1$. $PbSO_4$ is produced at both electrodes.
$2$. $H_2SO_4$ is consumed.
$3$. $H_2O$ is formed.
$SO_2$ is not evolved during the discharge process. Therefore,option $A$ is incorrect.

(D) The cell reaction is: $AgCl(s) + Br^-(aq) \rightleftharpoons AgBr(s) + Cl^-(aq)$.
Since the concentrations of $KCl$ and $KBr$ are both $1 \ M$,the reaction quotient $Q = \frac{[Cl^-]}{[Br^-]} = \frac{1}{1} = 1$.
At standard conditions,the standard cell potential $E_{cell}^{0}$ is determined by the difference in standard electrode potentials of the two half-cells.
However,for this specific concentration cell setup,the standard Gibbs free energy change $\Delta G_{cell}^{0}$ is not necessarily zero,and $E_{cell}^{0}$ depends on the standard reduction potentials of $Ag/AgCl/Cl^-$ and $Ag/AgBr/Br^-$.
Given the options,$E_{cell} \neq 0$ because the standard potentials of the two electrodes are different ($E_{Ag/AgCl/Cl^-}^{0} = 0.222 \ V$ and $E_{Ag/AgBr/Br^-}^{0} = 0.071 \ V$).

(A) The anode reaction for the oxidation of methane in a fuel cell is: $CH_4 + 3H_2O \to CO_3^{2-} + 10H^+ + 8e^-$.
From the stoichiometry,$1 \ mol$ of $CH_4$ releases $8 \ mol$ of electrons.
Calculate the total moles of electrons transferred: $n_{e^-} = \frac{I \times t}{F} = \frac{96.5 \times 10000}{96500} = 10 \ mol$.
Calculate the moles of $CH_4$ required: $n_{CH_4} = \frac{1}{8} \times n_{e^-} = \frac{10}{8} = 1.25 \ mol$.
At $STP$ ($1 \ atm$ and $273 \ K$),$1 \ mol$ of gas occupies $22.4 \ L$.
Therefore,the volume of $CH_4$ required is $1.25 \times 22.4 \ L = 28 \ L$.

(A) During electrolysis,the ion with the higher standard reduction potential $(E^{\circ})$ is reduced first at the cathode.
Comparing the given values:
$E^{\circ}_{A^{2+}/A} = 0.8 \ V$
$E^{\circ}_{B^{3+}/B} = 0.6 \ V$
Since $0.8 \ V > 0.6 \ V$,the ion $A^{2+}_{(aq.)}$ has a higher reduction potential.
Therefore,$A^{2+}_{(aq.)}$ will be deposited first at the cathode.

(A) In an electrochemical cell represented as $A | A^{+} || B^{+} | B$,the left side represents the anode where oxidation occurs,and the right side represents the cathode where reduction occurs.
At the anode: $A \to A^{+} + e^-$.
At the cathode: $B^{+} + e^- \to B$.
Adding these two half-reactions gives the overall cell reaction: $A + B^{+} \to A^{+} + B$.
Since the cell potential is positive $(+0.20 \ V)$,the reaction is spontaneous in the forward direction as written.

