Electrode potential and ECell Questions in English

(A) The strength of a reducing agent is determined by its standard electrode potential $(E^{\circ})$.
Lower (more negative) $E^{\circ}$ values indicate a stronger tendency to undergo oxidation,making the species a stronger reducing agent.
The standard reduction potentials are:
$K^+/K = -2.93 \ V$
$Mg^{2+}/Mg = -2.37 \ V$
$Al^{3+}/Al = -1.66 \ V$
$Ag^+/Ag = +0.80 \ V$
Since $K$ has the most negative $E^{\circ}$ value,it is the strongest reducing agent among the given options.

(A) For the given cell,$Zn$ acts as the anode and $Pb$ acts as the cathode.
$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$
$E_{cell}^{\circ} = E^{\circ}_{Pb^{2+}/Pb} - E^{\circ}_{Zn^{2+}/Zn}$
$E_{cell}^{\circ} = -0.126 \ V - (-0.763 \ V)$
$E_{cell}^{\circ} = -0.126 \ V + 0.763 \ V = 0.637 \ V$

(A) For the given cell reaction,the anode is $Cd$ and the cathode is $Ag$.
The standard cell potential is calculated using the formula: $E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ}$.
Substituting the given values: $E_{\text{cell}}^{\circ} = 0.799 \ V - (-0.403 \ V)$.
$E_{\text{cell}}^{\circ} = 0.799 + 0.403 = 1.202 \ V$.

(A) Among the given species,only iodine has a positive standard reduction potential $(E^{\circ})$ value.
Higher (more positive) $E^{\circ}$ value for a half-reaction indicates a greater tendency for the species to get reduced.
Since tin $(Sn)$ acts as a reducing agent,it will easily reduce the species with the highest reduction potential.
Therefore,iodine $(I_2)$ is reduced by tin $(Sn)$ easily.

(B) The standard cell potential is given by the formula: $E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ}$.
In the given cell,$Ag$ is the cathode and $Cu$ is the anode.
So,$E_{\text{cell}}^{\circ} = E_{Ag^{+}/Ag}^{\circ} - E_{Cu^{2+}/Cu}^{\circ}$.
Substituting the given values: $0.46 \ V = 0.80 \ V - E_{Cu^{2+}/Cu}^{\circ}$.
Rearranging the equation: $E_{Cu^{2+}/Cu}^{\circ} = 0.80 \ V - 0.46 \ V = 0.34 \ V$.

(B) The standard cell potential is calculated using the formula: $E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ}$.
In the given reaction,$Al$ is oxidized (anode) and $Ni^{2+}$ is reduced (cathode).
$E_{\text{cathode}}^{\circ} = E_{Ni^{2+}/Ni}^{\circ} = -0.25 \ V$.
$E_{\text{anode}}^{\circ} = E_{Al^{3+}/Al}^{\circ} = -1.66 \ V$.
$E_{\text{cell}}^{\circ} = -0.25 \ V - (-1.66 \ V) = -0.25 + 1.66 = 1.41 \ V$.

(B) The standard hydrogen electrode $(SHE)$ is represented as $Pt | H_2(g, 1 \ atm) | H^{+}(1 \ M)$ when it acts as an anode or $H^{+}(1 \ M) | H_2(g, 1 \ atm) | Pt$ when it acts as a cathode.
Given the options,the standard representation for the electrode is $Pt | H_2(g, 1 \ atm) | H^{+}(1 \ M)$.
Under standard conditions,the concentration of $H^{+}$ ions is $1 \ M$ and the pressure of $H_2$ gas is $1 \ atm$ (or $1 \ bar$).

(B) The standard cell potential is calculated using the formula: $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$.
In this cell,$Cd^{2+}/Cd$ acts as the cathode (higher reduction potential) and $Zn^{2+}/Zn$ acts as the anode (lower reduction potential).
Substituting the values: $E_{cell}^{\circ} = -0.403 \ V - (-0.763 \ V)$.
$E_{cell}^{\circ} = -0.403 + 0.763 = 0.360 \ V$.

