Electrode potential and ECell Questions in English

(A) The standard reduction potentials $(E^{\circ})$ for the given pairs are:
$E^{\circ}_{Cl_2 / Cl^{-}} = +1.36 \, V$
$E^{\circ}_{Ag^{+} / Ag} = +0.80 \, V$
$E^{\circ}_{I_2 / I^{-}} = +0.54 \, V$
$E^{\circ}_{Na^{+} / Na} = -2.71 \, V$
$E^{\circ}_{Li^{+} / Li} = -3.05 \, V$
Comparing these values,the order is $1.36 > 0.80 > 0.54 > -2.71 > -3.05$.
Therefore,the correct order is $A > C > B > D > E$.

(C) For a spontaneous reaction,the cell potential $E^{\circ}_{Cell}$ must be positive.
The reduction half-reactions are:
$Fe^{3+} + e^{-} \rightarrow Fe^{2+}$ $\quad$ $E^{\circ} = 0.77 \, V$ (Cathode)
$I_{2} + 2e^{-} \rightarrow 2I^{-}$ $\quad$ $E^{\circ} = 0.54 \, V$ (Anode)
The spontaneous cell reaction is:
$2Fe^{3+} + 2I^{-} \rightarrow 2Fe^{2+} + I_{2}$
$E^{\circ}_{Cell} = E^{\circ}_{Cathode} - E^{\circ}_{Anode}$
$E^{\circ}_{Cell} = 0.77 \, V - 0.54 \, V = 0.23 \, V$
Given $E^{\circ}_{Cell} = x \times 10^{-2} \, V$,we have:
$0.23 = x \times 10^{-2}$
$x = 23$.

(C) cell with less variation in $EMF$ with temperature is preferred as a reference electrode because it can be used over a wider range of temperatures without significant deviation from its standard value.
Therefore,a cell with the minimum value of $(\frac{\partial E}{\partial T})_P$ is preferred.
Comparing the given values: $A = 1 \times 10^{-4}$,$B = 2 \times 10^{-4}$,$C = 0.1 \times 10^{-4}$,$D = 0.2 \times 10^{-4}$.
The minimum value is $0.1 \times 10^{-4}$,which corresponds to cell $C$.

(C) The cell reaction is: $\frac{1}{2} H_{2(g)} + Ag^{+}_{(aq)} \rightarrow H^{+}_{(aq)} + Ag_{(s)}$
Here,the number of electrons involved in the balanced reaction is $n = 1$.
The standard Gibbs free energy change is given by: $\Delta_r G^0 = -n F E_{Cell}^0$
Substituting the values: $\Delta_r G^0 = -1 \times 96500 \ C \ mol^{-1} \times 0.5332 \ V$
$\Delta_r G^0 = -51453.8 \ J \ mol^{-1} = -51.45 \ kJ \ mol^{-1}$
Rounding to the nearest integer,the value is $-51 \ kJ \ mol^{-1}$.

(C) Given:
$(1) \ Sn^{2+} + 2e^{-}$ $\rightarrow Sn, \ E^{\circ}_{1} = -0.140 \ V, \ \Delta G^{\circ}_{1} = -nFE^{\circ}_{1} = -2 \times F \times (-0.140) = +0.280F$
$(2) \ Sn^{4+} + 4e^{-}$ $\rightarrow Sn, \ E^{\circ}_{2} = 0.010 \ V, \ \Delta G^{\circ}_{2} = -nFE^{\circ}_{2} = -4 \times F \times (0.010) = -0.040F$
We need the potential for $Sn^{4+} + 2e^{-} \rightarrow Sn^{2+}$.
This can be obtained by $(2) - (1)$:
$(Sn^{4+} + 4e^{-}) - (Sn^{2+} + 2e^{-}) \rightarrow Sn - Sn$
$Sn^{4+} + 2e^{-} \rightarrow Sn^{2+}$
$\Delta G^{\circ}_{3} = \Delta G^{\circ}_{2} - \Delta G^{\circ}_{1} = -0.040F - 0.280F = -0.320F$
Since $\Delta G^{\circ}_{3} = -nFE^{\circ}_{Sn^{4+}/Sn^{2+}}$,where $n=2$:
$-2FE^{\circ}_{Sn^{4+}/Sn^{2+}} = -0.320F$
$E^{\circ}_{Sn^{4+}/Sn^{2+}} = 0.160 \ V = 16 \times 10^{-2} \ V$
The value is $16$.

(A) The tendency to liberate $H_2$ gas from mineral acids depends on the standard reduction potential of the metal. Metals with a negative reduction potential (above $H$ in the electrochemical series) can displace $H_2$.
Copper $(Cu)$ has a positive standard reduction potential $(E^0 = +0.34 \ V)$,meaning it lies below $H$ in the reactivity series.
Therefore,$Cu$ cannot displace $H_2$ from mineral acids,while $Mn$,$Ni$,and $Zn$ have negative reduction potentials and can liberate $H_2$.

