Electrochemical cells Questions in English

(D) reference electrode is an electrode whose potential is known and stable,used to measure the potential of other electrodes.
The calomel electrode is a common secondary reference electrode.
It consists of mercury $(Hg)$ in contact with a saturated paste of mercurous chloride $(Hg_2Cl_2)$ and a potassium chloride $(KCl)$ solution.
Therefore,the substance present in the calomel electrode is $Hg_2Cl_2$.

(C) During the discharge of a lead storage battery,the following reactions occur:
At anode: $Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2e^-$
At cathode: $PbO_2(s) + SO_4^{2-}(aq) + 4H^+(aq) + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l)$
$1$. $Pb$ is oxidized from $0$ to $+2$ at the anode.
$2$. $PbO_2$ is reduced from $+4$ to $+2$ at the cathode.
$3$. The equivalent mass of $H_2SO_4$ is its molar mass divided by its valency factor ($n$-factor). Since $H_2SO_4$ provides $2H^+$ ions,its $n$-factor is $2$. Thus,equivalent mass = $98 / 2 = 49$. Therefore,the statement that the equivalent mass is $98$ is incorrect.
$4$. The extent of discharging depends on the concentration of $H_2SO_4$ as it is consumed during the reaction. Thus,the statement that it is independent is also incorrect. However,in standard multiple-choice contexts,the most fundamentally incorrect chemical property regarding the equivalent mass is often highlighted.

(D) The mercury cell is a primary cell,meaning it cannot be recharged and reused once the reactants are consumed. Therefore,the statement that it can be used as a secondary cell is incorrect. All other statements are correct: electrochemical reactions are generally efficient,fuel cells convert chemical energy directly to electrical energy,and hydrogen is a clean,renewable energy source.

(C) An intensive property is a bulk property,meaning that it is a physical property of a system that does not depend on the system size or the amount of material in the system.
An extensive property is a property that changes when the size of the sample changes.
From the equation $\Delta G = -nFE_{cell}$,we can observe the following:
$E_{cell}$ is an intensive property because it is independent of the amount of substance present in the system.
$\Delta G$ is an extensive property because it depends on the number of moles $(n)$ of the substance involved in the reaction.
Therefore,$E_{cell}$ is an intensive property while $\Delta G$ is an extensive property.

(B) In a $CH_4-O_2$ fuel cell,the oxidation reaction occurs at the anode.
The anode reaction is: $CH_4 + 2H_2O \to CO_2 + 8H^{+} + 8e^-$.
The cathode reaction is: $2O_2 + 8H^{+} + 8e^- \to 4H_2O$.
Therefore,the correct option is $B$.

(D) The standard free energy change is given by the formula: $\Delta G^o = -nFE^o$
Here,$n = 2$ (number of electrons transferred in the reaction).
$F = 96500 \ C \ mol^{-1}$ (Faraday's constant).
$E^o = 1.1 \ V$.
Substituting the values:
$\Delta G^o = -2 \times 96500 \times 1.1 \ J \ mol^{-1}$
$\Delta G^o = -212300 \ J \ mol^{-1}$
Converting to $kJ \ mol^{-1}$:
$\Delta G^o = -212.3 \ kJ \ mol^{-1}$

(B) For any chemical reaction,the spontaneity is determined by the change in Gibbs free energy $(\Delta G)$.
If $\Delta G < 0$,the reaction is spontaneous.
If $\Delta G > 0$,the reaction is non-spontaneous.
If $\Delta G = 0$,the reaction is at equilibrium.
Therefore,the correct answer is option $B$.

(B) During the charging process,electrical energy is supplied to the battery to reverse the chemical reaction,so it acts as an electrolytic cell.
During the discharging process,the battery converts chemical energy into electrical energy,acting as a galvanic cell.

