Electrochemical cells Questions in English

(A) The correct matches are as follows:
$A$. Transistors use Dry cells (Leclanché cell),where Anode is $Zn$ and Cathode is $Carbon$ $(A-III)$.
$B$. Hearing aids use Mercury cells,where Anode is $Zn/Hg$ and Cathode is $HgO + C$ $(B-I)$.
$C$. Inverters use Lead storage batteries,where Anode is $Pb$ and Cathode is $PbO_2$ $(C-IV)$.
$D$. Apollo space ships used Hydrogen fuel cells $(D-II)$.
Therefore,the correct sequence is $A-III, B-I, C-IV, D-II$.

(A) The cell reaction involves the oxidation of methanol at the anode and the reduction of oxygen at the cathode.
Given: $E_{\text{cell}}^{\ominus} = 1.21 \ V$ and $E_{\text{cathode}}^{\ominus} (O_2/H_2O) = 1.229 \ V$.
Using the formula: $E_{\text{cell}}^{\ominus} = E_{\text{cathode}}^{\ominus} - E_{\text{anode}}^{\ominus}$.
$1.21 \ V = 1.229 \ V - E_{\text{anode}}^{\ominus}$.
$E_{\text{anode}}^{\ominus} = 1.229 \ V - 1.21 \ V = 0.019 \ V = 19 \ mV$.
Since the anode reaction is the oxidation of methanol $(CH_3OH + H_2O \rightarrow CO_2 + 6H^+ + 6e^-)$,the reduction potential of the $CO_2/CH_3OH$ couple is $19 \ mV$.
Thus,option $A$ is correct.

(B) During the charging of a lead storage battery,the reverse reaction occurs: $2 PbSO_{4(s)} + 2 H_2 O_{(l)} \rightarrow Pb_{(s)} + PbO_{2(s)} + 2 H_2 SO_{4(aq)}$.
At the anode,$PbSO_4$ is reduced to $Pb$. The oxidation state of $Pb$ changes from $+2$ $(x_1)$ to $0$ $(y_1)$.
At the cathode,$PbSO_4$ is oxidized to $PbO_2$. The oxidation state of $Pb$ changes from $+2$ $(x_2)$ to $+4$ $(y_2)$.
Therefore,the values are $x_1 = +2, y_1 = 0, x_2 = +2, y_2 = +4$.

(A) For a spontaneous electrochemical reaction,the relationship between Gibbs free energy change $(\Delta G)$ and cell potential $(E_{cell})$ is given by the equation: $\Delta G = -nFE_{cell}$.
At equilibrium,the net reaction stops,meaning there is no further change in the concentration of reactants and products.
At this state,the Gibbs free energy change of the reaction becomes zero,i.e.,$\Delta G = 0$.
Substituting $\Delta G = 0$ into the equation $\Delta G = -nFE_{cell}$,we get $0 = -nFE_{cell}$,which implies $E_{cell} = 0$.

(D) Cells that can be recharged are known as secondary cells.
$(c)$ Lead storage cell and $(d)$ $Ni-Cd$ cell are examples of secondary cells.
$(a)$ Mercury cell and $(b)$ Dry cell are primary cells and cannot be recharged.

(A) $AgNO_{3}$ is a strong electrolyte. For strong electrolytes,the variation of molar conductivity $(\lambda_{M})$ with concentration $(C)$ is given by the Debye-$H$ückel-Onsager equation: $\lambda_{M} = \lambda_{M}^{\circ} - A\sqrt{C}$.
Here,$\lambda_{M}^{\circ}$ is the molar conductivity at infinite dilution and $A$ is a constant.
This equation represents a straight line with a negative slope $(-A)$ and an intercept of $\lambda_{M}^{\circ}$ on the y-axis when $\lambda_{M}$ is plotted against $\sqrt{C}$.
Therefore,the correct graph is a straight line with a negative slope.