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

(C) The standard cell potential is calculated using the formula: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
In the given cell $Ni|Ni^{2+} || Cu^{2+}| Cu$,$Cu$ acts as the cathode and $Ni$ acts as the anode.
$E^{\circ}_{cell} = E^{\circ}_{Cu^{2+}/Cu} - E^{\circ}_{Ni^{2+}/Ni}$.
Substituting the given values: $E^{\circ}_{cell} = 0.337 \ V - (-0.236 \ V)$.
$E^{\circ}_{cell} = 0.337 \ V + 0.236 \ V = 0.573 \ V$.

(B) The standard cell potential is given by the formula: $E_{cell}^{0} = E_{cathode}^{0} - E_{anode}^{0}$.
Here,the copper electrode acts as the anode and the silver electrode acts as the cathode.
Given: $E_{cell}^{0} = 0.463 \ V$ and $E_{Cu}^{0} = 0.337 \ V$.
Substituting the values: $0.463 \ V = E_{Ag}^{0} - 0.337 \ V$.
Therefore,$E_{Ag}^{0} = 0.463 \ V + 0.337 \ V = 0.800 \ V$.

(D) The given cell reaction is: $Fe^{2+} + Sn \longrightarrow Fe + Sn^{2+}$
In this reaction,$Fe^{2+}$ is reduced to $Fe$ (cathode) and $Sn$ is oxidized to $Sn^{2+}$ (anode).
The standard reduction potentials are given as: $E^o_{Fe^{2+}/Fe} = -0.44 \ V$ and $E^o_{Sn^{2+}/Sn} = -0.14 \ V$.
The standard emf of the cell $(E^o_{cell})$ is calculated as:
$E^o_{cell} = E^o_{cathode} - E^o_{anode}$
$E^o_{cell} = E^o_{Fe^{2+}/Fe} - E^o_{Sn^{2+}/Sn}$
$E^o_{cell} = -0.44 \ V - (-0.14 \ V)$
$E^o_{cell} = -0.44 + 0.14 = -0.30 \ V$

(D) The calomel electrode is a secondary reference electrode consisting of mercury $(Hg)$ and mercury$(I)$ chloride $(Hg_2Cl_2)$.
The standard reduction potential $(E^\circ_{red})$ of the saturated calomel electrode is approximately $+0.242 \ V$ to $+0.28 \ V$ depending on the concentration of $KCl$.
By convention,the standard oxidation potential $(E^\circ_{ox})$ is the negative of the standard reduction potential $(E^\circ_{ox} = -E^\circ_{red})$.
Therefore,for a standard calomel electrode,the oxidation potential is approximately $-0.28 \ V$.

(A) The given reaction is $3 Sn^{2+} + 2 Au^{3+} \longrightarrow 3 Sn^{4+} + 2 Au$.
In this reaction,$Sn^{2+}$ is oxidized to $Sn^{4+}$ (anode) and $Au^{3+}$ is reduced to $Au$ (cathode).
The standard reduction potential for the cathode is $E_{\text{cathode}}^{\circ} = E_{Au^{3+}/Au}^{\circ} = 1.5 \ V$.
The standard reduction potential for the anode is $E_{\text{anode}}^{\circ} = E_{Sn^{4+}/Sn^{2+}}^{\circ} = 0.15 \ V$.
The standard cell potential is calculated as:
$E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ} = 1.5 \ V - 0.15 \ V = 1.35 \ V$.

(D) The standard cell potential is calculated using the formula: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$
In the given reaction,$Mg$ is oxidized to $Mg^{2+}$ (anode) and $Cu^{2+}$ is reduced to $Cu$ (cathode).
Therefore,$E^{\circ}_{cell} = E^{\circ}_{(Cu^{2+}/Cu)} - E^{\circ}_{(Mg^{2+}/Mg)}$
Substituting the given values: $E^{\circ}_{cell} = 0.337 \ V - (-2.37 \ V)$
$E^{\circ}_{cell} = 0.337 + 2.37 = 2.707 \ V \approx 2.7 \ V$.

