Electrode potential and ECell Questions in English

(A) The tendency to lose $e^-$ is directly proportional to the oxidation potential of the metal.
Given oxidation potentials are:
$Zn = 0.76 \ V$
$Cu = -0.34 \ V$
$Ag = -0.80 \ V$
Since $0.76 > -0.34 > -0.80$,the order of tendency to lose $e^-$ is $Zn > Cu > Ag$.

(B) The Daniell cell consists of a zinc electrode in a zinc sulfate solution and a copper electrode in a copper sulfate solution.
The standard electrode potential $(E^{\circ}_{cell})$ is calculated as $E^{\circ}_{cathode} - E^{\circ}_{anode}$.
For a Daniell cell,$E^{\circ}_{Cu^{2+}/Cu} = 0.34 \ V$ and $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$.
Therefore,$E^{\circ}_{cell} = 0.34 \ V - (-0.76 \ V) = 1.10 \ V$.

(N/A) Definition: $A$ potential difference that develops between the electrode and the electrolyte is called electrode potential.
$(b)$ Standard electrode potential: When the concentrations of all the species involved in a half-cell are unity $(1 \ M)$,the electrode potential is known as standard electrode potential. According to $IUPAC$ convention,standard reduction potentials are now called standard electrode potentials.
$(c)$ Anode and cathode in electrode potential:
$(i)$ In a galvanic cell,the half-cell in which oxidation takes place is called the anode,and it has a negative potential with respect to the solution.
$(ii)$ The other half-cell in which reduction takes place is called the cathode,and it has a positive potential with respect to the solution.

(N/A) $(i)$ Cell Potential: The potential difference between the two electrodes of a galvanic cell is called the cell potential and is measured in volts.
$(ii)$ $emf$: The cell potential is the difference between the electrode potentials (reduction potentials) of the cathode and anode. It is called the cell electromotive force $(emf)$ of the cell when no current is drawn through the cell.
$(B)$ Formula and expression of cell potential:
Cell potential of a galvanic cell is positive,which can be calculated by the following formula:
$E_{cell} = E_{right} - E_{left} = E_{red(cathode)} - E_{red(anode)}$
The anode is on the left and the cathode is on the right while representing the galvanic cell. $A$ galvanic cell is generally represented by putting a vertical line between metal and electrolyte solution and putting a double vertical line between the two electrolytes connected by a salt bridge.

Anode half cell	Salt bridge	Cathode half cell
Metal of anode electrode	The product of oxidation of anode	Whose reduction is electrolyte	The product produced from reduction (metal)

$(C)$ Expression of Daniell cell:
Example: $Zn_{(s)} | Zn_{(aq)}^{2+} || Cu_{(aq)}^{2+} | Cu_{(s)}$
Oxidation on anode: $Zn_{(s)} \longrightarrow Zn_{(aq)}^{2+} + 2e^{-}$
Reduction on cathode: $Cu_{(aq)}^{2+} + 2e^{-} \longrightarrow Cu_{(s)}$
Total cell reaction: $Zn_{(s)} + Cu_{(aq)}^{2+} \longrightarrow Zn_{(aq)}^{2+} + Cu_{(s)}$
Where,$Zn$ electrode is the anode and copper electrode is the cathode.
Daniell cell potential:
$E_{cell} = E_{right} - E_{left}$
$E_{cell} = E_{Cu^{2+}|Cu} - E_{Zn^{2+}|Zn}$

(N/A) Selection of reference electrode: The potential of an individual half-cell cannot be measured directly. We can only measure the difference between the potentials of two half-cells,which gives the $EMF$ of the cell.
$(b)$ Use of Standard Hydrogen Electrode $(SHE)$: To determine the potential of a single half-cell,it is coupled with a reference electrode whose potential is known. By convention,the Standard Hydrogen Electrode $(SHE)$ is used as the reference,and its potential is arbitrarily assigned a value of $0.00 \ V$ at all temperatures.
$(c)$ Calculation: When the half-cell is connected to the $SHE$,the measured $EMF$ of the resulting cell corresponds to the potential of that half-cell. If the half-cell acts as the cathode,its potential is positive; if it acts as the anode,its potential is negative relative to the $SHE$.