(C) The oxidation reaction at the anode is: $H_2(g) \rightarrow 2H^+(aq) + 2e^-$.
Number of moles of $H_2 = \frac{67.2 \ L}{22.4 \ L/mol} = 3 \ mol$.
Since $2 \ mol$ of electrons are produced per mole of $H_2$,total moles of electrons $(n)$ = $3 \times 2 = 6 \ mol$.
Total charge $(Q)$ = $n \times F = 6 \times 96500 \ C = 579000 \ C$.
Time $(t)$ = $15 \ min = 15 \times 60 \ s = 900 \ s$.
Current $(I)$ = $\frac{Q}{t} = \frac{579000 \ C}{900 \ s} = 643.33 \ A$.

(B) The relationship between the standard Gibbs free energy change $(\Delta G^{\circ})$ and the standard cell potential $(E_{\text{cell}}^{\circ})$ is given by the equation: $\Delta G^{\circ} = -nFE_{\text{cell}}^{\circ}$.
Rearranging this for $E_{\text{cell}}^{\circ}$,we get: $E_{\text{cell}}^{\circ} = \frac{-\Delta G^{\circ}}{nF}$.
In a dry cell (Leclanché cell),the overall reaction involves the reduction of $MnO_2$ and the oxidation of $Zn$. The reaction is: $Zn(s) + 2MnO_2(s) + 2NH_4^+(aq) \rightarrow Zn^{2+}(aq) + Mn_2O_3(s) + 2NH_3(g) + H_2O(l)$.
Here,the number of electrons transferred $(n)$ is $2$.
Substituting $n = 2$ into the formula,we get $E_{\text{cell}}^{\circ} = \frac{-\Delta G^{\circ}}{2F}$.

(C) The cell reaction is: $Zn_{(s)} + 2Ag_{(1 \ M)}^{+1} \rightarrow Zn_{(1 \ M)}^{+2} + 2Ag_{(s)}$.
Here, the number of electrons transferred $(n)$ is $2$.
The standard $emf$ of the cell $(E_{cell}^o)$ is $1.55 \ V$.
The electrical work done under standard conditions is equal to the change in Gibbs free energy $(\Delta G^o)$.
$\Delta G^o = -nFE_{cell}^o$.
Substituting the values: $n = 2$, $F = 96500 \ C \ mol^{-1}$, $E_{cell}^o = 1.55 \ V$.
$\Delta G^o = -2 \times 96500 \times 1.55 = -299150 \ J = -299.150 \ kJ$.
Therefore, the electrical work done is $-299.150 \ kJ$.

(C) $1$. Standard cell potential $(E^{\circ}_{cell})$ is an intensive property because it does not depend on the amount of matter present in the system.
$2$. Electrode potential depends on the concentration of the electrolyte solution as described by the Nernst equation $(E = E^{\circ} - \frac{RT}{nF} \ln Q)$.
$3$. Standard free energy change $(\Delta G^{\circ} = -nFE^{\circ}_{cell})$ is an extensive property because it depends on the number of moles of electrons $(n)$ transferred,which is related to the amount of substance.
$4$. Electrical work done in a galvanic cell is equal to the decrease in Gibbs energy $(W_{elec} = -\Delta G)$.
$5$. Therefore,the statement that the standard free energy change is an intensive property is false.

(B) In a galvanic cell represented as $Anode | Anode_{electrolyte} || Cathode_{electrolyte} | Cathode$,the oxidation occurs at the anode and reduction occurs at the cathode.
For the given cell $A_{(s)} | A_{(aq)}^{+2} || B_{(aq)}^{+} | B_{(s)}$:
Anode reaction (Oxidation): $A_{(s)} \rightarrow A_{(aq)}^{+2} + 2e^-$.
Cathode reaction (Reduction): $B_{(aq)}^{+} + e^- \rightarrow B_{(s)}$.
To balance the electrons,multiply the cathode reaction by $2$: $2 B_{(aq)}^{+} + 2e^- \rightarrow 2 B_{(s)}$.
Adding both half-reactions: $A_{(s)} + 2 B_{(aq)}^{+} \rightarrow A_{(aq)}^{+2} + 2 B_{(s)}$.
Thus,the correct option is $B$.