(C) The strength of a reducing agent is inversely proportional to its standard reduction potential $(E^\circ)$.
Lower the standard reduction potential,stronger is the reducing agent.
The standard reduction potentials for the given metals are:
$E^\circ (Cu^{2+}/Cu) = +0.34 \ V$
$E^\circ (Ni^{2+}/Ni) = -0.25 \ V$
$E^\circ (Fe^{2+}/Fe) = -0.44 \ V$
$E^\circ (Zn^{2+}/Zn) = -0.76 \ V$
Comparing these values,$Zn$ has the lowest reduction potential,making it the strongest reducing agent among the given options.
Therefore,the correct option is $(C)$.

(B) Given:
$E_{Cu^{2+}/Cu}^{\circ} = 0.340 \ V$
$E_{Cu^{+}/Cu}^{\circ} = 0.522 \ V$
Step $1$: Write the half-cell reactions:
$(i) \ Cu^{2+} + 2e^{-} \longrightarrow Cu, \ E_{1}^{\circ} = 0.340 \ V, \ \Delta G_{1}^{\circ} = -2FE_{1}^{\circ}$
$(ii) \ Cu^{+} + e^{-} \longrightarrow Cu, \ E_{2}^{\circ} = 0.522 \ V, \ \Delta G_{2}^{\circ} = -1FE_{2}^{\circ}$
Step $2$: We need the reaction for $Cu^{2+} + e^{-} \longrightarrow Cu^{+}$:
This is obtained by $(i) - (ii)$:
$Cu^{2+} + e^{-} \longrightarrow Cu^{+}, \ E_{3}^{\circ} = ?, \ \Delta G_{3}^{\circ} = -1FE_{3}^{\circ}$
Step $3$: Use the relation $\Delta G_{3}^{\circ} = \Delta G_{1}^{\circ} - \Delta G_{2}^{\circ}$:
$-1FE_{3}^{\circ} = -2FE_{1}^{\circ} - (-1FE_{2}^{\circ})$
$E_{3}^{\circ} = 2E_{1}^{\circ} - E_{2}^{\circ}$
$E_{3}^{\circ} = 2(0.340) - 0.522$
$E_{3}^{\circ} = 0.680 - 0.522 = 0.158 \ V$

(B) The reducing strength of a metal is inversely proportional to its standard reduction potential. $A$ more negative standard reduction potential indicates a stronger reducing agent.
Comparing the given values:
$Cr^{3+}/Cr = -0.74 \ V$
$Pb^{2+}/Pb = -0.13 \ V$
$Cu^{2+}/Cu = +0.34 \ V$
$Ag^{+}/Ag = +0.80 \ V$
The order of increasing standard reduction potential is $Cr < Pb < Cu < Ag$. Therefore,the order of decreasing reducing strength is $Cr > Pb > Cu > Ag$.

(B)
For the reactions:
$Fe^{2+} + 2e^{-} \rightarrow Fe ; E_1^{\circ} = -0.440 \ V \quad (I)$
$Fe^{3+} + e^{-} \rightarrow Fe^{2+} ; E_2^{\circ} = 0.770 \ V \quad (II)$
To obtain the reaction $Fe^{3+} + 3e^{-} \rightarrow Fe ; E_3^{\circ} = ?$,we add Eq. $(I)$ and $(II)$.
Since $\Delta G^{\circ} = -nFE^{\circ}$,we have:
$\Delta G_3^{\circ} = \Delta G_1^{\circ} + \Delta G_2^{\circ}$
$-n_3 F E_3^{\circ} = -n_1 F E_1^{\circ} - n_2 F E_2^{\circ}$
$E_3^{\circ} = \frac{n_1 E_1^{\circ} + n_2 E_2^{\circ}}{n_3}$
$E_3^{\circ} = \frac{2(-0.440) + 1(0.770)}{3}$
$E_3^{\circ} = \frac{-0.880 + 0.770}{3} = \frac{-0.110}{3} = -0.0366 \ V \approx -0.037 \ V$

(D)
The reducing ability of metals is determined by their standard reduction potentials $(E^{\circ})$. $A$ metal with a more negative standard reduction potential is a stronger reducing agent.
The standard reduction potentials $(E^{\circ})$ for the given metals are:
$K^+ + e^- \rightarrow K$ $(E^{\circ} = -2.93 \ V)$
$Zn^{2+} + 2e^- \rightarrow Zn$ $(E^{\circ} = -0.76 \ V)$
$Pb^{2+} + 2e^- \rightarrow Pb$ $(E^{\circ} = -0.13 \ V)$
$Au^{3+} + 3e^- \rightarrow Au$ $(E^{\circ} = +1.50 \ V)$
Since the reducing ability increases as the standard reduction potential becomes more negative,the order of reducing ability is $K > Zn > Pb > Au$.