(B) The galvanometer has $150$ equal divisions.
Given: Current sensitivity $= 10 \text{ div/mA}$ and Voltage sensitivity $= 2 \text{ div/mV}$.
The maximum current $I_g$ for full-scale deflection is $I_g = \frac{150 \text{ div}}{10 \text{ div/mA}} = 15 \text{ mA} = 15 \times 10^{-3} \text{ A}$.
The maximum voltage $V_g$ for full-scale deflection is $V_g = \frac{150 \text{ div}}{2 \text{ div/mV}} = 75 \text{ mV} = 75 \times 10^{-3} \text{ V}$.
The resistance of the galvanometer $G$ is $G = \frac{V_g}{I_g} = \frac{75 \times 10^{-3} \text{ V}}{15 \times 10^{-3} \text{ A}} = 5 \ \Omega$.
To read $1 \text{ V}$ per division,the total range $V$ required is $150 \text{ divisions} \times 1 \text{ V/division} = 150 \text{ V}$.
The series resistance $R$ required is $R = \frac{V}{I_g} - G$.
$R = \frac{150}{15 \times 10^{-3}} - 5 = 10000 - 5 = 9995 \ \Omega$.

(C) salt bridge is used to maintain electrical neutrality in the two half-cells of an electrochemical cell.
For this purpose,the electrolyte used in the salt bridge must have ions with nearly equal ionic mobilities (or transport numbers).
$KNO_3$ is chosen because the ionic mobility of $K^{\oplus}$ $(73.5 \times 10^{-4} \ m^2 \ S \ mol^{-1})$ and $NO_3^{\Theta}$ $(71.4 \times 10^{-4} \ m^2 \ S \ mol^{-1})$ are almost the same.
This ensures that the liquid junction potential is minimized and the flow of ions balances the charge accumulation effectively.

(B) In a cell representation,the anode (oxidation) is written on the left and the cathode (reduction) is written on the right.
Here,$Mn$ is oxidized to $Mn^{+2}$ at the anode: $Mn \to Mn^{+2} + 2e^-$.
$Ag^+$ is reduced to $Ag$ at the cathode: $2Ag^+ + 2e^- \to 2Ag$.
The salt bridge is represented by $||$.
Thus,the cell representation is $Mn|Mn^{+2}(C_2) || Ag^{+}(C_1)|Ag$.

(D) The ability to act as a reducing agent is directly proportional to the reactivity of the metal.
$(1)$ Since only $C$ reacts with $HCl$ to evolve $H_{2(g)}$,$C$ is more reactive than $H_2$ and is the strongest reducing agent among the four.
$(2)$ When $A$ is added to solutions of other metal ions,it displaces $D$ but not $B$ or $C$. This implies that $A$ is a stronger reducing agent than $D$,but weaker than $B$ and $C$.
Combining these observations,the order of increasing reducing ability is $D < A < B < C$.
Therefore,the correct option is $D$.

(B) The electrolysis of brine ($NaCl$ solution) involves the oxidation of chloride ions to chlorine gas at the anode: $2Cl^- (aq) \rightarrow Cl_2 (g) + 2e^-$ $(E^\circ = -1.36 \ V)$.
At the cathode,water is reduced: $2H_2O (l) + 2e^- \rightarrow H_2 (g) + 2OH^- (aq)$ $(E^\circ = -0.83 \ V)$.
The overall cell reaction is $2Cl^- (aq) + 2H_2O (l) \rightarrow Cl_2 (g) + H_2 (g) + 2OH^- (aq)$.
The standard cell potential is $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = -0.83 \ V - 1.36 \ V = -2.19 \ V$.
Since the reaction is non-spontaneous,an external potential of at least $2.19 \ V$ (often approximated as $2.2 \ V$ or $2.18 \ V$ in textbooks) is required to drive the reaction.

(B) $Photodiode$ is a semiconductor device that converts light into an electrical current. When light falls on the $p-n$ junction of the photodiode,it generates electron-hole pairs,which results in the flow of current in the circuit.

(B) Nerve signals in animals are primarily generated by the electrical potential difference across the cell membrane.
This potential difference is maintained by the $(Na^{+}-K^{+})$ pump,which actively transports $Na^{+}$ ions out of the cell and $K^{+}$ ions into the cell.
The resting membrane potential is largely established by the concentration gradient of $K^{+}$ ions,and the rapid movement of $Na^{+}$ ions across the membrane during an action potential is responsible for the transmission of nerve signals.