(B) In a galvanic cell represented as $Anode | Anode \text{ } electrolyte || Cathode \text{ } electrolyte | Cathode$,the oxidation occurs at the anode and reduction occurs at the cathode.
Given the cell notation: $A_{(s)} | A^{+}_{(1 \ M)} || B^{+2}_{(1 \ M)} | B_{(s)}$.
At the anode (oxidation): $A_{(s)} \rightarrow A^{+}_{(aq)} + e^-$.
At the cathode (reduction): $B^{+2}_{(aq)} + 2e^- \rightarrow B_{(s)}$.
To balance the electrons,multiply the anode reaction by $2$: $2A_{(s)} \rightarrow 2A^{+}_{(aq)} + 2e^-$.
Adding the two half-reactions: $2A_{(s)} + B^{+2}_{(aq)} \rightarrow 2A^{+}_{(aq)} + B_{(s)}$.
Since the $EMF$ is positive,this reaction is spontaneous.

(B) In a galvanic cell,the electrode with the higher reduction potential acts as the cathode,and the one with the lower reduction potential acts as the anode.
Given $E^{\circ}(Cu^{2+}/Cu) = +0.34 \ V$ and $E^{\circ}(H^{+}/H_2) = 0.00 \ V$.
Since $0.34 \ V > 0.00 \ V$,the copper electrode acts as the cathode (reduction) and the standard hydrogen electrode $(SHE)$ acts as the anode (oxidation).
Anode reaction: $H_{2(g)} \longrightarrow 2H^{+}_{(aq)} + 2e^{-}$
Cathode reaction: $Cu^{2+}_{(aq)} + 2e^{-} \longrightarrow Cu_{(s)}$
Adding these two half-reactions gives the net cell reaction: $H_{2(g)} + Cu^{2+}_{(aq)} \longrightarrow 2H^{+}_{(aq)} + Cu_{(s)}$.

(A) The standard Gibbs free energy change is given by the formula: $\Delta G^{\circ} = -nFE_{cell}^{\circ}$
Here,$n$ is the number of moles of electrons transferred in the balanced redox reaction. For $Mg + Sn^{2+} \longrightarrow Mg^{2+} + Sn$,$n = 2$.
$F$ is the Faraday constant,approximately $96500 \ C \ mol^{-1}$.
$E_{cell}^{\circ} = 2.23 \ V$.
Substituting the values: $\Delta G^{\circ} = -2 \times 96500 \times 2.23 \ J \ mol^{-1}$.
$\Delta G^{\circ} = -430390 \ J \ mol^{-1}$.
Converting to $kJ \ mol^{-1}$ by dividing by $1000$: $\Delta G^{\circ} = -430.39 \ kJ \ mol^{-1} \approx -430.4 \ kJ \ mol^{-1}$.

(D) The electrochemical series is used to predict the spontaneity of redox reactions,the relative strength of oxidizing agents,and the relative strength of reducing agents.
However,the use of $SHE$ (Standard Hydrogen Electrode) in an automobile battery is not an application of the electrochemical series. $SHE$ is a reference electrode used to measure standard electrode potentials,whereas automobile batteries (like lead-acid batteries) rely on specific chemical reactions to store and release electrical energy.

(C) In an electrochemical cell,the species that undergoes oxidation acts as the reducing agent.
At the anode,oxidation occurs: $Ni_{(s)} \rightarrow Ni^{2+} + 2e^{-}$.
Since $Ni_{(s)}$ loses electrons,it is the reducing agent.
At the cathode,reduction occurs: $Ag^{+} + e^{-} \rightarrow Ag_{(s)}$.

(A) In an electrolytic cell,the anode is the positive electrode where oxidation takes place,and the cathode is the negative electrode where reduction takes place. Therefore,oxidation occurs at the positive electrode.

(A) Standard hydrogen electrode $(SHE)$ is a primary reference electrode.
$SHE$ is represented by $Pt_{(s)} | H_{2(g)} | H^{+}_{(aq)}$ and is assigned a potential of $0.00 \ V$ at all temperatures.
It serves as the fundamental standard for measuring electrode potentials of other half-cells.
The half-cell reaction is: $H^{+}_{(aq)} + e^- \rightarrow \frac{1}{2} H_{2(g)}$

(D) In a lead storage battery (lead accumulator),the positive electrode is made of $PbO_2$.
During discharge (when used as a source of electrical energy),the reduction reaction at the positive electrode is:
$PbO_{2\text{(s)}} + 4H^{+}{_{\text{(aq)}}} + SO_4^{2-}{_{\text{(aq)}}} + 2e^{-} \rightarrow PbSO_{4\text{(s)}} + 2H_2O_{\text{(l)}}$
In this reaction,the oxidation state of lead changes from $+4$ in $PbO_2$ to $+2$ in $PbSO_4$.
Therefore,the reduction process is represented by $Pb^{4+} \rightarrow Pb^{2+}$.