(C) For a reaction to be spontaneous,the cell potential $E^{\circ}_{cell}$ must be positive.
The reaction involves the oxidation of $I^-$ to $I_2$ $(E^{\circ}_{ox} = -0.54 \ V)$ and the reduction of the metal ion $(E^{\circ}_{red})$.
The condition for spontaneity is $E^{\circ}_{cell} = E^{\circ}_{red} + E^{\circ}_{ox} > 0$,which implies $E^{\circ}_{red} > 0.54 \ V$.
Comparing the given reduction potentials:
$(i)$ $E^{\circ} = 0.52 \ V < 0.54 \ V$ (Not spontaneous)
$(ii)$ $E^{\circ} = -0.41 \ V < 0.54 \ V$ (Not spontaneous)
$(iii)$ $E^{\circ} = 0.77 \ V > 0.54 \ V$ (Spontaneous)
$(iv)$ $E^{\circ} = -0.44 \ V < 0.54 \ V$ (Not spontaneous)
Therefore,only the transformation $(iii)$ is feasible.

(A) .
Metals that are placed below hydrogen in the electrochemical reactivity series cannot displace hydrogen from dilute acids.
Among the given metals,$Cu$ (copper) is less reactive than hydrogen $(H)$,meaning its standard reduction potential is higher than that of hydrogen.
Therefore,$Cu$ does not react with $HCl$ to produce $H_2$ gas.

(C) metal ion $M^{2+}$ can liberate $H_2$ from a dilute acid if it can reduce $H^{+}$ to $H_2$ while itself getting oxidized to $M^{3+}$.
This requires the reduction potential of the $M^{3+}/M^{2+}$ couple to be lower than the reduction potential of $H^{+}/H_2$ $(0.00 \ V)$.
Alternatively,the oxidation potential of $M^{2+}/M^{3+}$ must be positive.
Given $E^{\circ}(V^{3+}/V^{2+}) = -0.26 \ V$ and $E^{\circ}(Cr^{3+}/Cr^{2+}) = -0.41 \ V$,both are negative,meaning $V^{2+}$ and $Cr^{2+}$ are strong reducing agents that can reduce $H^{+}$ to $H_2$.
$Mn^{2+}$ and $Co^{2+}$ have positive reduction potentials,meaning they are stable and cannot reduce $H^{+}$ to $H_2$.

(C) For a cell to be feasible,$E^{\circ}_{cell}$ must be positive.
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$
Here,$Co^{3+}$ is reduced $(cathode)$ and $Al$ is oxidized $(anode)$.
$E^{\circ}_{cathode} = E^{\circ}_{Co^{3+}/Co^{2+}} = 1.81 \, V$
$E^{\circ}_{anode} = E^{\circ}_{Al^{3+}/Al} = -1.66 \, V$
$E^{\circ}_{cell} = 1.81 \, V - (-1.66 \, V) = 1.81 + 1.66 = 3.47 \, V$.

(B) The formula for standard Gibbs energy change is $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.
Here,$n = 2$ (number of electrons transferred in the reaction).
$E^{\circ}_{cell} = 1.1 \ V$.
$F = 96487 \ C \ mol^{-1}$.
Substituting the values: $\Delta G^{\circ} = -2 \times 96487 \times 1.1 \ C \cdot V \ mol^{-1}$.
$\Delta G^{\circ} = -212271.4 \ J \ mol^{-1}$.
Converting to $kJ \ mol^{-1}$: $\Delta G^{\circ} = \frac{-212271.4}{1000} \approx -212.27 \ kJ \ mol^{-1}$.

(A) The standard electrode potential $(E^{\circ})$ for the process $M^{+}(aq) + e^{-} \rightarrow M(s)$ is determined by the Born-Haber cycle involving the following steps:
$1$. Sublimation of solid metal: $M(s) \rightarrow M(g)$ (Enthalpy of sublimation,$\Delta H_{sub}$)
$2$. Ionisation of gaseous metal atom: $M(g) \rightarrow M^{+}(g) + e^{-}$ (Ionisation enthalpy,$\Delta H_{IE}$)
$3$. Hydration of gaseous metal ion: $M^{+}(g) + aq \rightarrow M^{+}(aq)$ (Hydration enthalpy,$\Delta H_{hyd}$)
Since the process involves the reduction of $M^{+}$ to $M(s)$,the ionisation of a solid metal atom directly is not a step in the cycle; rather,it is the ionisation of a gaseous metal atom that is considered.