(C) The cell reactions are:
Anode: $2Ag_{(s)} + C_2O_4^{2-} \rightarrow Ag_2C_2O_{4(s)} + 2e^-$
Cathode: $2Ag^+ + 2e^- \rightarrow 2Ag_{(s)}$
Overall reaction: $2Ag^+ + C_2O_4^{2-} \rightarrow Ag_2C_2O_{4(s)}$
$E_{cell} = E^0_{cell} - \frac{0.06}{2} \log \frac{1}{[Ag^+]^2 [C_2O_4^{2-}]}$
$E^0_{cell} = E^0_{Ag^+/Ag} - E^0_{Ag^+/Ag} = 0 \ V$
$0.264 = 0 - 0.03 \log \frac{1}{(0.5)^2 (0.02)}$
$0.264 = -0.03 \log \frac{1}{0.25 \times 0.02} = -0.03 \log \frac{1}{0.005} = -0.03 \log(200)$
$0.264 = -0.03 (\log 2 + 2) = -0.03 (0.3 + 2) = -0.03 (2.3) = -0.069 \ V$
Wait,the $E^0_{cell}$ is related to $K_{sp}$ as $E^0_{cell} = \frac{0.06}{2} \log K_{sp}$.
$0.264 = \frac{0.06}{2} \log \frac{1}{[Ag^+]^2 [C_2O_4^{2-}]} = 0.03 \log \frac{1}{K_{sp}}$
$0.264 = 0.03 \log \frac{1}{[Ag^+]^2 [C_2O_4^{2-}]}$
$8.8 = \log \frac{1}{[Ag^+]^2 [C_2O_4^{2-}]}$
Using $K_{sp} = [Ag^+]^2 [C_2O_4^{2-}]$,we have $E_{cell} = \frac{0.06}{2} \log \frac{1}{K_{sp}}$
$0.264 = 0.03 \log \frac{1}{K_{sp}} \Rightarrow 8.8 = \log \frac{1}{K_{sp}}$
$K_{sp} = 10^{-8.8} = 10^{0.2} \times 10^{-9} \approx 1.58 \times 10^{-9}$.
Re-evaluating: $E_{cell} = E^0_{cell} - 0.03 \log Q$. $E^0_{cell} = 0.03 \log K_{sp}$.
$0.264 = 0.03 \log K_{sp} - 0.03 \log \frac{1}{[Ag^+]^2 [C_2O_4^{2-}]}$
$0.264 = 0.03 \log (K_{sp} \times [Ag^+]^2 [C_2O_4^{2-}])$
$8.8 = \log (K_{sp} / (0.25 \times 0.02)) = \log (K_{sp} / 0.005)$
$10^{8.8} = K_{sp} / 0.005$ $\Rightarrow K_{sp} = 0.005 \times 10^{8.8} = 5 \times 10^{-3} \times 6.3 \times 10^8 = 3.15 \times 10^6$ (Incorrect approach).
Correct approach: $E_{cell} = 0.03 \log \frac{[Ag^+]^2 [C_2O_4^{2-}]}{K_{sp}}$
$0.264 = 0.03 \log \frac{(0.5)^2 (0.02)}{K_{sp}} = 0.03 \log \frac{0.005}{K_{sp}}$
$8.8 = \log \frac{0.005}{K_{sp}} \Rightarrow 10^{8.8} = \frac{0.005}{K_{sp}}$
$K_{sp} = 0.005 / 6.3 \times 10^8 \approx 8 \times 10^{-12}$.

(B) In an alkaline dry cell,the anode reaction involves the oxidation of the metal $(M \rightarrow M^{2+} + 2e^-)$. The standard cell potential $(E^o_{cell})$ is given by $E^o_{cell} = E^o_{cathode} - E^o_{anode}$.
Since the cathode remains the same,the change in $E^o_{cell}$ depends on the change in the anode potential $(E^o_{anode})$.
For $Zn$,$E^o_{Zn^{2+}/Zn} = -0.76 \ V$.
For $Cd$,$E^o_{Cd^{2+}/Cd} = -0.40 \ V$.
Replacing $Zn$ with $Cd$ changes the anode potential from $-0.76 \ V$ to $-0.40 \ V$.
The new $E^o_{cell}$ will be $E^o_{cathode} - (-0.40 \ V) = E^o_{cathode} + 0.40 \ V$.
The original $E^o_{cell}$ was $E^o_{cathode} - (-0.76 \ V) = E^o_{cathode} + 0.76 \ V$.
The change in emf is $(E^o_{cathode} + 0.40) - (E^o_{cathode} + 0.76) = -0.36 \ V$.
Therefore,the standard emf of the cell decreases by $0.36 \ V$.

(B) The standard Gibbs free energy change is given by the formula: $\Delta_r G^o = -n F E^o_{cell}$
First,calculate the standard cell potential: $E^o_{cell} = E^o_{cathode} - E^o_{anode} = 0.34 \, V - (-0.76 \, V) = 1.10 \, V$
Here,$n = 2$ (number of electrons transferred) and $F = 96500 \, C \, mol^{-1}$ (Faraday's constant).
Substituting the values: $\Delta_r G^o = -2 \times 96500 \, C \, mol^{-1} \times 1.10 \, V$
$\Delta_r G^o = -212300 \, J \, mol^{-1} = -2.12 \times 10^5 \, J \, mol^{-1}$

(C) Anode (Oxidation): $2Cl^{-}_{(aq)} \to Cl_{2(g)} + 2e^-$
Cathode (Reduction): $2Ag^{+}_{(aq)} + 2e^- \to 2Ag_{(s)}$
Net cell reaction: $2Cl^{-}_{(aq)} + 2Ag^{+}_{(aq)} \to 2Ag_{(s)} + Cl_{2(g)}$