(A) In a dry cell,the container made of $Zn$ acts as the negative electrode (anode),while the graphite rod acts as the positive electrode (cathode).

(A) The cell reaction is $Pb_{(s)} + 2Ag_{(aq)}^{+} \longrightarrow Pb_{(aq)}^{2+} + 2Ag_{(s)}$.
In this reaction,$Pb$ loses electrons to form $Pb^{2+}$,which is an oxidation process.
The species that undergoes oxidation acts as the reducing agent.
Therefore,$Pb$ is the reducing agent.

(A) The given reaction is a disproportionation reaction: $2 Cu^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + Cu_{(s)}$.
This can be split into two half-reactions:
$1$. Oxidation: $Cu^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + e^{-}$,where $E^{\circ}_{ox} = -E^{\circ}_{Cu^{2+}/Cu^{+}} = -0.15 \ V$.
$2$. Reduction: $Cu^{+}_{(aq)} + e^{-} \rightarrow Cu_{(s)}$,where $E^{\circ}_{red} = E^{\circ}_{Cu^{+}/Cu}$.
To find $E^{\circ}_{Cu^{+}/Cu}$,we use the relation: $\Delta G^{\circ}_{Cu^{2+}/Cu} = \Delta G^{\circ}_{Cu^{2+}/Cu^{+}} + \Delta G^{\circ}_{Cu^{+}/Cu}$.
$-(2)F(E^{\circ}_{Cu^{2+}/Cu}) = -(1)F(E^{\circ}_{Cu^{2+}/Cu^{+}}) + -(1)F(E^{\circ}_{Cu^{+}/Cu})$.
$2(0.34) = 0.15 + E^{\circ}_{Cu^{+}/Cu} \implies E^{\circ}_{Cu^{+}/Cu} = 0.68 - 0.15 = 0.53 \ V$.
Now,$E^{\circ}_{cell} = E^{\circ}_{ox} + E^{\circ}_{red} = -0.15 \ V + 0.53 \ V = +0.38 \ V$.

(C) metal-metal ion electrode consists of a metal rod dipped in a solution containing its own ions.
In the option $Zn^{2+} (aq.) \mid Zn (s)$,a zinc metal rod is in contact with $Zn^{2+}$ ions in the solution.
Therefore,it represents a metal-metal ion electrode.

(B) metal-sparingly soluble salt electrode consists of a metal in contact with its sparingly soluble salt and a solution containing the anion of that salt.
In the electrode $Cl^{-} (aq) | AgCl_{(s)} | Ag_{(s)}$,the metal $Ag$ is in contact with its sparingly soluble salt $AgCl$ and the anion $Cl^{-}$.
The half-cell reaction is: $AgCl_{(s)} + e^{-} \rightarrow Ag_{(s)} + Cl^{-} (aq)$.
This is a classic example of a metal-sparingly soluble salt electrode,often used as a reference electrode.

(B) salt bridge is a device used in an electrochemical cell to connect its oxidation and reduction half-cells. Its primary functions are:
$1$. To complete the electrical circuit by providing electrical contact between the two solutions.
$2$. To maintain electrical neutrality in both half-cells by allowing the migration of ions.
$3$. To prevent the direct mixing of the two electrolyte solutions.
It does not convert electrical energy into chemical energy; rather,it facilitates the flow of current in a galvanic cell where chemical energy is converted into electrical energy.

(A) In a hydrogen-oxygen fuel cell,the following reactions occur at the electrodes:
At the anode: $2H_2(g) + 4OH^-(aq) \rightarrow 4H_2O(l) + 4e^-$
At the cathode: $O_2(g) + 2H_2O(l) + 4e^- \rightarrow 4OH^-(aq)$
In the anode reaction,$H_2$ undergoes oxidation (loss of electrons),which means it acts as a reducing agent.
Therefore,$H_2$ is the species that acts as the reducing agent.