(C) metal will be oxidized by $NO_3^{-}$ if the standard reduction potential $(SRP)$ of the metal is lower than the $SRP$ of the $NO_3^{-}$ half-cell $(E^0 = 0.97 \ V)$.
Comparing the given values:
$1$. $V_{(s)} \rightarrow V^{2+} + 2e^{-}$ $(E^0 = -1.19 \ V < 0.97 \ V)$: Oxidized.
$2$. $Fe_{(s)} \rightarrow Fe^{3+} + 3e^{-}$ $(E^0 = -0.04 \ V < 0.97 \ V)$: Oxidized.
$3$. $Ag_{(s)} \rightarrow Ag^{+} + e^{-}$ $(E^0 = 0.80 \ V < 0.97 \ V)$: Oxidized.
$4$. $Au_{(s)} \rightarrow Au^{3+} + 3e^{-}$ $(E^0 = 1.40 \ V > 0.97 \ V)$: Not oxidized.
Therefore,$3$ metals $(V, Fe, Ag)$ will be oxidized by $NO_3^{-}$.

(B) The reaction $FeO_4^{2-} \rightarrow Fe^{2+}$ involves a change in oxidation state from $+6$ to $+2$,so $n = 4$.
Using the relation $\Delta G^{\theta} = -nFE^{\theta}$,for the overall reaction:
$n_{total} E_{total} = n_1 E_1 + n_2 E_2$
Here,$n_1 = 3$ $(FeO_4^{2-} \rightarrow Fe^{3+})$,$E_1 = 2.2 \ V$,$n_2 = 1$ $(Fe^{3+} \rightarrow Fe^{2+})$,$E_2 = 0.70 \ V$,and $n_{total} = 4$.
$4 \times E_{FeO_4^{2-} / Fe^{2+}}^{\theta} = (3 \times 2.2) + (1 \times 0.70)$
$4 \times E_{FeO_4^{2-} / Fe^{2+}}^{\theta} = 6.6 + 0.70 = 7.3 \ V$
$E_{FeO_4^{2-} / Fe^{2+}}^{\theta} = \frac{7.3}{4} = 1.825 \ V$
$1.825 \ V = 1825 \times 10^{-3} \ V$
Therefore,$x = 1825$.

(B) The oxidising power of a species is directly proportional to its standard reduction potential $(E^{\circ})$.
Higher the value of $E^{\circ}$,greater is the tendency to undergo reduction,and thus stronger is the oxidising power.
Comparing the given values:
$BrO_4^{-} (1.74 \ V) > IO_4^{-} (1.65 \ V) > ClO_4^{-} (1.19 \ V)$.
Therefore,the correct order of oxidising power is $BrO_4^{-} > IO_4^{-} > ClO_4^{-}$.

(A, B, D) The reduction potential of $NO_3^{-}$ is $E^0 = +0.96 \ V$.
$A$ metal is oxidized by $NO_3^{-}$ if its standard oxidation potential is greater than the reduction potential of $NO_3^{-}$,or equivalently,if its standard reduction potential $E^0_{red}$ is less than $E^0_{red}(NO_3^{-}) = +0.96 \ V$.
Comparing the given values:
$V: E^0 = -1.19 \ V < +0.96 \ V$ (Oxidized)
$Fe: E^0 = -0.04 \ V < +0.96 \ V$ (Oxidized)
$Hg: E^0 = +0.86 \ V < +0.96 \ V$ (Oxidized)
$Au: E^0 = +1.40 \ V > +0.96 \ V$ (Not oxidized)
Thus,$V, Fe,$ and $Hg$ are oxidized by $NO_3^{-}$.
The pairs containing these metals are $(A) V$ and $Hg$,$(B) Hg$ and $Fe$,and $(D) Fe$ and $V$.

(B) The overall reaction is $MnO_4^{-} + 7e^{-} \rightarrow Mn$. The total change in Gibbs free energy is the sum of the changes for the individual steps:
$\Delta G^{\circ}_{total} = \Delta G^{\circ}_{1} + \Delta G^{\circ}_{2} + \Delta G^{\circ}_{3}$
Since $\Delta G^{\circ} = -nFE^{\circ}$,we have:
$-7 \times F \times E^{\circ} = -(3 \times F \times 1.68) + (-2 \times F \times 1.21) + (-2 \times F \times (-1.03))$
$7 \times E^{\circ} = (3 \times 1.68) + (2 \times 1.21) + (2 \times (-1.03))$
$7 \times E^{\circ} = 5.04 + 2.42 - 2.06$
$7 \times E^{\circ} = 5.40$
$E^{\circ} = \frac{5.40}{7} \approx 0.7714 \ V$
Thus,the reduction potential is approximately $0.77 \ V$.