(N/A) According to the convention,a half-cell called the standard hydrogen electrode $(SHE)$,represented by $Pt_{(s)} | H_{2_{(g)}} (1 \ bar) | H^{+}_{(aq)} (1 \ M)$,is assigned a zero potential at all temperatures.
Construction of hydrogen electrode:
The standard hydrogen electrode consists of a platinum electrode coated with platinum black. The electrode is dipped in an acidic solution,and pure hydrogen gas is bubbled through it at $1 \ bar$ pressure.
The concentration of hydrogen ions in the solution is maintained at $1 \ M$.
The half-cell reactions are:
Oxidation: $\frac{1}{2} H_{2_{(g)}} \longrightarrow H^{+}_{(aq)} + e^{-}$
Reduction: $H^{+}_{(aq)} + e^{-} \longrightarrow \frac{1}{2} H_{2_{(g)}}$
Uses:
The standard hydrogen electrode is used as a reference electrode to measure the electrode potential of any other half-cell. By constructing a galvanic cell with the $SHE$ and another electrode,the measured cell potential at $298 \ K$ directly gives the potential of that half-cell,as the potential of the $SHE$ is taken as zero.

(N/A) Reduction potential: At $298 \ K$,the $emf$ of a cell constructed by taking the standard hydrogen electrode as the anode (reference half-cell) and the other half-cell as the cathode gives the reduction potential of the other half-cell.
Standard hydrogen electrode half-cell $\mid$ another half-cell
(anode half-cell) $\mid$ (cathode half-cell)
Cell potential $=$ (Reduction potential of other cell) $=$ $emf$ value of the other half-cell.
$(b)$ Explanation with a general example:
$(i)$ If the concentrations of the oxidized and reduced forms of the species in the right-hand half-cell are unity,the cell potential equals the standard electrode potential.
General cell: $Pt \mid H_{2(g)} (1 \ bar) \mid H^+_{(aq)} (1 \ M) \| M^{n+}_{(aq)} (1 \ M) \mid M_{(s)}$
$(ii)$ $E^{\ominus} = (E^{\ominus}_R - E^{\ominus}_L)$:
Where $E^{\ominus}$ is the standard reduction potential of the cell,$E^{\ominus}_R$ is the standard reduction potential of the right-hand electrode,and $E^{\ominus}_L$ is the standard reduction potential of the left-hand electrode.
Since $E^{\ominus}_L = 0.0 \ V$ for the standard hydrogen electrode,$\therefore E^{\ominus} = (E^{\ominus}_R - 0.0) = E^{\ominus}_R$.
$(iii)$ Hydrogen half-cell $\mid$ Copper half-cell:
$Pt \mid H_{2(g)} (1 \ bar) \mid H^+_{(aq)} (1 \ M) \| Cu^{2+}_{(aq)} (1 \ M) \mid Cu_{(s)}$
The measured $emf$ of the cell is $+0.34 \ V$,which is the standard electrode potential for the reaction:
Oxidation (left): $H_{2(g)} \rightarrow 2H^+_{(aq)} + 2e^- \ (E^{\ominus}_L = 0.0 \ V)$
Reduction (right): $Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}$
$\therefore E^{\ominus}_{cell} = (E^{\ominus}_R - E^{\ominus}_L) = 0.34 \ V \implies E^{\ominus}_R = 0.34 \ V$.
Thus,the standard reduction potential for $Cu^{2+} \mid Cu$ is $+0.34 \ V$.

(N/A) At $298 \ K$,the $EMF$ of a cell constructed by taking the standard hydrogen electrode $(SHE)$ as the cathode (reference half-cell) and the other half-cell as the anode,gives the reduction potential of the other half-cell.
General cell representation:
$M_{(s)} | M_{(aq)}^{n+} (1 \ M) || H_{(aq)}^{+} (1 \ M) | \frac{1}{2} H_{2_{(g)}} (1 \ bar) | Pt_{(s)}$
Using the formula $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$,since $E^{\circ}_{cathode} = E^{\circ}_{H^{+}/H_2} = 0.0 \ V$,we get $E^{\circ}_{cell} = -E^{\circ}_{anode}$.
Example: Zinc half-cell as anode and $SHE$ as cathode:
$Zn_{(s)} | Zn_{(aq)}^{2+} (1 \ M) || H_{(aq)}^{+} (1 \ M) | H_{2_{(g)}} (1 \ bar) | Pt_{(s)}$
The measured $E^{\circ}_{cell}$ is $-0.76 \ V$. Therefore,the standard reduction potential of $Zn^{2+}/Zn$ is $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$.
When the $SHE$ is on the right side (cathode),the other half-cell acts as the anode,and the cell potential is negative if the metal is more reactive than hydrogen.
Cathode reaction: $H^{+}_{(aq, 1 \ M)} + e^{-} \rightarrow \frac{1}{2} H_{2_{(g)}} (1 \ bar)$
Overall reaction: $Zn_{(s)} + 2H^{+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + H_{2_{(g)}}$