(C) In a dry cell (Leclanché cell),the anode is the zinc container,where oxidation occurs: $Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^-$.
The cathode (positive electrode) is a graphite rod surrounded by powdered $MnO_2$ and carbon.
At the cathode,the reduction reaction occurs: $MnO_{2\text{(s)}} + NH_4^+{_{\text{(aq)}}} + e^{-} \rightarrow MnO(OH)_{\text{(s)}} + NH_{3\text{(g)}}$.
This reaction involves the reduction of $Mn$ from the $+4$ oxidation state in $MnO_2$ to the $+3$ oxidation state in $MnO(OH)$,which is often represented as $Mn_2O_3$.
Thus,$MnO_{2_{(s)}}$ is reduced to $Mn_2O_{3_{(s)}}$.

(B) During the discharging of a lead storage battery,the positive electrode $(PbO_2)$ is reduced to $PbSO_4$.
During the recharging process,the chemical reactions are reversed.
At the positive electrode (anode during recharging),$PbSO_4$ is oxidised to $PbO_2$ according to the following reaction:
$PbSO_4(s) + 2H_2O(l) \rightarrow PbO_2(s) + SO_4^{2-}(aq) + 4H^+(aq) + 2e^-$.

(B) During the discharging of a lead accumulator,the following overall cell reaction occurs:
$Pb_{(s)} + PbO_{2_{(s)}} + 2 H_2 SO_{4_{(aq)}} \rightarrow 2 PbSO_{4_{(s)}} + 2 H_2 O_{(l)}$
In this process,$Pb$ is oxidized at the anode and $PbO_2$ is reduced at the cathode.
As the reaction proceeds,$H_2 SO_4$ is consumed,leading to a decrease in the concentration of the electrolyte.

(C) In an $H_2-O_2$ fuel cell,the electrodes are typically porous carbon electrodes impregnated with catalysts like platinum,palladium,or nickel to facilitate the reactions.
The use of platinum wires as electrodes is not standard practice because:
$1$. Platinum is expensive,and using it as solid wires would not be cost-effective.
$2$. Porous electrodes increase the surface area for the reaction and improve efficiency.
Conclusion: The statement that platinum wires are used as anode and cathode is $NOT$ correct.

(C) During the discharging of a lead storage battery (lead accumulator),the cathode is made of lead dioxide $(PbO_2)$.
At the cathode,$PbO_2$ undergoes reduction by gaining electrons in the presence of sulfuric acid $(H_2SO_4)$.
The balanced chemical equation for the reduction reaction at the cathode is:
$PbO_{2(s)} + 4 H_{(aq)}^{+} + SO_{4(aq)}^{2-} + 2 e^{-} \longrightarrow PbSO_{4(s)} + 2 H_2 O_{(l)}$
Therefore,the correct option is $(C)$.

(C) In a dry cell,the cathode is an inert graphite rod placed in the center,surrounded by an electrolyte paste.
The electrolyte consists of a moist paste of $NH_4Cl$ and $ZnCl_2$.
Starch is added to this paste to increase its viscosity,which prevents the leakage of the electrolyte from the cell.

(B) In a dry cell,at the cathode,$NH_4^{+}$ ions undergo reduction to produce ammonia and hydrogen gas.
The reaction is: $2NH_{4(aq)}^{+} + 2e^{-} \longrightarrow 2NH_{3(aq)} + H_{2(g)}$.
Thus,the gas produced at the cathode is hydrogen.

(C) When the lead storage cell provides current (i.e.,during discharge),spongy lead $(Pb)$ is oxidized to $Pb^{2+}$ ions at the anode. The $Pb^{2+}$ ions so formed combine with $SO_4^{2-}$ ions from $H_2SO_4$ to form insoluble $PbSO_4$. The overall oxidation half-reaction is the sum of these two processes.