(A) The reaction $FeO_4^{2-} \rightarrow Fe^{2+}$ involves a change in oxidation state from $+6$ to $+2$,so the number of electrons involved is $n_4 = 4$.
This process can be broken down into two steps:
$1) FeO_4^{2-} + 3e^- \rightarrow Fe^{3+} \quad (E_1^{\Theta} = 2.0 \ V, n_1 = 3)$
$2) Fe^{3+} + 1e^- \rightarrow Fe^{2+} \quad (E_2^{\Theta} = 0.8 \ V, n_2 = 1)$
The total Gibbs free energy change is given by $\Delta G_4^{\Theta} = \Delta G_1^{\Theta} + \Delta G_2^{\Theta}$.
Using $\Delta G^{\Theta} = -nFE^{\Theta}$,we get:
$-n_4 F E_4^{\Theta} = -n_1 F E_1^{\Theta} - n_2 F E_2^{\Theta}$
$n_4 E_4^{\Theta} = n_1 E_1^{\Theta} + n_2 E_2^{\Theta}$
$4 \times E_4^{\Theta} = (3 \times 2.0) + (1 \times 0.8)$
$4 E_4^{\Theta} = 6.0 + 0.8 = 6.8$
$E_4^{\Theta} = \frac{6.8}{4} = 1.7 \ V$.

(B) The cell reaction is $Fe^{2+} + Ag^{+} \rightarrow Fe^{3+} + Ag$.
This can be split into two half-reactions:
Oxidation: $Fe^{2+} \rightarrow Fe^{3+} + e^-$
Reduction: $Ag^+ + e^- \rightarrow Ag$
We are given:
$(1)$ $Ag^+ + e^- \rightarrow Ag, \ E^0 = x \ V, \ \Delta G_1^0 = -F(x)$
$(2)$ $Fe^{2+} + 2e^- \rightarrow Fe, \ E^0 = y \ V, \ \Delta G_2^0 = -2F(y)$
$(3)$ $Fe^{3+} + 3e^- \rightarrow Fe, \ E^0 = z \ V, \ \Delta G_3^0 = -3F(z)$
We need the potential for $Fe^{2+} \rightarrow Fe^{3+} + e^-$. Let this be $E^0_{ox}$.
Using Hess's Law for $\Delta G^0$:
$
\Delta G^0(Fe^{2+}$ $\rightarrow Fe^{3+} + e^-) = \Delta G^0(Fe^{2+}$ $\rightarrow Fe) - \Delta G^0(Fe^{3+}$ $\rightarrow Fe)
$
$-1F(E^0_{ox}) = -2F(y) - (-3F(z))$
$-F(E^0_{ox}) = -2Fy + 3Fz$
$E^0_{ox} = 2y - 3z$
Now,$E^0_{cell} = E^0_{red} + E^0_{ox} = x + (2y - 3z) = x + 2y - 3z$.

(B) 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 value,$Cr$ is the strongest reducing agent.

(D) The formula for standard Gibbs energy is $\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}$.
Here,$n = 2$ (number of electrons transferred),
$F = 96500 \ C \ mol^{-1}$ (Faraday constant),
$E^{\circ}_{\text{cell}} = 1.1 \ V$.
Substituting the values:
$\Delta G^{\circ} = -2 \times 96500 \times 1.1 \ J \ mol^{-1}$.
$\Delta G^{\circ} = -212300 \ J \ mol^{-1}$.
Converting to $kJ \ mol^{-1}$:
$\Delta G^{\circ} = -212.3 \ kJ \ mol^{-1}$.

(A) The standard reduction potential $(\text{SRP})$ for $Cu^{2+} + 2e^{-} \rightarrow Cu$ is $+0.34 \ V$.
The $\text{SRP}$ for $2H^{+} + 2e^{-} \rightarrow H_2$ is $0.00 \ V$.
Since the $\text{SRP}$ of $Cu^{2+}$ $(+0.34 \ V)$ is greater than the $\text{SRP}$ of $H^{+}$ $(0.00 \ V)$,$Cu^{2+}$ ions are reduced more easily than $H^{+}$ ions.
Therefore,both the Assertion and the Reason are true,and the Reason correctly explains the Assertion.

(B) The half-cell reaction is: $\frac{1}{2} Hg_2Cl_{2(s)} + e^- \rightarrow Hg(\ell) + Cl^-_{(aq)}$
Using the Nernst equation: $E = E^o - \frac{0.0591}{n} \log Q$
Here,$n = 1$ and $Q = [Cl^-]$
So,$E = E^o - 0.0591 \log [Cl^-]$
To increase the electrode potential $E$,the term $-0.0591 \log [Cl^-]$ must increase,which means $\log [Cl^-]$ must decrease.
Therefore,decreasing the concentration of $[Cl^-]$ will increase the electrode potential.

(A) The reducing power of a metal is directly proportional to its Standard Oxidation Potential $(SOP)$.
Given $SOP$ values:
$P = +2.87 \ V$
$Q = +3.05 \ V$
$R = -0.80 \ V$
$S = +0.25 \ V$
Comparing these values,the order of $SOP$ is $Q > P > S > R$.
Therefore,the order of reducing power is $Q > P > S > R$.