(A) The reducing power of a substance is inversely proportional to its standard reduction potential $(E^{\circ}_{red})$.
$A$ more negative $E^{\circ}_{red}$ value indicates a greater tendency to undergo oxidation (loss of electrons),making the species a stronger reducing agent.
Comparing the given values:
$E^{\circ}_{Zn^{2+}/Zn} = -0.762 \ V$
$E^{\circ}_{Cr^{3+}/Cr} = -0.74 \ V$
$E^{\circ}_{H^{+}/H_2} = 0.00 \ V$
$E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.77 \ V$
Since $Zn_{(s)}$ has the most negative standard reduction potential $(-0.762 \ V)$,it is the strongest reducing agent among the given options.

(C) For a spontaneous cell reaction,the Gibbs free energy change $(\Delta G)$ must be negative.
According to the relation $\Delta G = -nFE^{\circ}_{cell}$,for $\Delta G$ to be negative,$E^{\circ}_{cell}$ must be positive.
Thus,both the condition of $\Delta G^{\circ} < 0$ and $E^{\circ}_{cell} > 0$ indicate spontaneity.
However,in the context of standard thermodynamic criteria for spontaneity,$\Delta G^{\circ}$ being negative is the fundamental condition.

(D) The reducing power of a substance is inversely proportional to its standard reduction potential $(E^{\circ}_{red})$.
Substances with more negative $E^{\circ}_{red}$ values are stronger reducing agents.
Given values are:
$A: +0.68 \ V$
$C: -0.50 \ V$
$B: -2.54 \ V$
Comparing the values: $-2.54 < -0.50 < +0.68$.
Therefore,the order of reducing power is $B > C > A$.

(C) The strength of a reducing agent is inversely proportional to its standard reduction potential $(E^0_{red})$. $A$ lower (more negative) $E^0_{red}$ value indicates a stronger reducing agent.
Comparing the given values:
$E^0_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}$
$E^0_{Cl_2/Cl^{-}} = 1.36 \text{ V}$
$E^0_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \text{ V}$
$E^0_{Cr^{3+}/Cr} = -0.74 \text{ V}$
Since $E^0_{Cr^{3+}/Cr}$ has the lowest (most negative) value of $-0.74 \text{ V}$,the species $Cr$ acts as the strongest reducing agent.

(D) The strength of a reducing agent is inversely proportional to its standard reduction potential $(E^0_{red})$.
$A$ lower (more negative) reduction potential indicates that the substance is more easily oxidized and is therefore a stronger reducing agent.
Comparing the given values:
$1. E^0_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}$
$2. E^0_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \text{ V}$
$3. E^0_{Br_2/Br^{-}} = 1.09 \text{ V}$
$4. E^0_{Zn^{2+}/Zn} = -0.76 \text{ V}$
Since $E^0_{Zn^{2+}/Zn}$ has the lowest value $(-0.76 \text{ V})$,$Zn$ is the strongest reducing agent.

(C) To store an aqueous solution of $CuSO_4$,the metal of the container must not react with $Cu^{2+}$ ions.
$A$ reaction will occur if the reduction potential of the container metal is lower than that of $Cu^{2+}/Cu$ $(0.34 \ V)$.
The reduction potentials $(E^0_{M^{n+}/M})$ are:
$E^0_{Fe^{2+}/Fe} = -0.44 \ V$
$E^0_{Ni^{2+}/Ni} = -0.25 \ V$
$E^0_{Al^{3+}/Al} = -1.66 \ V$
$E^0_{Ag^{+}/Ag} = 0.80 \ V$
For the reaction $M + Cu^{2+} \rightarrow M^{n+} + Cu$ to be non-spontaneous,the $E^0_{cell}$ must be negative.
$E^0_{cell} = E^0_{cathode} - E^0_{anode} = E^0_{Cu^{2+}/Cu} - E^0_{M^{n+}/M}$.
For $E^0_{cell} < 0$,we require $E^0_{M^{n+}/M} > E^0_{Cu^{2+}/Cu}$.
Comparing values,only $Ag$ has a reduction potential $(0.80 \ V)$ greater than $Cu$ $(0.34 \ V)$.
Therefore,$CuSO_4$ can be stored in an $Ag$ container.