(C) In a Daniel cell,$Zn | Zn^{2+} || Cu^{2+} | Cu$,the standard cell potential is $E_{cell} = 1.1 \ V$.
The oxidation half-reaction at the anode is $Zn \to Zn^{2+} + 2e^-$.
The reduction half-reaction at the cathode is $Cu^{2+} + 2e^- \to Cu$.
When an external potential greater than $1.1 \ V$ is applied,the cell reaction is reversed,and the cell acts as an electrolytic cell.
In this state,electrons flow from the cathode to the anode,and the chemical processes are reversed: $Zn^{2+}$ is reduced to $Zn$ at the anode,and $Cu$ is oxidized to $Cu^{2+}$ at the cathode.
Therefore,the Assertion is correct,but the Reason is incorrect because $Zn$ is not deposited at the anode during the reverse process; rather,$Zn^{2+}$ is reduced to $Zn$ at the anode.

(B) The cell reaction is: $Zn_{(s)} + HgO_{(s)} \to ZnO_{(s)} + Hg_{(l)}$.
The cell potential remains constant during its life because the overall reaction does not involve any ions in the solution whose concentration changes during its operation.
The electrolyte used in a mercury cell is a paste of $KOH$ and $ZnO$,which is a correct statement,but it does not explain why the cell potential remains constant. The constancy of the potential is due to the fact that the concentration of the electrolyte does not change.

(A) According to Kohlrausch law of independent migration of ions,the limiting molar conductivity of an electrolyte is the sum of the individual contributions of the anion and cation of the electrolyte.
Thus,for $NaCl$,$\Lambda^o_{NaCl} = \lambda^o_{Na^{+}} + \lambda^o_{Cl^{-}}$.
Both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(A) The standard reduction potential of $Zn^{2+}/Zn$ is $-0.76 \ V$,while that of $Fe^{2+}/Fe$ is $-0.44 \ V$.
Since zinc has a more negative electrode potential than iron,it acts as a sacrificial anode.
Zinc gets oxidized in preference to iron,thereby protecting the iron from rusting.
Thus,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(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.
$1. \lambda_{m(H_{2}SO_{4})}^{0} = 2\lambda_{H^{+}}^{0} + \lambda_{SO_{4}^{2-}}^{0} = x$
$2. \lambda_{m(K_{2}SO_{4})}^{0} = 2\lambda_{K^{+}}^{0} + \lambda_{SO_{4}^{2-}}^{0} = y$
$3. \lambda_{m(CH_{3}COOK)}^{0} = \lambda_{CH_{3}COO^{-}}^{0} + \lambda_{K^{+}}^{0} = z$
We need to find $\lambda_{m(CH_{3}COOH)}^{0} = \lambda_{CH_{3}COO^{-}}^{0} + \lambda_{H^{+}}^{0}$.
From the given equations:
$\lambda_{m(CH_{3}COOH)}^{0} = \lambda_{m(CH_{3}COOK)}^{0} + \frac{1}{2}\lambda_{m(H_{2}SO_{4})}^{0} - \frac{1}{2}\lambda_{m(K_{2}SO_{4})}^{0}$
Substituting the values:
$\lambda_{m(CH_{3}COOH)}^{0} = z + \frac{x}{2} - \frac{y}{2} = \frac{x-y}{2} + z \ S \ cm^{2} mol^{-1}$.

(N/A) The galvanic cell corresponding to the given redox reaction is represented as:
$Zn_{(s)} | Zn^{2+}_{(aq)} || Ag^{+}_{(aq)} | Ag_{(s)}$
$(i)$ The $Zn$ electrode is negatively charged because oxidation occurs here $(Zn \rightarrow Zn^{2+} + 2e^-)$,leaving behind electrons.
$(ii)$ The carriers of current in the cell are ions in the electrolyte and electrons in the external circuit.
$(iii)$ The individual reactions are:
At anode: $Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^-$
At cathode: $Ag^{+}_{(aq)} + e^- \rightarrow Ag_{(s)}$

(N/A) When a $Zn$ rod is placed in a copper nitrate solution,a displacement reaction occurs. $Zn$ is more reactive than $Cu$,so it displaces $Cu^{2+}$ ions from the solution.
The chemical reaction is:
$Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$
Observations:
$1$. The blue colour of the copper nitrate solution gradually fades and eventually disappears as $Cu^{2+}$ ions are reduced to metallic $Cu$.
$2$. $A$ reddish-brown deposit of metallic $Cu$ forms on the surface of the $Zn$ rod.
$3$. The solution becomes colourless due to the formation of $Zn^{2+}$ ions.