(A) According to Kohlrausch's law of independent migration of ions,the equivalent conductivity at infinite dilution is the sum of the equivalent conductivities of the individual ions.
$\lambda_{eq}^\infty (Fe_2(SO_4)_3) = \lambda_{eq}^\infty (Fe^{3+}) + \lambda_{eq}^\infty (SO_4^{2-})$
Given that $\lambda_{eq}^\infty (Fe^{3+}) = \frac{1}{3} \lambda_m^\infty (Fe^{3+}) = 68 \ \Omega^{-1} \ cm^2 \ eq^{-1}$ and $\lambda_{eq}^\infty (SO_4^{2-}) = \frac{1}{2} \lambda_m^\infty (SO_4^{2-}) = 80 \ \Omega^{-1} \ cm^2 \ eq^{-1}$.
Therefore,$\lambda_{eq}^\infty (Fe_2(SO_4)_3) = 68 + 80 = 148 \ \Omega^{-1} \ cm^2 \ eq^{-1}$.

(C) For a good salt bridge,the ionic mobilities (velocities) of the cation and anion should be nearly the same.
If the velocities are significantly different,the ions would not be able to neutralize the charge buildup in both half-cells at the same rate.
This would create a liquid junction potential that restricts the flow of current.
For $KNO_3$,the ionic velocities of $K^{+}$ and $NO_3^-$ are nearly identical,making it an ideal electrolyte for a salt bridge.

(D) In the given cell representation,the left side represents the anode where oxidation occurs: $H_{2(g)} \rightarrow 2H^+_{(aq)} + 2e^-$.
The right side represents the cathode where reduction occurs: $Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}$.
Therefore,$H_2$ acts as the anode and $Cu$ acts as the cathode.

(B) During the charging of a lead storage battery,the cell acts as an electrolytic cell. The chemical reactions are the reverse of the discharging process.
At the cathode: $PbSO_{4} + 2H_{2}O \rightarrow PbO_{2} + SO_{4}^{2-} + 4H^{+} + 2e^{-}$
At the anode: $PbSO_{4} + 2e^{-} \rightarrow Pb + SO_{4}^{2-}$
Overall reaction: $2PbSO_{4} + 2H_{2}O \rightarrow Pb + PbO_{2} + 2H_{2}SO_{4}$
As shown in the overall reaction,$H_{2}SO_{4}$ is produced (regenerated) during the charging process.

(A) In a galvanic cell,oxidation occurs at the anode. The metal atoms of the anode electrode lose electrons to form metal cations,which enter the solution. Therefore,the concentration of cations in the anode solution increases as electricity flows.

(D) According to Kohlrausch's law,the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent ions.
$\Lambda^{\circ}_{NaCl} = \lambda^{\circ}_{Na^+} + \lambda^{\circ}_{Cl^-} = 126 \ S \ cm^2 \ mol^{-1}$
$\Lambda^{\circ}_{KBr} = \lambda^{\circ}_{K^+} + \lambda^{\circ}_{Br^-} = 152 \ S \ cm^2 \ mol^{-1}$
$\Lambda^{\circ}_{KCl} = \lambda^{\circ}_{K^+} + \lambda^{\circ}_{Cl^-} = 150 \ S \ cm^2 \ mol^{-1}$
To find $\Lambda^{\circ}_{NaBr} = \lambda^{\circ}_{Na^+} + \lambda^{\circ}_{Br^-}$,we use the expression:
$\Lambda^{\circ}_{NaBr} = \Lambda^{\circ}_{NaCl} + \Lambda^{\circ}_{KBr} - \Lambda^{\circ}_{KCl}$
$\Lambda^{\circ}_{NaBr} = 126 + 152 - 150 = 128 \ S \ cm^2 \ mol^{-1}$

(B) During the charging process of a lead storage battery,the chemical reactions are the reverse of the discharging process.
At the anode,$PbSO_4$ is converted to $PbO_2$ by oxidation:
$PbSO_4(s) + 2H_2O(l) \to PbO_2(s) + SO_4^{2-}(aq) + 4H^{+}(aq) + 2e^-$
At the cathode,$PbSO_4$ is reduced to $Pb$:
$PbSO_4(s) + 2e^- \to Pb(s) + SO_4^{2-}(aq)$
Therefore,the reaction at the anode is $PbSO_4 + 2H_2O \to PbO_2 + SO_4^{2-} + 4H^{+} + 2e^-$.