(C) The reducing power of a metal is inversely proportional to its standard reduction potential $(E^0)$.
Lower (more negative) $E^0$ values indicate a stronger tendency to lose electrons,making the metal a better reducing agent.
The given values are: $E^0_A = 0.34 \ V$,$E^0_B = -0.80 \ V$,and $E^0_C = -0.46 \ V$.
Comparing these values: $-0.80 \ V < -0.46 \ V < 0.34 \ V$.
Therefore,the order of reducing ability is $B > C > A$.

(C) metal can displace $H_2$ gas from $HCl$ solution if its standard reduction potential $(E^0_{M^{n+}/M})$ is less than the standard reduction potential of the hydrogen electrode,which is $0.00 \ V$.
Metals with negative reduction potentials act as reducing agents and can reduce $H^+$ ions to $H_2$ gas.
Comparing the given values:
$E^0_{Zn^{2+}/Zn} = -0.76 \ V$ (Negative,can produce $H_2$)
$E^0_{Fe^{2+}/Fe} = -0.44 \ V$ (Negative,can produce $H_2$)
$E^0_{Ni^{2+}/Ni} = -0.25 \ V$ (Negative,can produce $H_2$)
$E^0_{Cu^{2+}/Cu} = +0.34 \ V$ (Positive,cannot produce $H_2$)
Therefore,$Cu$ cannot produce $H_2$ gas.

(C) The standard Gibbs energy change $\Delta G^{\circ}$ is given by the formula: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.
Here,the number of electrons transferred in the reaction is $n = 2$.
The standard electrode potential is $E^{\circ}_{cell} = 1.1 \ V$.
The Faraday constant is $F = 96487 \ C \ mol^{-1}$.
Substituting these values: $\Delta G^{\circ} = -(2) \times (96487 \ C \ mol^{-1}) \times (1.1 \ V)$.
$\Delta G^{\circ} = -212271.4 \ J \ mol^{-1}$.
Converting to $kJ \ mol^{-1}$: $\Delta G^{\circ} = -212.27 \ kJ \ mol^{-1}$.

(D) We use the relationship $\Delta G^0 = -nFE^0$.
For the reaction $Fe^{3+} + 3e^- \rightarrow Fe$,$\Delta G^0_1 = -3Fx$.
For the reaction $Fe^{2+} + 2e^- \rightarrow Fe$,$\Delta G^0_2 = -2Fy$.
We want the potential for $Fe^{3+} + e^- \rightarrow Fe^{2+}$.
This can be obtained by subtracting the second reaction from the first: $(Fe^{3+} + 3e^-$ $\rightarrow Fe) - (Fe^{2+} + 2e^-$ $\rightarrow Fe)$ $\Rightarrow Fe^{3+} + e^-$ $\rightarrow Fe^{2+}$.
Therefore,$\Delta G^0_3 = \Delta G^0_1 - \Delta G^0_2 = -3Fx - (-2Fy) = -3Fx + 2Fy$.
Since $\Delta G^0_3 = -nFE^0_{Fe^{3+}/Fe^{2+}}$ where $n=1$,we have $-1FE^0_{Fe^{3+}/Fe^{2+}} = -3Fx + 2Fy$.
Thus,$E^0_{Fe^{3+}/Fe^{2+}} = 3x - 2y$.

(C) The standard cell potential $E_{cell}^0$ is calculated using the formula: $E_{cell}^0 = E_{cathode}^0 - E_{anode}^0$.
In the given cell $Zn | Zn^{2+} || Cu^{2+} | Cu$,$Zn$ acts as the anode and $Cu$ acts as the cathode.
Given: $E_{Cu^{2+}/Cu}^0 = 0.34 \ V$ and $E_{Zn^{2+}/Zn}^0 = -0.76 \ V$.
Substituting the values: $E_{cell}^0 = 0.34 \ V - (-0.76 \ V) = 0.34 \ V + 0.76 \ V = 1.10 \ V$.