$Pb_{(s)} \longrightarrow Pb_{(aq)}^{2+} + 2 e^{-}$	(Oxidation)
$Pb_{(aq)}^{2+} + SO_{4(aq)}^{2-} \longrightarrow PbSO_{4(s)}$	(Precipitation)
$Pb_{(s)} + SO_{4(aq)}^{2-} \longrightarrow PbSO_{4(s)} + 2 e^{-}$	(Overall oxidation)

(C) voltaic cell (or galvanic cell) is an electrochemical cell that converts chemical energy into electrical energy through spontaneous redox reactions.
Therefore,the statement that it converts electrical energy into chemical energy is incorrect.
In a voltaic cell,the anode is the negative electrode and the cathode is the positive electrode.
$A$ dry cell is a common type of voltaic cell.

(B) secondary voltaic cell is a rechargeable battery.
$A$,$C$,and $D$ are examples of secondary cells or rechargeable systems.
$B$ (Dry cell) is a primary cell because it cannot be recharged.

(A) In a Galvanic cell,the anode is the negative electrode where oxidation occurs,and the cathode is the positive electrode where reduction occurs.
In a Galvanic cell,chemical energy is converted into electrical energy.

(C) In a lead accumulator (lead-acid battery),the cathode is made of lead dioxide $(PbO_{2_{(s)}})$.
During the discharging process,the reaction at the cathode is: $PbO_{2\text{(s)}} + 4H^{+}{_{\text{(aq)}}} + SO_4^{2-}{_{\text{(aq)}}} + 2e^{-} \rightarrow PbSO_{4\text{(s)}} + 2H_2O_{\text{(l)}}$.
In this reaction,the oxidation state of $Pb$ in $PbO_{2}$ changes from $+4$ to $+2$ in $PbSO_{4}$.
Since the oxidation state decreases,$PbO_{2_{(s)}}$ is reduced to $Pb^{2+}$ ions (which then form $PbSO_{4_{(s)}}$).

(B) hydrogen-oxygen fuel cell is an electrochemical cell that converts the chemical energy of hydrogen,which acts as a fuel,and oxygen,which acts as an oxidising agent,into electricity through a pair of redox reactions.
In this cell,the cathode reaction involves the reduction of oxygen: $O_2(g) + 2H_2O(l) + 4e^- \rightarrow 4OH^-(aq)$.
Since oxygen gains electrons,it acts as the oxidising agent.
Thus,option $B$ is correct.

(D) In a Standard Hydrogen Electrode $(SHE)$,a platinum plate coated with platinum black is used as the electrode.
Platinum black is highly effective at adsorbing large quantities of $H_2$ gas.
This property facilitates the reversible conversion between gaseous $H_2$ and aqueous $H^+$ ions,meaning the reaction can easily proceed in both directions.
Therefore,bringing the reaction in the reverse direction is not a difficulty.

(B) In lead storage batteries (lead accumulators),the electrolyte used is an aqueous solution of sulphuric acid $(H_2SO_4)$.
Its concentration is typically maintained at approximately $38\%$ by mass.
The density of this sulphuric acid solution is approximately $1.2 \ g \ mL^{-1}$.

(D) The relationship between Gibbs free energy and cell potential is given by ${\Delta}G = -nFE_{cell}$. For a spontaneous reaction,${\Delta}G$ must be negative,which implies that $E_{cell}$ must be positive. Therefore,the statement that $E^{\circ}_{cell}$ must be negative for a spontaneous reaction is incorrect.

(A) In a nickel-cadmium cell,the chemical reactions are as follows:
At the anode: $Cd(s) + 2OH^-(aq) \rightarrow Cd(OH)_2(s) + 2e^-$
At the cathode: $NiO_2(s) + 2H_2O(l) + 2e^- \rightarrow Ni(OH)_2(s) + 2OH^-(aq)$
In the cathode reaction,$NiO_2$ (often represented as $Ni(OH)_3$ in some contexts or hydrated nickel oxide) undergoes reduction,meaning it acts as the oxidising agent.
Therefore,the correct option is $A$.