(N/A) When a $Cu$ rod is dipped into an $AgNO_{3}$ solution,$Ag^{+}$ ions are reduced to $Ag$ metal,which gets deposited on the copper rod.
The oxidation and reduction half-reactions are:
Oxidation: $Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^{-}$
Reduction: $2Ag^{+}_{(aq)} + 2e^{-} \rightarrow 2Ag_{(s)}$
The overall cell reaction is:
$Cu_{(s)} + 2AgNO_{3(aq)} \rightarrow Cu(NO_{3})_{2(aq)} + 2Ag_{(s)}$
As the reaction proceeds,the concentration of $Cu^{2+}$ ions in the solution increases,which imparts a blue colour to the solution.

(B) Zinc is more reactive than copper,as indicated by their positions in the electrochemical series.
Therefore,zinc can displace copper from its salt solution.
If copper sulphate solution is stored in a zinc pot,the following displacement reaction occurs:
$Zn(s) + CuSO_{4}(aq) \longrightarrow ZnSO_{4}(aq) + Cu(s)$
Hence,copper sulphate solution cannot be stored in a zinc pot.

(N/A) lead storage battery consists of an anode of lead $(Pb)$,a cathode of a grid of lead packed with lead dioxide $(PbO_2)$,and a $38\%$ $H_2SO_4$ solution as an electrolyte.
When the battery is in use (discharging),the reactions are:
Anode: $Pb_{\text{(s)}} + SO_4^{2-}{_{\text{(aq)}}} \rightarrow PbSO_{4\text{(s)}} + 2e^{-}$
Cathode: $PbO_{2\text{(s)}} + SO_4^{2-}{_{\text{(aq)}}} + 4H^+{_{\text{(aq)}}} + 2e^- \rightarrow PbSO_{4\text{(s)}} + 2H_2O_{\text{(l)}}$
Overall reaction: $Pb_{(s)} + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O_{(l)}$
On charging the battery,the reverse reaction takes place,where $PbSO_4$ deposited on the electrodes is converted back to $Pb$ and $PbO_2$,and $H_2SO_4$ is regenerated:
$2PbSO_4(s) + 2H_2O_{(l)} \rightarrow Pb_{(s)} + PbO_2(s) + 2H_2SO_4(aq)$

(N/A) Apart from hydrogen,$CH_4$ (methane) and $CH_3OH$ (methanol) are two common materials that can be used as fuels in fuel cells.

(A) The displacement of metals from their salt solutions is determined by their position in the electrochemical series. $A$ metal with a more negative standard reduction potential can displace a metal with a more positive standard reduction potential from its salt solution.
The standard reduction potentials $(E^\circ)$ are:
$Mg^{2+} + 2e^- \rightarrow Mg$ $(-2.37 \ V)$
$Al^{3+} + 3e^- \rightarrow Al$ $(-1.66 \ V)$
$Zn^{2+} + 2e^- \rightarrow Zn$ $(-0.76 \ V)$
$Fe^{2+} + 2e^- \rightarrow Fe$ $(-0.44 \ V)$
$Cu^{2+} + 2e^- \rightarrow Cu$ $(+0.34 \ V)$
Since the reactivity (displacing power) increases as the reduction potential becomes more negative,the order of displacement is $Mg > Al > Zn > Fe > Cu$.

(N/A) The galvanic cell in which the given reaction takes place is represented as:
$Zn_{(s)} | Zn^{2+}_{(aq)} || Ag^{+}_{(aq)} | Ag_{(s)}$
$(i)$ The $Zn$ electrode (anode) is negatively charged.
$(ii)$ Ions are the carriers of current inside the cell,while electrons are the carriers in the external circuit.
$(iii)$ The reaction at the anode (oxidation) is:
$Zn_{(s)} \longrightarrow Zn^{2+}_{(aq)} + 2e^-$
The reaction at the cathode (reduction) is:
$Ag^{+}_{(aq)} + e^- \longrightarrow Ag_{(s)}$