(D) Reducing power is inversely proportional to the standard reduction potential $(E^\circ)$. The lower (more negative) the $E^\circ$ value,the stronger the reducing agent.
Given standard reduction potentials:
$E^\circ_{Ag^{+}/Ag} = 0.80 \ V$
$E^\circ_{Hg^{2+}/Hg} = 0.79 \ V$
$E^\circ_{Cr^{3+}/Cr} = -0.74 \ V$
$E^\circ_{Mg^{2+}/Mg} = -2.37 \ V$
Comparing these values,the increasing order of reducing power is: $Ag < Hg < Cr < Mg$.

(B) For a reaction to be spontaneous in an electrochemical cell,the cell potential $E_{cell}^0$ must be positive.
$E_{cell}^0 = E_{cathode}^0 - E_{anode}^0$.
Given $E_{Cl_2 \mid 2Cl^-}^0 = 1.36 \ V$ and $E_{Br_2 \mid 2Br^-}^0 = 1.09 \ V$.
For option $B$: $2Br^- + Cl_2 \rightarrow Br_2 + 2Cl^-$.
Here,$Cl_2$ is reduced to $Cl^-$ (cathode) and $Br^-$ is oxidized to $Br_2$ (anode).
$E_{cell}^0 = E_{Cl_2 \mid 2Cl^-}^0 - E_{Br_2 \mid 2Br^-}^0 = 1.36 \ V - 1.09 \ V = 0.27 \ V$.
Since $E_{cell}^0 > 0$,the reaction is spontaneous.

(A) reducing agent is a substance that loses electrons and gets oxidized.
The strength of a reducing agent is inversely proportional to its standard reduction potential $(E^{\circ})$.
The more negative the $E^{\circ}$ value,the greater the tendency to lose electrons,making it a stronger reducing agent.
Comparing the given values:
$E^{\circ} (Zn^{2+}/Zn) = -0.762 \ V$
$E^{\circ} (Cr^{3+}/Cr) = -0.740 \ V$
$E^{\circ} (H^{+}/H_2) = 0.0 \ V$
$E^{\circ} (F_2/F^-) = 2.87 \ V$
Since $-0.762 \ V$ is the most negative value,$Zn_{(s)}$ is the strongest reducing agent.

(C) For a cell reaction to be spontaneous,the Gibbs free energy change $(\Delta G)$ must be negative.
Since $\Delta G = -nFE_{\text{cell}}$,a negative $\Delta G$ implies that the cell potential $(E_{\text{cell}})$ must be positive.

(A) For the reaction $Mn^{+7} + 5e^{-} \rightarrow Mn^{2+}$,$\Delta G^{0}_{1} = -nFE^{0} = -5 \times F \times 1.5 = -7.5F$.
For the reaction $Mn^{+4} + 2e^{-} \rightarrow Mn^{2+}$,$\Delta G^{0}_{2} = -nFE^{0} = -2 \times F \times 1.2 = -2.4F$.
To obtain the reaction $Mn^{+7} + 3e^{-} \rightarrow Mn^{4+}$,we subtract the second reaction from the first:
$(Mn^{+7} + 5e^{-}$ $\rightarrow Mn^{2+}) - (Mn^{+4} + 2e^{-}$ $\rightarrow Mn^{2+})$ $\Rightarrow Mn^{+7} + 3e^{-}$ $\rightarrow Mn^{4+}$.
Thus,$\Delta G^{0}_{3} = \Delta G^{0}_{1} - \Delta G^{0}_{2} = -7.5F - (-2.4F) = -5.1F$.
Using $\Delta G^{0}_{3} = -nFE^{0}_{3}$ where $n=3$:
$-3FE^{0}_{3} = -5.1F
\therefore E^{0}_{3} = \frac{5.1}{3} = 1.7 \ V$.

(C) The standard reduction potentials are $E^o_{Zn^{2+}/Zn} = -0.76 \ V$ and $E^o_{Fe^{2+}/Fe} = -0.41 \ V$.
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.
$E^o_{cell} = E^o_{cathode} - E^o_{anode}$
$E^o_{cell} = (-0.41 \ V) - (-0.76 \ V) = -0.41 \ V + 0.76 \ V = +0.35 \ V$.