(A) To find the standard electrode potential for the reaction $Fe^{3+} + e^{-} \rightarrow Fe^{2+}$, we use the relationship between Gibbs free energy change $(\Delta G^{\circ})$ and standard electrode potential $(E^{\circ})$, given by $\Delta G^{\circ} = -nFE^{\circ}$.
Let the reactions be:
$(1) Fe^{3+} + 3e^{-} \rightarrow Fe ; \Delta G^{\circ}_{1} = -3 \times F \times (-0.036) = 0.108F$
$(2) Fe^{2+} + 2e^{-} \rightarrow Fe ; \Delta G^{\circ}_{2} = -2 \times F \times (-0.44) = 0.88F$
The target reaction is $(1) - (2)$:
$Fe^{3+} + e^{-} \rightarrow Fe^{2+} ; \Delta G^{\circ}_{3} = \Delta G^{\circ}_{1} - \Delta G^{\circ}_{2} = 0.108F - 0.88F = -0.772F$
Since $\Delta G^{\circ}_{3} = -nFE^{\circ}_{3}$ and $n=1$:
$-0.772F = -1 \times F \times E^{\circ}_{3}$
$E^{\circ}_{3} = 0.772 \ V$.

(B) The standard Gibbs free energy change is given by the formula: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.
Here,the number of electrons transferred $(n)$ is $2$.
The Faraday constant $(F)$ is approximately $96487 \ C \ mol^{-1}$.
The standard cell potential $(E^{\circ}_{cell})$ is $1.1 \ V$.
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 kilojoules: $\Delta G^{\circ} \approx -212.27 \ kJ \ mol^{-1}$.

(D) In a Galvanic cell,the electrode with the higher reduction potential acts as the cathode (positive electrode),and the electrode with the lower reduction potential acts as the anode (negative electrode).
Given $E^{\circ}(Zn^{+2}/Zn) = -0.76 \ V$ and $E^{\circ}(H^{+}/H_2) = 0.00 \ V$.
Since $0.00 \ V > -0.76 \ V$,the standard hydrogen electrode acts as the cathode (positive electrode).
Reduction occurs at the cathode: $2H^{+}_{(aq)} + 2e^{-} \longrightarrow H_{2(g)}$.

(D) The standard electrode potential $(E^\circ)$ is an intensive property,meaning it does not depend on the amount of substance present or the stoichiometric coefficients of the reaction.
For the reduction reaction: $Zn^{2+}_{(aq)} + 2e^- \rightarrow Zn_{(s)}$,$E^\circ = -0.76 \ V$.
The given reaction is the reverse of the reduction reaction multiplied by a factor of $2$: $2Zn_{(s)} \rightarrow 2Zn^{2+}_{(aq)} + 4e^-$.
Reversing the reaction changes the sign of the potential: $Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^-$,$E^\circ = -(-0.76 \ V) = +0.76 \ V$.
Since the potential is an intensive property,multiplying the reaction by a coefficient does not change the value of $E^\circ$.
Therefore,the standard potential for the reaction $2Zn_{(s)} \rightarrow 2Zn^{2+}_{(aq)} + 4e^-$ remains $+0.76 \ V$.

(C) The relationship between standard Gibbs free energy change and standard cell potential is given by the formula: $\Delta G^{\circ} = -nFE_{\text{cell}}^{\circ}$.
Given: $\Delta G^{\circ} = -386 \ kJ = -386000 \ J$,$n = 2$,and $F \approx 96500 \ C \ mol^{-1}$.
Substituting the values: $-386000 \ J = -(2) \times (96500 \ C \ mol^{-1}) \times E_{\text{cell}}^{\circ}$.
$E_{\text{cell}}^{\circ} = \frac{386000}{2 \times 96500} = \frac{386000}{193000} = 2 \ V$.
Therefore,the correct option is $C$.

(C) The relationship between standard Gibbs free energy change and standard cell potential is given by the formula: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.
Here,the reaction is $2 Al + 3 Cu^{2+} \rightarrow 2 Al^{3+} + 3 Cu$.
The number of electrons transferred $(n)$ is $6$ (since $Al \rightarrow Al^{3+} + 3e^-$ and $Cu^{2+} + 2e^- \rightarrow Cu$,multiplied by $2$ and $3$ respectively).
Given $\Delta G^{\circ} = -1158 \ kJ = -1158000 \ J$.
$F = 96500 \ C \ mol^{-1}$.
Substituting the values: $-1158000 = -(6) \times (96500) \times E^{\circ}_{cell}$.
$E^{\circ}_{cell} = \frac{1158000}{6 \times 96500} = \frac{1158000}{579000} = 2 \ V$.

(D) The standard cell potential is given by $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
From the Nernst equation at equilibrium,$\Delta G^{\circ} = -nFE^{\circ}_{cell}$,which implies $E^{\circ}_{cell} = \frac{-\Delta G^{\circ}}{nF}$.
Also,$E^{\circ}_{cell} = \frac{RT}{nF} \ln K_{c} = \frac{0.0592}{n} \log_{10} K_{c}$ at $298 \ K$.
Option $D$ is false because the standard cell potential is the difference between the reduction potentials of the cathode and the anode,not the sum.