(C) No reaction will occur because $Zn$ is placed above $Cu$ in the electrochemical series.
Since $Cu$ is a weaker reducing agent than $Zn$,it cannot displace $Zn^{2+}$ ions from the $ZnSO_4$ solution.
Therefore,no displacement reaction takes place.

(C) The $EMF$ of the given galvanic cell is $1.1 \ V$.
When $E_{ext} < 1.1 \ V$,the cell functions as a galvanic cell,and electrons flow from the anode $(Zn)$ to the cathode $(Cu)$.
When $E_{ext} > 1.1 \ V$,the external potential overcomes the cell potential,forcing the reaction to reverse; thus,the cell acts as an electrolytic cell and electrons flow from the cathode $(Cu)$ to the anode $(Zn)$.

(D) The reaction is $\frac{2}{3}Al_2O_3 \to \frac{4}{3}Al + O_2$.
In $Al_2O_3$,the oxidation state of $Al$ is $+3$. The reduction half-reaction is $Al^{3+} + 3e^- \to Al$.
For the overall reaction,the number of electrons transferred $(n)$ is $4$ (since $\frac{4}{3} \times 3 = 4$).
Using the formula $\Delta_rG = -nFE^o$,where $\Delta_rG = 940 \times 10^3\,J\,mol^{-1}$,$n = 4$,and $F = 96500\,C\,mol^{-1}$:
$E^o = -\frac{\Delta_rG}{nF} = -\frac{940 \times 10^3}{4 \times 96500} \approx -2.43\,V$.
The potential difference required for the electrolytic reduction is the magnitude of this value,which is approximately $2.5\,V$.

(B) Metals that are more reactive than hydrogen can displace it from dilute acids. According to the Electrochemical Series $(ECS)$,$Mg$ is placed above hydrogen,meaning it has a higher oxidation potential and can reduce $H^+$ ions to $H_2$ gas. Metals like $Cu$,$Hg$,and $Ag$ are placed below hydrogen in the $ECS$ and cannot displace hydrogen from dilute $HCl$.

(B) The electrode reaction for the $Ag/AgCl_{(s)}/KCl$ electrode is given by:
$AgCl_{(s)} + e^- \to Ag_{(s)} + Cl^-_{(aq)}$
Since the potential of this electrode depends on the concentration of $Cl^-$ ions,it is reversible with respect to the $Cl^-$ ion.

(B) In a dry cell,the cathode reaction is $MnO_2 + NH_4^+ + e^- \to MnO(OH) + NH_3$.
Here,the oxidation state of $Mn$ changes from $+4$ in $MnO_2$ to $+3$ in $MnO(OH)$.
$NH_3$ gas is not liberated out; instead,it reacts with $Zn^{2+}$ ions to form a complex ion $[Zn(NH_3)_4]^{2+}$.
Therefore,the statement that $NH_3$ gas is liberated out is incorrect.

(D) The efficiency of a cell is given by the ratio of Gibbs free energy change to the enthalpy change: $\eta = \frac{\Delta G^o}{\Delta H^o}$.
Given $\eta = 0.70$ and $\Delta H^o = -551.5 \ kJ \ mol^{-1}$.
$\Delta G^o = \eta \times \Delta H^o = 0.70 \times (-551.5 \ kJ \ mol^{-1}) = -386.05 \ kJ \ mol^{-1} = -386050 \ J \ mol^{-1}$.
The cell reaction is $A_{(s)} + B^{\oplus} \longrightarrow A^{\oplus} + B_{(s)}$,so $n = 1$.
Using the relation $\Delta G^o = -nFE^o_{cell}$:
$-386050 = -1 \times 96500 \times E^o_{cell}$.
$E^o_{cell} = \frac{386050}{96500} = 4.0 \ V$.