(D) The cell reaction is $Fe^{2+} + Zn \longrightarrow Zn^{2+} + Fe$.
Here,$Zn$ is oxidized to $Zn^{2+}$ (anode) and $Fe^{2+}$ is reduced to $Fe$ (cathode).
The formula for the $emf$ of the cell is $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
Substituting the given values: $E^{\circ}_{cell} = E^{\circ}_{Fe^{2+}/Fe} - E^{\circ}_{Zn^{2+}/Zn}$.
$E^{\circ}_{cell} = -0.44 \ V - (-0.76 \ V)$.
$E^{\circ}_{cell} = -0.44 \ V + 0.76 \ V = +0.32 \ V$.

(A) The metals that are present below hydrogen in the electrochemical series cannot displace hydrogen from dilute acids.
Among the given metals, only $Ag$ is present below hydrogen in the electrochemical series.
Therefore, $Ag$ does not react with dilute $HCl$ to evolve hydrogen gas.
$Ag + \text{dil } HCl \longrightarrow \text{No reaction}$

(B) The given cell reaction is $Fe_{(s)} + 2 Fe^{3+}_{(aq)} \longrightarrow 3 Fe^{2+}_{(aq)}$.
This reaction can be split into two half-reactions:
Anode (Oxidation): $Fe_{(s)} \longrightarrow Fe^{2+}_{(aq)} + 2e^-$,$E^0_{ox} = -E^0_{Fe^{2+}/Fe} = -(-0.441 \ V) = 0.441 \ V$.
Cathode (Reduction): $2 Fe^{3+}_{(aq)} + 2e^- \longrightarrow 2 Fe^{2+}_{(aq)}$,$E^0_{red} = E^0_{Fe^{3+}/Fe^{2+}} = 0.771 \ V$.
The standard cell potential is given by $E^0_{cell} = E^0_{ox} + E^0_{red}$.
$E^0_{cell} = 0.441 \ V + 0.771 \ V = 1.212 \ V$.

(D) The stability of an oxidation state can be inferred from the standard electrode potential $(E^0)$. $A$ more negative $E^0$ value for the reduction half-reaction $(M^n+ + ne^- \rightarrow M)$ indicates that the metal is more easily oxidized,meaning the higher oxidation state is more stable relative to the metal.
For $Al$: $E^0(Al^{3+}/Al) = -1.66 \ V$ and $E^0(Al^{+}/Al) = +0.55 \ V$. Since $E^0(Al^{3+}/Al) < E^0(Al^{+}/Al)$,$Al^{3+}$ is much more stable than $Al^{+}$.
For $Tl$: $E^0(Tl^{3+}/Tl) = +1.26 \ V$ and $E^0(Tl^{+}/Tl) = -0.34 \ V$. Since $E^0(Tl^{+}/Tl) < E^0(Tl^{3+}/Tl)$,$Tl^{+}$ is more stable than $Tl^{3+}$.
Comparing $Tl^{+}$ and $Al^{+}$: $Tl^{+}$ has a negative reduction potential $(-0.34 \ V)$,making it relatively stable,whereas $Al^{+}$ has a positive reduction potential $(+0.55 \ V)$,making it highly unstable and a strong oxidizing agent.
Therefore,$Tl^{+}$ is more stable than $Al^{+}$.

(C) $Al^{3+}$ is much more stable than $Al$ and due to its high electropositive nature,the standard reduction potential of $Al^{3+}/Al$ is $-1.66 \ V$.
$Tl^{3+}$ is less stable than $Tl$ due to the inert pair effect,making $Tl^{3+}$ a strong oxidizing agent that is easily reduced to $Tl$. Thus,the standard reduction potential of $Tl^{3+}/Tl$ is $+1.26 \ V$.

(A) The formula for standard Gibbs energy change is $\Delta G^{\circ} = -nFE_{\text{cell}}^{\circ}$.
Given values are $n = 2$ (number of electrons transferred),$E_{\text{cell}}^{\circ} = 1.1 \ V$,and Faraday's constant $F = 96500 \ C \ mol^{-1}$.
Substituting these values into the equation:
$\Delta G^{\circ} = -2 \times 96500 \times 1.1 \ J \ mol^{-1}$.
$\Delta G^{\circ} = -212300 \ J \ mol^{-1}$.
Converting to $kJ \ mol^{-1}$ by dividing by $1000$:
$\Delta G^{\circ} = -212.3 \ kJ \ mol^{-1}$.