(C) The cell reaction is $2 \ Al_{(s)} + 3 \ Ni_{(aq)}^{+2} \rightarrow 2 \ Al_{(aq)}^{+3} + 3 \ Ni_{(s)}$.
Here,$Al$ is oxidized to $Al^{+3}$ (anode) and $Ni^{+2}$ is reduced to $Ni$ (cathode).
The standard cell potential is given by $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
Given $E^{\circ}_{Ni^{+2}/Ni} = -0.25 \ V$ and $E^{\circ}_{Al^{+3}/Al} = -1.66 \ V$.
$E^{\circ}_{cell} = (-0.25 \ V) - (-1.66 \ V) = -0.25 \ V + 1.66 \ V = +1.41 \ V$.

(C) The standard reduction potential $(E^{\circ})$ values for the given half-reactions are as follows:
$1$. For $2 H_{(aq)}^{+} + 2 e^{-} \rightarrow H_{2_{(g)}}$,$E^{\circ} = 0.00 \ V$.
$2$. For $F_{2_{(g)}} + 2 e^{-} \rightarrow 2 F_{(aq)}^{-}$,$E^{\circ} = +2.87 \ V$.
$3$. For $Li_{(aq)}^{+} + e^{-} \rightarrow Li_{(s)}$,$E^{\circ} = -3.05 \ V$.
$4$. For $Cl_{2_{(g)}} + 2 e^{-} \rightarrow 2 Cl_{(aq)}^{-}$,$E^{\circ} = +1.36 \ V$.
Comparing these values,the reaction involving lithium has the most negative value,which is $-3.05 \ V$. Therefore,it exhibits the minimum standard reduction potential.

(A) The standard $emf$ of a cell is calculated using the formula: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
For a spontaneous cell reaction,the cathode is the electrode with the higher reduction potential.
Here,$E^{\circ}(Sn^{+2} \mid Sn) = -0.14 \ V$ and $E^{\circ}(Fe^{+2} \mid Fe) = -0.44 \ V$.
Since $-0.14 \ V > -0.44 \ V$,the $Sn$ electrode acts as the cathode and the $Fe$ electrode acts as the anode.
Therefore,$E^{\circ}_{cell} = (-0.14 \ V) - (-0.44 \ V) = -0.14 \ V + 0.44 \ V = +0.30 \ V$.

(D) The given reaction is the reduction of $Cu^{2+}$ to $Cu_{(s)}$: $Cu^{2+}_{(aq)} + 2e^{-} \rightarrow Cu_{(s)}$ with $E^{\circ} = +0.34 \ V$.
The requested reaction is the oxidation of $Cu_{(s)}$ to $Cu^{2+}_{(aq)}$: $2Cu_{(s)} \rightarrow 2Cu^{2+}_{(aq)} + 4e^{-}$.
This is the reverse of the reduction reaction multiplied by a factor of $2$.
Since the standard electrode potential $(E^{\circ})$ is an intensive property,it does not depend on the stoichiometric coefficients of the reaction.
Therefore,reversing the reaction changes the sign of the potential,but multiplying the coefficients does not change the value of $E^{\circ}$.
Thus,the potential for the oxidation reaction $Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^{-}$ is $-0.34 \ V$.
The potential for $2Cu_{(s)} \rightarrow 2Cu^{2+}_{(aq)} + 4e^{-}$ remains $-0.34 \ V$.

(C) The standard Gibbs energy change $\Delta G^{\circ}$ is related to the standard cell potential $E^{\circ}_{\text{cell}}$ by the equation: $\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}$.
In the given reaction,$Zn$ is oxidized to $Zn^{2+}$ $(n=2)$ and $Ni^{2+}$ is reduced to $Ni$ $(n=2)$,so the number of electrons transferred $n = 2$.
The Faraday constant $F \approx 96500 \ C \ mol^{-1}$.
Given $E^{\circ}_{\text{cell}} = 0.5 \ V$.
Substituting the values: $\Delta G^{\circ} = -2 \times 96500 \ C \ mol^{-1} \times 0.5 \ V = -96500 \ J \ mol^{-1}$.
Converting to $kJ \ mol^{-1}$: $\Delta G^{\circ} = -96.5 \ kJ \ mol^{-1}$.

(C) reaction is spontaneous if the standard cell potential $(E^{\circ}_{cell})$ is positive.
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
For option $(C)$: $Zn_{(s)} + Cu^{2+}_{(aq)} \longrightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$.
Here,$Zn$ is oxidized (anode) and $Cu^{2+}$ is reduced (cathode).
$E^{\circ}_{cell} = E^{\circ}_{Cu^{2+}/Cu} - E^{\circ}_{Zn^{2+}/Zn} = 0.34 \ V - (-0.76 \ V) = +1.10 \ V$.
Since $E^{\circ}_{cell} > 0$,the reaction is spontaneous.