(N/A) The standard cell potential $E^\Theta$ is given as $1.104 \ V$.
The reaction involves the transfer of $n = 2$ electrons.
Using the formula $\Delta_r G^\Theta = -nFE^\Theta$,where $F = 96487 \ C \ mol^{-1}$:
$\Delta_r G^\Theta = -2 \times 96487 \ C \ mol^{-1} \times 1.104 \ V$
$= -213043.296 \ J \ mol^{-1}$
$= -213.04 \ kJ \ mol^{-1}$

(N/A) reaction is feasible if the standard cell potential $E^{\circ}_{cell}$ is positive.
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
$(i)$ $2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2$: $E^{\circ}_{cell} = 0.77V - 0.54V = +0.23V$ (Feasible).
$(ii)$ $2Ag^+ + Cu \rightarrow 2Ag + Cu^{2+}$: $E^{\circ}_{cell} = 0.80V - 0.34V = +0.46V$ (Feasible).
$(iii)$ $2Fe^{3+} + 2Br^- \rightarrow 2Fe^{2+} + Br_2$: $E^{\circ}_{cell} = 0.77V - 1.09V = -0.32V$ (Not feasible).
$(iv)$ $Ag + Fe^{3+} \rightarrow Ag^+ + Fe^{2+}$: $E^{\circ}_{cell} = 0.77V - 0.80V = -0.03V$ (Not feasible).
$(v)$ $Br_2 + 2Fe^{2+} \rightarrow 2Br^- + 2Fe^{3+}$: $E^{\circ}_{cell} = 1.09V - 0.77V = +0.32V$ (Feasible).

(N/A) Electrochemistry is the study of the production of electricity from energy released during spontaneous chemical reactions and the use of electrical energy to bring about non-spontaneous chemical transformations.
Applications: The theoretical and practical utilization of electrochemistry is as follows:
$(i)$ Many metals,sodium hydroxide,chlorine,fluorine,and many other chemicals are produced by electrochemical methods.
$(ii)$ Batteries and fuel cells convert chemical energy into electrical energy and are used on a large scale in various instruments and devices. The reactions carried out electrochemically are more energy-efficient and less polluting.
$(iii)$ Therefore,the study of electrochemistry is important for creating new technologies that are ecofriendly.
$(iv)$ The transmission of sensory signals through cells to the brain and vice versa,and communication between cells,are known to have an electrochemical origin.
Electrochemistry is,therefore,a very vast and interdisciplinary subject.

(N/A) Electrochemistry is the study of the production of electricity from energy released during spontaneous chemical reactions and the use of electrical energy to bring about non-spontaneous chemical transformations.
Applications: The theoretical and practical utilization of electrochemistry is as follows:
$(i)$ Many metals,sodium hydroxide,chlorine,fluorine,and many other chemicals are produced by electrochemical methods.
$(ii)$ Batteries and fuel cells convert chemical energy into electrical energy and are used on a large scale in various instruments and devices. The reactions carried out electrochemically are more energy-efficient and less polluting.
$(iii)$ Therefore,the study of electrochemistry is important for creating new technologies that are eco-friendly.
$(iv)$ The transmission of sensory signals through cells to the brain and vice versa,and communication between cells,are known to have an electrochemical origin.
Electrochemistry is,therefore,a very vast and interdisciplinary subject.

(A) Batteries and fuel cells are types of electrochemical cells that function as galvanic cells.
In these devices,spontaneous chemical reactions occur,which result in the flow of electrons through an external circuit.
Therefore,they convert chemical energy directly into electrical energy.

(N/A) Daniell cell is a type of electrochemical cell (specifically a galvanic or voltaic cell) that converts chemical energy into electrical energy.
Diagram: Refer to the provided image of the Daniell cell,which consists of a zinc electrode in a zinc salt solution and a copper electrode in a copper salt solution,connected by a salt bridge.
Chemical reaction: The Daniell cell operates based on the following spontaneous redox reaction:
$Zn_{(s)} + Cu_{(aq)}^{2+} \longrightarrow Zn_{(aq)}^{2+} + Cu_{(s)}$
Cell potential: When the concentration of both $Zn^{2+}$ and $Cu^{2+}$ ions is unity $(1 \ mol \ dm^{-3})$,the standard cell potential is $1.1 \ V$.
Type of cell: It is a galvanic or voltaic cell.