(A) The cell potential $(E^o_{cell})$ is calculated using the formula: $E^o_{cell} = E^o_{cathode} - E^o_{anode}$.
In a Daniell cell,the $Cu$ electrode acts as the cathode and the $Zn$ electrode acts as the anode.
Given: $E^o_{Cu^{2+}|Cu} = 0.34 \ V$ and $E^o_{Zn^{2+}|Zn} = -0.76 \ V$.
Substituting the values: $E^o_{cell} = 0.34 \ V - (-0.76 \ V)$.
$E^o_{cell} = 0.34 \ V + 0.76 \ V = 1.10 \ V$.

(N/A) Standard electrode (half-cell) potential: Under standard conditions,which include a temperature of $298 \ K$,a gas pressure of $1 \ bar$,and a solution concentration of $1 \ M$,the reduction potential of an electrode (half-cell) is defined as the standard electrode potential,measured in volts $(V)$.
Uses:
$(i)$ Construction of the electrochemical series ($EMF$ series): When half-cell reactions are arranged in decreasing order of their standard reduction potential,the resulting series is known as the electrochemical series. This series helps in predicting the spontaneity of redox reactions and the relative strength of oxidizing and reducing agents.

(D) The standard reduction potential of gold $(E^o = 1.40 \ V)$ is higher than that of silver $(E^o = 0.80 \ V)$.
This indicates that gold is a weaker reducing agent than silver and cannot displace silver ions $(Ag^+)$ from the $AgNO_3$ solution.
Therefore,no chemical reaction occurs when a gold ring is placed in $AgNO_3$ solution.

(A) The cell reaction is: $Mg_{(s)} + 2H^+_{(aq)} \rightarrow Mg^{2+}_{(aq)} + H_{2(g)}$.
Given $E^\circ_{cell} = 2.36 \ V$ at standard conditions ($1 \ M$ concentration and $1 \ bar$ pressure).
The cell potential is given by $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$.
Here,the cathode is the Standard Hydrogen Electrode $(SHE)$,so $E^\circ_{H^+/H_2} = 0.00 \ V$.
Therefore,$2.36 \ V = 0.00 \ V - E^\circ_{Mg^{2+}/Mg}$.
$E^\circ_{Mg^{2+}/Mg} = -2.36 \ V$.

(0.34 V) The cell reaction is as follows:
Anode (Oxidation): $H_{2(g)} \rightarrow 2H^+_{(aq)} + 2e^-$
Cathode (Reduction): $Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}$
Overall reaction: $H_{2(g)} + Cu^{2+}_{(aq)} \rightarrow 2H^+_{(aq)} + Cu_{(s)}$
The cell potential is given by: $E^o_{cell} = E^o_{cathode} - E^o_{anode}$
Given $E^o_{cell} = 0.34 \, V$ and $E^o_{H^+|H_2} = 0.00 \, V$ (Standard Hydrogen Electrode).
$0.34 \, V = E^o_{Cu^{2+}|Cu} - 0.00 \, V$
Therefore,$E^o_{Cu^{2+}|Cu} = 0.34 \, V$.

The standard cell potential is calculated using the formula: $E_{cell}^{o} = E_{cathode}^{o} - E_{anode}^{o}$.
$(i)$ For $Al|Al^{3+}||Cu^{2+}|Cu$:
$E_{cell}^{o} = E_{Cu^{2+}|Cu}^{o} - E_{Al^{3+}|Al}^{o} = 0.34 \ V - (-1.66 \ V) = 2.00 \ V$.
$(ii)$ For $Al|Al^{3+}||Zn^{2+}|Zn$:
$E_{cell}^{o} = E_{Zn^{2+}|Zn}^{o} - E_{Al^{3+}|Al}^{o} = -0.76 \ V - (-1.66 \ V) = 0.90 \ V$.
$(iii)$ For $Al|Al^{3+}||Ag^{+}|Ag$:
$E_{cell}^{o} = E_{Ag^{+}|Ag}^{o} - E_{Al^{3+}|Al}^{o} = 0.80 \ V - (-1.66 \ V) = 2.46 \ V$.

(A) To determine the potential of any half cell,it must be connected to a reference electrode whose potential is known. The most commonly used reference electrode is the Standard Hydrogen Electrode $(SHE)$,which is assigned a potential of $0.00 \ V$ at all temperatures. By connecting the unknown half cell to the $SHE$,the potential difference measured by the voltmeter corresponds to the electrode potential of the unknown half cell.