(A) In a cell representation, the anode (oxidation) is written on the left and the cathode (reduction) on the right.
$1$. Oxidation half-reaction: $Mg_{(s)} \rightarrow Mg^{2+}_{(aq)} + 2e^-$. This occurs at the anode: $Mg_{(s)} \mid Mg^{2+}_{(aq)}$.
$2$. Reduction half-reaction: $Cl_{2(g)} + 2e^- \rightarrow 2Cl^-_{(aq)}$. This occurs at the cathode. Since $Cl_2$ is a gas, an inert electrode like $Pt$ is used: $Cl_{2(g)} \mid Cl^-_{(aq)} \mid Pt_{(s)}$.
$3$. Combining these with the salt bridge $(\parallel)$, the representation is $Mg_{(s)} \mid Mg^{2+}_{(aq)} \parallel Cl_{2(g)} \mid Cl^-_{(aq)} \mid Pt_{(s)}$.

(D) The given cell is $Zn_{(s)} | Zn_{(aq)}^{2+}(1 \ M) || Ni_{(aq)}^{2+}(1 \ M) | Ni_{(s)}$.
This is an electrochemical cell,also known as a voltaic or galvanic cell,where chemical energy is converted into electrical energy.
$A$ $Daniel$ cell is a specific type of galvanic cell that uses a $Zn|Zn^{2+}$ anode and a $Cu|Cu^{2+}$ cathode.
Since this cell uses a $Ni|Ni^{2+}$ cathode instead of $Cu|Cu^{2+}$,it is not a $Daniel$ cell.
Therefore,the incorrect statement is $D$.

(B) The given cell is a concentration cell: $Pt \mid Cl_{2(g, 1 \ bar)} \mid Cl_{(C_1)}^{-} \parallel Cl_{(C_2)}^{-} \mid Cl_{2(g, 1 \ bar)} \mid Pt$.
For a concentration cell,$E_{cell}^{0} = 0$.
The cell reaction is $2Cl_{(C_1)}^{-} \longrightarrow 2Cl_{(C_2)}^{-}$.
The Nernst equation is $E_{cell} = E_{cell}^{0} - \frac{0.0591}{n} \log \frac{C_2}{C_1}$.
Since $E_{cell}^{0} = 0$ and $n = 2$,$E_{cell} = -0.0591 \log \frac{C_2}{C_1} = 0.0591 \log \frac{C_1}{C_2}$.
For the cell to be spontaneous,$E_{cell} > 0$,which implies $\log \frac{C_1}{C_2} > 0$,so $C_1 > C_2$.
If $C_1 > C_2$,then $\Delta G = -nFE_{cell} < 0$ (negative).
Therefore,the option $C_2 > C_1$ is incorrect.

(A) In an electrochemical cell representation,the left side represents the anode (oxidation) and the right side represents the cathode (reduction).
Anode reaction (oxidation): $2Br_{(aq)}^{-} \rightarrow Br_{2(g)} + 2e^{-}$
Cathode reaction (reduction): $Cl_{2(g)} + 2e^{-} \rightarrow 2Cl_{(aq)}^{-}$
Adding these two half-reactions gives the overall cell reaction:
$2Br_{(aq)}^{-} + Cl_{2(g)} \rightarrow Br_{2(g)} + 2Cl_{(aq)}^{-}$
Therefore,the correct option is $A$.

(C) In a dry cell (Leclanché cell),the cathode reaction involves the reduction of manganese dioxide $(MnO_2)$.
The reaction is: $MnO_2 + NH_4^{+} + e^{-} \rightarrow MnO(OH) + NH_3$.
Here,the oxidation state of $Mn$ changes from $+4$ to $+3$,indicating reduction.
Therefore,$MnO_2$ is the substance that gets reduced.

(D) In a dry cell (Leclanché cell),the anode is made of zinc $(Zn)$ and the cathode is a carbon rod surrounded by manganese dioxide $(MnO_2)$.
At the anode,zinc undergoes oxidation: $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$.
Since zinc loses electrons,it acts as the reducing agent.
Therefore,the correct option is $D$.

(C) In a Mercury cell,the anode is a zinc-mercury amalgam $(Zn(Hg))$ and the cathode is a paste of mercuric oxide $(HgO)$ and carbon.
The electrolyte is a moist paste of potassium hydroxide $(KOH)$ and zinc oxide $(ZnO)$.
Therefore,the correct option is $C$.