(N/A) In a Daniell cell,a zinc rod is dipped in a $ZnSO_4$ solution and a copper rod is dipped in a $CuSO_4$ solution.
At the interface of the metal and its salt solution,both the reduced and oxidized forms of the same species are present. $A$ redox couple is defined as the oxidized and reduced forms of a substance taking part in an oxidation or reduction half-reaction.
This is represented by separating the oxidized form from the reduced form by a vertical line or a slash. For example,the two redox couples are represented as $Zn^{2+}/Zn$ and $Cu^{2+}/Cu$.
The two solutions are connected by a salt bridge,which is a $U$-tube containing an electrolyte like $KCl$ or $NH_4NO_3$ in an agar-agar jelly.
When the zinc and copper rods are connected by a metallic wire,the following observations are made:
$(i)$ Electrons flow from the $Zn$ rod to the $Cu$ rod through the external metallic wire.
$(ii)$ The circuit is completed by the migration of ions through the salt bridge,maintaining electrical neutrality.
The potential associated with each electrode is known as electrode potential. When the concentration of each species is $1 \ M$ and the temperature is $298 \ K$,it is called the standard electrode potential $(E^{\ominus})$.
By convention,the standard electrode potential of the hydrogen electrode is $0.0 \ V$. $A$ negative $E^{\ominus}$ value indicates that the redox couple is a stronger reducing agent than the $H^{+}/H_2$ couple.

(A) The given redox reaction is: $Zn_{(s)} + 2Ag_{(aq)}^{+} \to Zn_{(aq)}^{2+} + 2Ag_{(s)}$
Representation of the galvanic cell: $Zn_{(s)} | Zn_{(aq)}^{2+} || Ag_{(aq)}^{+} | Ag_{(s)}$
$(i)$ The $Zn$ electrode acts as the anode where oxidation occurs $(Zn \to Zn^{2+} + 2e^-)$,so it is the negatively charged electrode.
$(ii)$ In the external circuit,electrons are the carriers of current (flowing from $Zn$ to $Ag$),while in the internal circuit (electrolyte),ions are the carriers of current.
$(iii)$ Individual reactions:
Anode (Oxidation): $Zn_{(s)} \to Zn_{(aq)}^{2+} + 2e^-$
Cathode (Reduction): $2Ag_{(aq)}^{+} + 2e^- \to 2Ag_{(s)}$

(B) When a $Cu$ rod is placed in an $AgNO_3$ solution,$Cu$ acts as a reducing agent and $Ag^{+}$ ions act as an oxidizing agent.
$Cu$ undergoes oxidation by losing $2$ electrons to form $Cu^{2+}$ ions.
$Ag^{+}$ ions undergo reduction by gaining $1$ electron each to form $Ag$ metal.
The balanced redox reaction is: $Cu_{(s)} + 2 Ag_{(aq)}^{+} \rightarrow Cu_{(aq)}^{2+} + 2 Ag_{(s)}$.

(B) In an electrochemical cell,a salt bridge is used to maintain electrical neutrality by allowing the migration of ions between the two half-cell solutions.

(A) salt bridge is typically filled with a gel-like substance containing an inert electrolyte such as $KCl$,$KNO_3$,or $NH_4NO_3$. These electrolytes are chosen because the ionic mobilities of the cation and anion are approximately equal.

(A) In a $Daniell$ cell,the oxidation half-reaction occurs at the zinc electrode,which acts as the anode: $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$.
The reduction half-reaction occurs at the copper electrode,which acts as the cathode: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$.
Therefore,$Zn$ acts as the anode and $Cu$ acts as the cathode.

(D) The functions of a salt bridge in a Daniell cell are as follows:
$(i)$ It connects the two half-cells,thereby completing the electrical circuit.
$(ii)$ It maintains electrical neutrality in the electrolyte solutions of both half-cells by allowing the migration of ions.
$(iii)$ It minimizes or eliminates the liquid junction potential.