(A) The potential of a single half-cell cannot be measured in isolation.
To determine the potential of any half-cell,it is connected to a Standard Hydrogen Electrode $(SHE)$ to form a complete electrochemical cell.
The potential of the $SHE$ is arbitrarily assigned as $0.00 \ V$.
The cell potential $(E_{cell})$ is measured using a voltmeter.
Since $E_{cell} = E_{cathode} - E_{anode}$,if the half-cell acts as the cathode,$E_{half-cell} = E_{cell} + E_{SHE}$.
If it acts as the anode,$E_{half-cell} = E_{SHE} - E_{cell}$.

(N/A) In a galvanic cell,if the hydrogen half-cell $(Pt, H_{2(g)} | H^{+}_{(aq)})$ is placed on the anode (left side),its oxidation potential is $0.00 \ V$.
Since $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$ and $E_{anode}^{\circ} = 0 \ V$,then $E_{cell}^{\circ} = E_{cathode}^{\circ}$.
If the cathode has a positive standard reduction potential (like $Cu^{2+}/Cu$ where $E^{\circ} = +0.34 \ V$),the cell potential will be positive.
For the cell: $Pt, H_{2(g)} | H^{+}(1 \ M) || Cu^{2+}(1 \ M) | Cu_{(s)}$,the standard cell potential is $E_{cell}^{\circ} = +0.34 \ V - 0.00 \ V = +0.34 \ V$.

(A) The standard hydrogen electrode $(SHE)$ is used as a reference electrode.
By convention,the standard electrode potential of $SHE$ is assigned a value of $0.00 \ V$ at all temperatures.
Since the oxidation and reduction processes are reverse of each other,the potential values are equal in magnitude but opposite in sign.
However,for $SHE$,both the standard oxidation potential and standard reduction potential are defined as $0.00 \ V$.

(N/A) In a galvanic cell,the cell potential is calculated as $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
Given that the hydrogen half-cell is the cathode,$E^{\circ}_{cathode} = E^{\circ}_{H^+/H_2} = 0.00 \ V$.
Therefore,$E^{\circ}_{cell} = 0.00 \ V - E^{\circ}_{anode} = -E^{\circ}_{anode}$.
If the anode is a metal with a negative standard reduction potential (e.g.,$Zn/Zn^{2+}$ where $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$),then $E^{\circ}_{cell} = 0.00 \ V - (-0.76 \ V) = +0.76 \ V$,which is positive.
However,if the anode is a metal with a positive standard reduction potential (e.g.,$Cu/Cu^{2+}$ where $E^{\circ}_{Cu^{2+}/Cu} = +0.34 \ V$),then $E^{\circ}_{cell} = 0.00 \ V - (+0.34 \ V) = -0.34 \ V$,which is negative.
Thus,the sign of the standard potential depends on the standard reduction potential of the metal electrode used as the anode.

(A) The cell potential $E_{cell}$ is given by $E_{cell} = E_{cathode} - E_{anode}$.
Given that the standard hydrogen electrode $(SHE)$ has a potential of $0.00 \ V$,if the cell potential is negative,it implies that the potential of the other half-cell is lower than the $SHE$.
$(i)$ Since the other half-cell has a lower potential,it acts as the $Anode$ (oxidation occurs here).
$(ii)$ The $Anode$ in a galvanic cell is the $Negative$ electrode.
$(iii)$ By convention,the $Anode$ is placed on the $Left$ side of the cell diagram.

(N/A) The standard hydrogen electrode $(SHE)$ is used as a reference electrode to measure the standard electrode potential of other half-cells.
$1$. It serves as a primary reference electrode with an assigned potential of $0.00 \ V$ at all temperatures.
$2$. It is used to determine the standard reduction potential of various metal/metal-ion couples by connecting them to the $SHE$ in an electrochemical cell.
$3$. It helps in constructing the electrochemical series by providing a baseline for comparison.

(A) The term $emf$ stands for $Electromotive \ force$.
It is defined as the potential difference between the two electrodes of a galvanic cell when no current is flowing through the external circuit.
It is the maximum potential difference that the cell can provide.

(A) In electrochemistry,the standard electrode potential $(E^o)$ is measured relative to the Standard Hydrogen Electrode $(SHE)$,which is assigned a value of $0.00 \ V$.
$1$. $A$ negative $E^o$ value indicates that the species is a stronger reducing agent than $H_2$ gas.
$2$. $A$ positive $E^o$ value indicates that the species is a weaker reducing agent than $H_2$ gas (or a stronger oxidizing agent).