(C) Given reactions are:
$(i) M^{2+}_{(aq)} + e^- \rightarrow M^{+}_{(aq)}, E_1^\circ = +0.15 \text{ V}$
$(ii) M^{+}_{(aq)} + e^- \rightarrow M_{(s)}, E_2^\circ = +0.50 \text{ V}$
The target reaction is:
$(iii) M^{2+}_{(aq)} + 2e^- \rightarrow M_{(s)}, E_3^\circ = ?$
Using the relation $\Delta G_3^\circ = \Delta G_1^\circ + \Delta G_2^\circ$:
$-n_3FE_3^\circ = -n_1FE_1^\circ - n_2FE_2^\circ$
$n_3E_3^\circ = n_1E_1^\circ + n_2E_2^\circ$
$2 \cdot E_3^\circ = (1 \cdot 0.15) + (1 \cdot 0.50)$
$2 \cdot E_3^\circ = 0.65$
$E_3^\circ = \frac{0.65}{2} = 0.325 \text{ V}$.

(C) metal with a more negative standard reduction potential can displace a metal with a more positive standard reduction potential from its salt solution.
Given potentials:
$E^{\circ}_{Fe^{2+}/Fe} = -0.44 \ V$
$E^{\circ}_{Cu^{2+}/Cu} = +0.34 \ V$
$E^{\circ}_{Ag^{+}/Ag} = +0.80 \ V$
Analysis:
$(i)$ $Cu$ $(+0.34 \ V)$ cannot displace $Fe$ $(-0.44 \ V)$. Statement $(i)$ is incorrect.
$(ii)$ $Fe$ $(-0.44 \ V)$ can displace $Cu$ $(+0.34 \ V)$. Statement $(ii)$ is correct.
$(iii)$ $Ag$ $(+0.80 \ V)$ cannot displace $Cu$ $(+0.34 \ V)$. Statement $(iii)$ is incorrect.
$(iv)$ $Fe$ $(-0.44 \ V)$ can displace $Ag$ $(+0.80 \ V)$. Statement $(iv)$ is correct.
Therefore,statements $(ii)$ and $(iv)$ are correct.

(C) For hydrogen electrode: $E = E^0 - 0.059 \ pH = 0 - 0.059 \times 10 = -0.591 \ V$. Thus,$(A)$ matches with $(III)$.
$(B)$ The standard reduction potential of $Cu^{2+} | Cu$ is $E^0_{Cu^{2+}/Cu} = 0.337 \ V$. Thus,$(B)$ matches with $(IV)$.
$(C)$ The standard oxidation potential of $Zn | Zn^{2+}$ is $E^0_{Zn/Zn^{2+}} = 0.76 \ V$. Thus,$(C)$ matches with $(I)$.
$(D)$ The value of $\frac{2.303 RT}{F}$ at $298 \ K$ is $\frac{2.303 \times 8.314 \times 298}{96500} \approx 0.059$. Thus,$(D)$ matches with $(II)$.
Therefore,the correct matching is $A-III, B-IV, C-I, D-II$,which corresponds to option $(c)$.

(A) The standard reduction potentials are: $E^{\circ}_{Li^+/Li} = -3.05 \ V$,$E^{\circ}_{Mg^{2+}/Mg} = -2.36 \ V$,$E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$,and $E^{\circ}_{Ni^{2+}/Ni} = -0.25 \ V$.
$A$ metal with a more negative reduction potential acts as a stronger reducing agent and can displace a metal with a less negative (or more positive) reduction potential from its salt solution.
Comparing the values: $E^{\circ}_{Mg^{2+}/Mg} (-2.36 \ V) < E^{\circ}_{Zn^{2+}/Zn} (-0.76 \ V)$.
Since $Mg$ has a lower reduction potential than $Zn$,$Mg$ can displace $Zn$ from its solution.
Therefore,the statement '$Mg$ displaces $Zn$ from its solution' is correct.