(N/A) Condition-$I$: (Condition in which the flow of current continues through the galvanic cell or $E_{\text{ext}} < 1.1 \ V$): According to diagram-$(a)$,if an external opposite potential is applied to the galvanic cell and increased slowly,we find that the reaction continues to take place until the opposing voltage reaches the value $1.1 \ V$. The redox reaction continues in the forward direction and the reaction does not stop.
$(b)$ Condition-$II$: (Condition in which the galvanic cell is stopped or $E_{\text{ext}} = 1.1 \ V$): According to diagram-$(b)$,when the external potential reaches $1.1 \ V$,the reaction stops altogether and no current flows through the cell. Any further increase in the external potential starts the reaction in the opposite direction (electrolytic cell).
$(c)$ Condition-$III$: (Condition in which the cell acts as an electrolytic cell or $E_{\text{ext}} > 1.1 \ V$): If the external potential is increased beyond $1.1 \ V$,the reaction proceeds in the reverse direction,and the cell acts as an electrolytic cell. Electrons flow from the copper electrode to the zinc electrode,and current flows from the zinc electrode to the copper electrode.

(N/A) The redox reaction occurring in a Daniell cell involves the oxidation of zinc and the reduction of copper ions:
$Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^-$ (Oxidation half-reaction)
$Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}$ (Reduction half-reaction)
Overall redox reaction:
$Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$

(A) In a Daniell cell,a spontaneous redox reaction occurs between $Zn$ and $Cu^{2+}$ ions.
This chemical reaction releases energy,which is converted into electrical energy.
Therefore,the energy conversion is from chemical energy to electrical energy.

(N/A) In a Daniell cell,the standard cell potential is $1.1 \ V$.
When an external potential $E_{ext} < 1.1 \ V$ is applied,the cell continues to function as a galvanic cell.
$(1)$ The $Zn$ electrode acts as the anode (negative terminal) and the $Cu$ electrode acts as the cathode (positive terminal).
$(2)$ Electrons flow from the $Zn$ anode to the $Cu$ cathode through the external circuit,while the conventional current flows from the $Cu$ cathode to the $Zn$ anode.

(N/A) When an external potential $E_{ext} < 1.1 \ V$ is applied to a Daniell cell,the cell continues to function as a galvanic cell.
$(1)$ The reactions are:
Anode reaction: $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$
Cathode reaction: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$
Redox reaction: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$
$(2)$ Since $Zn$ is the anode,$Zn$ metal dissolves into the solution as $Zn^{2+}$ ions.

(N/A) The standard potential of a Daniell cell is $1.1 \ V$.
$(1)$ When an external potential $E_{ext} < 1.1 \ V$ is applied,the cell continues to function as a galvanic cell,and the cell potential is given by $E_{cell} = 1.1 \ V - E_{ext}$.
$(2)$ Since the cell is still producing electrical energy from a chemical reaction,it acts as a galvanic (or voltaic) cell.

(C) galvanic cell converts chemical energy into electrical energy.
When an external opposing potential $(E_{ext})$ is applied to the galvanic cell,the flow of electrons and current is affected.
$1$. If $E_{ext} < E_{cell}$,electrons flow from anode to cathode,and the cell continues to function as a galvanic cell.
$2$. If $E_{ext} = E_{cell}$,the reaction stops,and no current flows.
$3$. If $E_{ext} > E_{cell}$,the direction of the current is reversed,and the external source forces the reaction to occur in the opposite direction. In this state,the device acts as an electrolytic cell.

(F, T, T, F) $(1)$ False $(F)$: Galvanic cells convert chemical energy to electrical energy,while electrolytic cells convert electrical energy to chemical energy.
$(2)$ True $(T)$: In an electrolytic cell,the polarity is reversed compared to a galvanic cell; the electrode connected to the positive terminal is the anode,and the one connected to the negative terminal is the cathode.
$(3)$ True $(T)$: In a galvanic cell,electrons flow from anode to cathode,whereas in an electrolytic cell,an external power source forces electrons to flow in the opposite direction.
$(4)$ False $(F)$: The standard potential of a Daniell cell is $1.1 \ V$. If an external voltage of $2.0 \ V$ (which is greater than $1.1 \ V$) is applied,the cell reaction reverses,and the cell acts as an electrolytic cell.