(A) In the electrochemical series,the standard electrode potential $(E^{\circ}_{red})$ values are arranged based on the reduction potential of elements.
Fluorine $(F_2)$ has the highest positive standard reduction potential $(+2.87 \ V)$,making it the strongest oxidizing agent.
Lithium $(Li)$ has the lowest (most negative) standard reduction potential $(-3.05 \ V)$,making it the strongest reducing agent.
Therefore,Fluorine has the highest and Lithium has the lowest standard electrode potential.

(N/A) The symbolic representation of half-cells follows the $IUPAC$ convention where the electrode is written on the left and the electrolyte on the right for oxidation,and vice versa for reduction.
$(i)$ $Zn_{(s)} | Zn^{2+}_{(aq)} (1M)$ represents an oxidation half-cell (Anode).
$(ii)$ $Cu^{2+}_{(aq)} (1M) | Cu_{(s)}$ represents a reduction half-cell (Cathode).
$(iii)$ $Pt_{(s)} | H_{2(g)} (1 \text{ bar}) | H^+_{(aq)} (1M)$ represents a standard hydrogen electrode $(SHE)$ acting as an oxidation half-cell.

(C) The reactivity of a metal towards an acidic solution depends on its standard reduction potential $(E^\circ)$.
For the reaction $2Ag(s) + 2H^+(aq) \rightarrow 2Ag^+(aq) + H_2(g)$,the standard cell potential $(E^\circ_{cell})$ is calculated as $E^\circ_{cathode} - E^\circ_{anode}$.
Here,$E^\circ_{H^+/H_2} = 0.00 \ V$ and $E^\circ_{Ag^+/Ag} = +0.80 \ V$.
$E^\circ_{cell} = 0.00 \ V - 0.80 \ V = -0.80 \ V$.
Since the $E^\circ_{cell}$ is negative,the reaction is non-spontaneous.
Therefore,silver does not react with acidic solutions to evolve hydrogen gas.

(N/A) The cell potential formula is given by $E^o_{cell} = E^o_{cathode} - E^o_{anode}$.
$1$. For the Daniell cell ($Zn-Cu$ cell):
$E^o_{cell} = E^o_{Cu^{2+}/Cu} - E^o_{Zn^{2+}/Zn}$
$2$. For the Copper-Silver cell ($Cu-Ag$ cell):
$E^o_{cell} = E^o_{Ag^{+}/Ag} - E^o_{Cu^{2+}/Cu}$

(N/A) $1$. The strongest reducing agent in aqueous solution is $Li$ (Lithium) due to its high hydration energy and low sublimation energy.
$2$. The strongest oxidizing agent among gases is $F_2$ (Fluorine) gas.
$3$. The standard reduction potential of any electrode is measured with reference to the Standard Hydrogen Electrode $(SHE)$,which is assigned a potential of $0.00 \ V$.

(A) At the anode,oxidation occurs,which involves the loss of electrons.
For a metal electrode represented as $M \,|\, M^{n+}$,the metal atom $M$ loses $n$ electrons to form the metal ion $M^{n+}$.
The half-reaction is: $M(s) \rightarrow M^{n+}(aq) + ne^-$.

(A) The electrode potential of any electrode cannot be measured in isolation. To measure it,a reference electrode is required. The Standard Hydrogen Electrode $(SHE)$ is universally accepted as the primary reference electrode,and its potential is arbitrarily assigned a value of $0.00 \ V$ at all temperatures. By connecting the unknown electrode to the $SHE$ to form a galvanic cell,the potential difference measured corresponds to the electrode potential of the unknown electrode.

(A) The relationship between the Gibbs energy change $(\Delta G)$ and the cell potential $(E_{cell})$ is given by the equation:
$\Delta G = -nFE_{cell}$
Where:
$n$ is the number of moles of electrons transferred,
$F$ is the Faraday constant $(96487 \ C \ mol^{-1})$,
$E_{cell}$ is the cell potential of the reaction.

(C) The standard electrode potential $(E^o)$ of a metal is determined by the sum of three energy factors: enthalpy of sublimation (atomization),ionization enthalpy,and enthalpy of hydration.
For $Cu$,the sum of the enthalpy of sublimation and the first two ionization enthalpies is very high.
This high energy requirement is not fully compensated by the hydration enthalpy of $Cu^{2+}$ ions.
As a result,the overall $E^o$ value for $Cu^{2+}/Cu$ is positive $(+0.34 \, V)$.
In contrast,for $Zn$,the energy required to convert $Zn(s)$ to $Zn^{2+}(aq)$ is relatively low and is well-compensated by the high hydration enthalpy of $Zn^{2+}$,leading to a negative $E^o$ value $(-0.76 \, V)$.