(C) The sensitivity of a metal towards oxidation is determined by its standard oxidation potential. Metals with higher positive standard oxidation potentials are more easily oxidized.
Comparing the standard oxidation potentials $(E^o_{ox})$:
$Ca \rightarrow Ca^{2+} + 2e^-$,$E^o_{ox} = +2.87 \ V$
$Al \rightarrow Al^{3+} + 3e^-$,$E^o_{ox} = +1.66 \ V$
$Cr \rightarrow Cr^{3+} + 3e^-$,$E^o_{ox} = +0.74 \ V$
$Fe \rightarrow Fe^{2+} + 2e^-$,$E^o_{ox} = +0.44 \ V$
Since $Ca$ has the highest positive oxidation potential,it is the most sensitive towards oxidation.

(A) The standard electrode potential $(E^{\circ})$ represents the tendency of an electrode to undergo reduction.
For $Li^{+} / Li$,the standard electrode potential is $-3.04 \ V$.
For $Na^{+} / Na$,the standard electrode potential is $-2.714 \ V$.
Therefore,the values are $-3.04 \ V$ and $-2.714 \ V$ respectively.

(C) The half-cell reactions are:
$(1)$ $Mn^{7+} + 5e^- \rightarrow Mn^{2+}$,$\Delta G^{\circ}_1 = -5FE^{\circ}_{Mn^{7+}/Mn^{2+}}$
$(2)$ $Mn^{4+} + 2e^- \rightarrow Mn^{2+}$,$\Delta G^{\circ}_2 = -2FE^{\circ}_{Mn^{4+}/Mn^{2+}}$
$(3)$ $Mn^{7+} + 3e^- \rightarrow Mn^{4+}$,$\Delta G^{\circ}_3 = -3FE^{\circ}_{Mn^{7+}/Mn^{4+}}$
Since reaction $(3)$ = $(1)$ - $(2)$,we have $\Delta G^{\circ}_3 = \Delta G^{\circ}_1 - \Delta G^{\circ}_2$.
Substituting the values: $-3FE^{\circ}_3 = -5F(1.51) - (-2F(1.23))$.
$3E^{\circ}_3 = 5(1.51) - 2(1.23) = 7.55 - 2.46 = 5.09$.
$E^{\circ}_3 = 5.09 / 3 = 1.696 \ V \approx 1.70 \ V$.

(B) The reaction is: $Fe^{3+} + Ce^{3+} \rightarrow Fe^{2+} + Ce^{4+}$
Here,$Fe^{3+}$ is reduced to $Fe^{2+}$ (cathode) and $Ce^{3+}$ is oxidised to $Ce^{4+}$ (anode).
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$
$E^{\circ}_{cell} = E^{\circ}_{Fe^{3+}/Fe^{2+}} - E^{\circ}_{Ce^{4+}/Ce^{3+}}$
$E^{\circ}_{cell} = 0.77 \ V - 1.6 \ V = -0.83 \ V$

(A) The electrode with the higher standard reduction potential acts as the cathode.
Given standard reduction potentials: $E_{Ag^+/Ag}^{\circ} = +0.80 \ V$ and $E_{Cu^{2+}/Cu}^{\circ} = +0.34 \ V$.
Since $E_{Ag^+/Ag}^{\circ} > E_{Cu^{2+}/Cu}^{\circ}$,the silver electrode acts as the cathode and the copper electrode acts as the anode.
The standard cell potential is calculated as:
$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$
$E_{cell}^{\circ} = +0.80 \ V - (+0.34 \ V) = +0.46 \ V$.

(B) The galvanic cell can be represented as: $Mg | Mg^{2+} || Cu^{2+} | Cu$.
The standard cell potential is given by the formula: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
Here,$E^{\circ}_{cathode} = E^{\circ}_{Cu^{2+}/Cu} = 0.34 \ V$ and $E^{\circ}_{cell} = 2.71 \ V$.
Substituting these values: $2.71 \ V = 0.34 \ V - E^{\circ}_{Mg^{2+}/Mg}$.
Therefore,$E^{\circ}_{Mg^{2+}/Mg} = 0.34 \ V - 2.71 \ V = -2.37 \ V$.