(A) The cell reaction is: $Al(s) + 3H^+(aq) \rightarrow Al^{3+}(aq) + \frac{3}{2}H_2(g)$.
For this cell,the anode is $Al$ and the cathode is $Pt, H_2$.
The standard electrode potential of the standard hydrogen electrode $(SHE)$ is $E^{\circ}_{H^+/H_2} = 0.00 \ V$.
The standard cell potential is given by the formula: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
Substituting the values: $E^{\circ}_{cell} = E^{\circ}_{H^+/H_2} - E^{\circ}_{Al^{3+}/Al} = 0.00 \ V - (-1.66 \ V) = 1.66 \ V$.

(A) The given cell is composed of a copper electrode as the anode and a silver electrode as the cathode.
The standard cell potential is given by the formula:
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$
Substituting the given values:
$E^{\circ}_{cell} = E^{\circ}_{Ag^{+}|Ag} - E^{\circ}_{Cu^{2+}|Cu}$
$0.647 \ V = E^{\circ}_{Ag^{+}|Ag} - 0.153 \ V$
$E^{\circ}_{Ag^{+}|Ag} = 0.647 \ V + 0.153 \ V = 0.8 \ V$

(A) The higher the standard reduction potential $(E^\circ)$ value,the greater the tendency of the species to accept electrons and undergo reduction.
Order of $E^\circ_{red}: Ag^{+} (0.80 \ V) > Cu^{2+} (0.337 \ V) > Sn^{2+} (-0.136 \ V) > Cd^{2+} (-0.403 \ V)$.
Therefore,the decreasing order of deposition of metal on the electrode is $Ag > Cu > Sn > Cd$.

(A) The standard cell potential is given by the formula:
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$
Here,the calomel electrode acts as the cathode and the zinc electrode acts as the anode.
$E^{\circ}_{cell} = E^{\circ}_{calomel} - E^{\circ}_{Zn^{2+}/Zn}$
Substituting the given values:
$1.007 \ V = 0.242 \ V - E^{\circ}_{Zn^{2+}/Zn}$
$E^{\circ}_{Zn^{2+}/Zn} = 0.242 \ V - 1.007 \ V$
$E^{\circ}_{Zn^{2+}/Zn} = -0.765 \ V$

(A) The standard Gibbs free energy change is given by the formula: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.
Here,the cell reaction is: $Sn_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Sn^{2+}_{(aq)} + 2Ag_{(s)}$.
The number of electrons transferred,$n = 2$.
The Faraday constant,$F = 96500 \ C \ mol^{-1}$.
Given $E^{\circ}_{cell} = 0.90 \ V$.
Substituting the values: $\Delta G^{\circ} = -2 \times 96500 \times 0.90 \ J$.
$\Delta G^{\circ} = -173700 \ J = -173.7 \ kJ$.

(A) The standard cell potential is calculated using the formula:
$E^0_{cell} = E^0_{cathode} - E^0_{anode}$
In the given reaction,$Ni$ is oxidized to $Ni^{2+}$ (anode) and $Cu^{2+}$ is reduced to $Cu$ (cathode).
Therefore,$E^0_{cell} = E^0_{Cu^{2+}/Cu} - E^0_{Ni^{2+}/Ni}$
$E^0_{cell} = 0.337 \ V - (-0.257 \ V)$
$E^0_{cell} = 0.337 \ V + 0.257 \ V = 0.594 \ V$

(B) In the given cell reaction,$Mg$ undergoes oxidation at the anode and $Ag^{+}$ undergoes reduction at the cathode.
$E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ}$
$E_{\text{cell}}^{\circ} = E_{Ag^{+}/Ag}^{\circ} - E_{Mg^{2+}/Mg}^{\circ}$
$E_{\text{cell}}^{\circ} = 0.8 \ V - (-2.37 \ V) = 3.17 \ V$

(A) The cell reaction is represented as $Zn_{(s)}|Zn^{2+}_{(1M)}||Cd^{2+}_{(1M)}|Cd_{(s)}$.
In this cell,$Zn$ acts as the anode (oxidation) and $Cd$ acts as the cathode (reduction).
The standard cell potential is calculated using the formula:
$E_{cell}^o = E_{cathode}^{\circ} - E_{anode}^{\circ}$
Substituting the given values:
$E_{cell}^o = E_{Cd^{2+}/Cd}^{\circ} - E_{Zn^{2+}/Zn}^{\circ}$
$E_{cell}^o = -0.403 \ V - (-0.763 \ V)$
$E_{cell}^o = -0.403 \ V + 0.763 \ V = 0.36 \ V$.