(N/A) The standard electrode potential $(E^{\circ})$ of a metal depends on the sum of its enthalpy of sublimation,ionization enthalpy,and hydration enthalpy.
For $Cu$,the sum of sublimation and ionization enthalpies is very high,which is not compensated by the hydration enthalpy of $Cu^{2+}$. Thus,$Cu$ has a positive $E^{\circ}$ value $(+0.34 \ V)$.
For $Zn$,the second ionization enthalpy is relatively low because the removal of electrons from the $4s$-orbital leads to a stable $3d^{10}$ configuration. This low energy requirement makes the overall process favorable,resulting in a negative $E^{\circ}$ value $(-0.76 \ V)$.

(F, T, F, F) $(i)$ False: $E_{cell}$ depends on the concentration of ions and temperature,whereas $E_{cell}^o$ is a constant for a given cell at $298 \ K$.
$(ii)$ True: Under standard conditions (concentration $= 1 \ M$,pressure $= 1 \ bar$,temperature $= 298 \ K$),the cell potential $E_{cell}$ is equal to the standard cell potential $E_{cell}^o$.
$(iii)$ False: When a cell attains equilibrium,$E_{cell} = 0$,not $E_{cell} = E_{cell}^o$.
$(iv)$ False: The correct relationship is $\Delta_r G^o = -nFE_{cell}^o$,which implies $E_{cell}^o = -\frac{\Delta_r G^o}{nF}$.

(N/A) Reduction reactions occur on the surface of the cathode. If more than one species are present near the cathode,the species with the higher $E^{\theta}$ value undergoes reduction.
For example,in an aqueous $NaCl$ solution,both $Na^{+}$ ions from $NaCl$ and $H^{+}$ ions from $H_{2}O$ are available near the cathode.
$NaCl_{(aq)} \rightarrow Na^{+}_{(aq)} + Cl^{-}_{(aq)}$
$H_{2}O_{(l)} \rightleftharpoons H^{+}_{(aq)} + OH^{-}_{(aq)}$
The reduction half-reactions are:
$(i) Na^{+}_{(aq)} + e^{-} \rightarrow Na_{(s)} \quad E^{\theta} = -2.71 \ V$
$(ii) H^{+}_{(aq)} + e^{-} \rightarrow \frac{1}{2} H_{2(g)} \quad E^{\theta} = 0.00 \ V$
Comparing the $E^{\theta}$ values,reaction $(ii)$ has a higher value than reaction $(i)$. Therefore,reaction $(ii)$ occurs preferentially at the cathode. Consequently,$H^{+}$ ions from water are reduced to produce $H_{2}$ gas. The overall reaction at the cathode during the electrolysis of aqueous $NaCl$ is:
$(iii) H_{2}O_{(l)} + e^{-} \rightarrow \frac{1}{2} H_{2(g)} + OH^{-}_{(aq)}$
The $Na^{+}$ ions remain in the solution as spectator ions,forming $NaOH$ with the produced $OH^{-}$ ions. The presence of $NaOH$ is confirmed experimentally by the pink color observed upon adding phenolphthalein indicator near the cathode.

(N/A) Reduction reactions occur on the surface of the cathode. If more than one species is present near the cathode,the species with the higher $E^{\ominus}$ value undergoes reduction.
For example,in an aqueous $NaCl$ solution,both $Na^{+}$ ions from $NaCl$ and $H^{+}$ ions from $H_{2}O$ are present near the cathode.
$NaCl_{(aq)} \rightarrow Na^{+}_{(aq)} + Cl^{-}_{(aq)}$
$H_{2}O_{(l)} \rightleftharpoons H^{+}_{(aq)} + OH^{-}_{(aq)}$
$(i) \ Na^{+}_{(aq)} + e^{-} \rightarrow Na_{(s)} \quad E^{\ominus} = -2.71 \ V$
$(ii) \ H^{+}_{(aq)} + e^{-} \rightarrow \frac{1}{2} H_{2(g)} \quad E^{\ominus} = 0.00 \ V$
Comparing the $E^{\ominus}$ values,reaction $(ii)$ has a higher value than reaction $(i)$. Therefore,$H^{+}$ ions from water are reduced at the cathode to produce $H_{2}$ gas.
The overall cathodic reaction is:
$H_{2}O_{(l)} + e^{-} \rightarrow \frac{1}{2} H_{2(g)} + OH^{-}_{(aq)}$
As $Na^{+}$ ions do not participate in the reaction,they remain in the solution as spectator ions,combining with $OH^{-}$ to form $NaOH$. The presence of $NaOH$ is confirmed by the pink color observed when phenolphthalein is added near the cathode.