(C) The given cell reaction is $Zn_{(s)} + 2AgCl_{(s)} \to Zn^{2+}_{(aq)} + 2Ag_{(s)} + 2Cl^{-}_{(aq)}$.
In this reaction,$Zn$ is oxidized to $Zn^{2+}$ at the anode: $Zn_{(s)} \to Zn^{2+}_{(aq)} + 2e^-$.
$AgCl$ is reduced to $Ag$ and $Cl^-$ at the cathode: $2AgCl_{(s)} + 2e^- \to 2Ag_{(s)} + 2Cl^{-}_{(aq)}$.
The cell notation is written as $\text{Anode} | \text{Anode Electrolyte} || \text{Cathode Electrolyte} | \text{Cathode}$.
For the cathode,the electrode consists of $Ag$ metal in contact with $AgCl$ solid and $Cl^-$ ions in the electrolyte.
Thus,the correct representation is $Zn_{(s)} | Zn^{2+}_{(aq)} || Cl^{-}_{(aq)} | AgCl_{(s)} | Ag_{(s)}$.

(A) reaction is spontaneous if the standard cell potential $E^o_{cell}$ is positive.
$E^o_{cell} = E^o_{cathode} - E^o_{anode}$.
For option $A$: $Ni^{2+} + Cu \longrightarrow Ni + Cu^{2+}$.
$E^o_{cell} = E^o_{Ni^{2+}|Ni} - E^o_{Cu^{2+}|Cu} = -0.25 - 0.34 = -0.59 \ V$.
Since $E^o_{cell} < 0$,this reaction is non-spontaneous and will not take place in the specified direction.
For option $B$: $2Ag^{+} + Cu \longrightarrow 2Ag + Cu^{2+}$.
$E^o_{cell} = 0.80 - 0.34 = +0.46 \ V$.
Since $E^o_{cell} > 0$,this reaction is spontaneous.
For option $C$: $Zn + Cu^{2+} \longrightarrow Zn^{2+} + Cu$.
$E^o_{cell} = 0.34 - (-0.76) = +1.10 \ V$.
Since $E^o_{cell} > 0$,this reaction is spontaneous.
For option $D$: $2H^{+} + Zn \longrightarrow H_2 + Zn^{2+}$.
$E^o_{cell} = 0.00 - (-0.76) = +0.76 \ V$.
Since $E^o_{cell} > 0$,this reaction is spontaneous.

(C) At anode: $2Cl^{-}_{(1 \, M)} \rightarrow Cl_{2(P_1)} + 2e^-$
At cathode: $Cl_{2(P_2)} + 2e^- \rightarrow 2Cl^{-}_{(1 \, M)}$
Net reaction: $Cl_{2(P_2)} \rightarrow Cl_{2(P_1)}$
$E_{cell} = E^0_{cell} - \frac{0.0591}{2} \log \frac{P_1}{P_2}$
Since $E^0_{cell} = 0$ for a concentration cell,$E_{cell} = - \frac{0.0591}{2} \log \frac{P_1}{P_2} = \frac{0.0591}{2} \log \frac{P_2}{P_1}$
For a spontaneous process,$E_{cell} > 0$,which implies $\log \frac{P_2}{P_1} > 0$,so $\frac{P_2}{P_1} > 1$,which means $P_2 > P_1$ or $P_1 < P_2$.

(B) Let us analyze each statement:
$A$: Mercury cell is a primary cell that provides a constant potential of approximately $1.35 \ V$ throughout its life. This is a correct statement.
$B$: In an electrolytic cell,reduction occurs at the cathode and oxidation occurs at the anode. In a galvanic cell,oxidation also occurs at the anode. However,the statement claims reduction occurs at the anode in an electrolytic cell,which is incorrect.
$C$: Galvanised iron is coated with a layer of zinc. Zinc acts as a sacrificial anode and prevents the iron from rusting. This is a correct statement.
$D$: During the recharging process,a lead storage battery acts as an electrolytic cell where electrical energy is converted into chemical energy. This is a correct statement.
Therefore,the incorrect statement is $B$.

(C) An electrochemical reaction is unfavourable if the standard cell potential $(E^0_{cell})$ is negative,which corresponds to a non-spontaneous reaction.
In option $C$,the reaction is $2KCl + I_2 \to 2KI + Cl_2$.
Chlorine $(Cl_2)$ is a stronger oxidizing agent than iodine $(I_2)$ because it has a higher reduction potential.
Therefore,$Cl_2$ can displace $I^-$ ions from $KI$,but $I_2$ cannot displace $Cl^-$ ions from $KCl$.
Thus,the reaction $KCl + I_2 \to KI + Cl_2$ is non-spontaneous and unfavourable.