(N/A) When multiple species are available for reduction at the cathode,the species with the higher (more positive) standard electrode potential $(E^\circ)$ will be reduced preferentially.
This is because a higher $E^\circ$ value indicates a greater tendency to gain electrons.
Example: In an aqueous solution of $NaCl$,both $Na^+$ ions and $H_2O$ molecules are present near the cathode.
The reduction potentials are:
$Na^+ + e^- \rightarrow Na$ $(E^\circ = -2.71 \ V)$
$2H_2O + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$ $(E^\circ = -0.83 \ V)$
Since $-0.83 \ V > -2.71 \ V$,water molecules are reduced at the cathode instead of $Na^+$ ions,resulting in the evolution of $H_2$ gas.

(N/A) The standard electrode potentials are given as:
$E_{(H^+|O_2|H_2O)}^o = 1.23 \ V$ (Cathode reaction: $O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$)
$E_{(Fe^{2+}|Fe)}^o = -0.44 \ V$ (Anode reaction: $Fe \rightarrow Fe^{2+} + 2e^-$)
To construct the cell,the reaction with the higher reduction potential acts as the cathode and the one with the lower reduction potential acts as the anode.
Cell Reaction:
Anode: $2Fe(s) \rightarrow 2Fe^{2+}(aq) + 4e^-$
Cathode: $O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l)$
Overall: $2Fe(s) + O_2(g) + 4H^+(aq) \rightarrow 2Fe^{2+}(aq) + 2H_2O(l)$
Calculation of $E_{cell}^o$:
$E_{cell}^o = E_{cathode}^o - E_{anode}^o$
$E_{cell}^o = 1.23 \ V - (-0.44 \ V)$
$E_{cell}^o = 1.67 \ V$

(B) No,only the difference in potential between two electrodes can be measured.
Absolute electrode potential cannot be measured because a single electrode represents only half of a redox reaction (either oxidation or reduction). To complete the circuit and measure a potential difference,two different electrodes are required to form a complete electrochemical cell.

(B) No,$E_{cell}^o$ or $\Delta_r G^o$ for a cell reaction cannot be equal to zero.
If $E_{cell}^o = 0$,then the standard Gibbs free energy change $\Delta_r G^o = -nFE_{cell}^o$ would also be zero.
This condition implies that the reaction is at equilibrium under standard conditions,meaning there is no net flow of electrons and the cell cannot perform any useful work.
Therefore,such a cell has no practical utility.

(A) The negative value of the standard reduction potential $(E^o)$ indicates that the metal is more reactive than hydrogen $(H_2)$.
Since $E_{Zn^{2+}|Zn}^o = -0.76 \ V$ is negative,$Zn$ has a greater tendency to lose electrons compared to $H_2$.
Therefore,$Zn$ acts as a stronger reducing agent than $H_2$ and can displace $H_2$ from acidic solutions.
In a galvanic cell,$Zn$ will act as the anode and undergo oxidation $(Zn \rightarrow Zn^{2+} + 2e^-)$.

(A) The potential difference that develops between a metal electrode and its surrounding electrolyte solution is known as electrode potential.

(A) The Standard Hydrogen Electrode $(SHE)$ is used as the primary reference electrode.
The reduction potential of the standard hydrogen electrode is arbitrarily assigned a value of $0.00 \ V$ at all temperatures.
Because of this defined value,it serves as the universal reference point to measure the electrode potential of any other half-cell.

(N/A) The relationship between the Gibbs free energy change $(\Delta_{r}G)$ and the cell emf $(E_{cell})$ is given by the equation: $\Delta_{r}G = -nFE_{cell}$.
Here,$n$ is the number of moles of electrons transferred,$F$ is the Faraday constant $(96487 \ C \ mol^{-1})$,and $E_{cell}$ is the cell potential.
Under standard conditions,this is expressed as: $\Delta_{r}G^{\circ} = -nFE^{\circ}_{cell}$.
Maximum work is obtained from a galvanic cell when the cell reaction is carried out reversibly.

The symbolic representation of a half-cell is written as $\text{Electrode} | \text{Electrolyte}$ for oxidation and $\text{Electrolyte} | \text{Electrode}$ for reduction.
$(i)$ $2H^{+}_{(aq)} + 2e^- \to H_{2_{(g)}}$ represents a reduction half-cell involving a gas electrode: $H^{+}_{(aq)} | H_{2_{(g)}} | Pt_{(s)}$.
$(ii)$ $Br_{2_{(aq)}} + 2e^- \to 2Br^{-}_{(aq)}$ represents a reduction half-cell involving a non-metal electrode: $Br_{2_{(aq)}} | Br^{-}_{(aq)} | Pt_{(s)}$.
$(iii)$ $2Br^{-}_{(aq)} \to Br_{2_{(aq)}} + 2e^-$ represents an oxidation half-cell involving a non-metal electrode: $Pt_{(s)} | Br^{-}_{(aq)} | Br_{2_{(aq)}}$.

(N/A) The strength of a reductant or oxidant can be determined by measuring the standard electrode potential $(E^{\circ})$ of the half-cell by connecting it with a Standard Hydrogen Electrode $(SHE)$.
$1$. Construct an electrochemical cell by connecting the half-cell of interest with a $SHE$ (where $E^{\circ}_{SHE} = 0.00 \ V$).
$2$. Measure the standard cell potential $(E^{\circ}_{cell})$.
$3$. Using the formula $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$,calculate the standard electrode potential of the half-cell.
$4$. $A$ more negative $E^{\circ}$ value indicates a stronger reducing agent,while a more positive $E^{\circ}$ value indicates a stronger oxidizing agent.
Example: Determining the standard electrode potential of $Zn^{2+}/Zn$.
When $Zn$ electrode is connected to $SHE$ as an anode,the measured $E^{\circ}_{cell} = 0.76 \ V$.
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$
$0.76 \ V = 0.00 \ V - E^{\circ}_{Zn^{2+}/Zn}$
$E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$
Since the value is negative,$Zn$ acts as a strong reducing agent.

(D) The standard reduction potentials $(E^{0})$ at $298 \; K$ are as follows:
$Au^{3+} + 3e^{-} \rightarrow Au_{(s)}, E^{0} = +1.40 \; V$
$Fe^{2+} + 2e^{-} \rightarrow Fe_{(s)}, E^{0} = -0.44 \; V$
$Mg^{2+} + 2e^{-} \rightarrow Mg_{(s)}, E^{0} = -2.36 \; V$
$K^{+} + 1e^{-} \rightarrow K_{(s)}, E^{0} = -2.93 \; V$
According to the electrochemical series,the species with the highest positive standard reduction potential occupies the top position. Therefore,$Au^{3+} + 3e^{-} \rightarrow Au_{(s)}$ is at the top.

(C) The reduction potential of an element in aqueous solution is determined by the energy changes involved in the conversion of a solid metal to its aqueous ion.
The process is represented by the Born-Haber cycle:
$M(s) \rightarrow M(g)$ (Sublimation enthalpy)
$M(g) \rightarrow M^+(aq) + e^-$ (Ionisation enthalpy)
$M^+(g) + H_2O \rightarrow M^+(aq)$ (Hydration enthalpy)
Therefore,sublimation enthalpy,ionisation enthalpy,and hydration enthalpy all contribute to the standard electrode potential $(E^\circ)$.
Electron gain enthalpy is not involved in the oxidation of a metal to its cation.
Thus,the total number of properties that affect the reduction potential is $3$.

(N/A) The standard electrode potential $(E^{\circ})$ for the reduction of $Cu^{2+}$ to $Cu$ is positive $(+0.34 \ V)$.
This indicates that $Cu$ is a weaker reducing agent than $H_{2}$.
Therefore,copper cannot reduce $H^{+}$ ions present in acids to $H_{2}$ gas.

(D) For the reaction to be spontaneous,the cell potential $E_{cell}^{o}$ must be positive.
Reduction (Cathode): $2MnO_{4}^{-} + 16H^{+} + 10e^{-} \rightarrow 2Mn^{2+} + 8H_{2}O$ $(E^{o} = +1.510 \, V)$
Oxidation (Anode): $5H_{2}O \rightarrow \frac{5}{2} O_{2} + 10H^{+} + 10e^{-}$ $(E^{o} = +1.223 \, V)$
Overall reaction: $2MnO_{4}^{-} + 6H^{+} \rightarrow 2Mn^{2+} + \frac{5}{2} O_{2} + 3H_{2}O$
$E_{cell}^{o} = E_{cathode}^{o} - E_{anode}^{o}$
$E_{cell}^{o} = 1.510 \, V - 1.223 \, V = +0.287 \, V$
Since $E_{cell}^{o} > 0$,the reaction is spontaneous,and $MnO_{4}^{-}$ will liberate $O_{2}$ from water in the presence of an acid.