Electrode potential and ECell Questions in English

(C) The standard reduction potential $(E^o)$ indicates the tendency of a species to get reduced. $A$ higher $E^o$ value means a stronger oxidizing agent and a weaker reducing agent.
Comparing the given potentials:
$E^o(PbO_2/Pb^{2+}) = +1.45 \ V$
$E^o(Hg^{2+}/Hg_2^{2+}) = +0.92 \ V$
$E^o(Sn^{4+}/Sn^{2+}) = +0.15 \ V$
$1$. Reducing strength is the reverse of reduction potential. Since $E^o(Sn^{4+}/Sn^{2+}) < E^o(Hg^{2+}/Hg_2^{2+})$,$Sn^{2+}$ is a stronger reducing agent than $Hg_2^{2+}$.
$2$. Oxidizing strength follows the order of $E^o$ values: $PbO_2 > Hg^{2+} > Sn^{4+}$.
Thus,$Sn^{2+}$ is a stronger reducing agent than $Hg_2^{2+}$.

(A) The reduction potential order is $E^\circ_{Z/Z^-} > E^\circ_{Y/Y^-} > E^\circ_{X/X^-}$.
Species with higher reduction potential act as stronger oxidizing agents.
Therefore,$Z$ is the strongest oxidizing agent and $X$ is the weakest.
$Y^-$ can oxidize species with a lower reduction potential than itself.
Since $E^\circ_{Y/Y^-} > E^\circ_{X/X^-}$,$Y^-$ can oxidize $X$ to $X^+$.
Since $E^\circ_{Y/Y^-} < E^\circ_{Z/Z^-}$,$Y^-$ cannot oxidize $Z^-$.
Thus,$Y^-$ oxidizes $X$ but does not oxidize $Z^-$.

(D) The strength of an oxidizing agent is directly proportional to its standard reduction potential $(E^o)$.
Comparing the given values:
$E_{Li^+|Li}^o = -3.03 \ V$
$E_{Ba^{2+}|Ba}^o = -2.73 \ V$
$E_{Na^+|Na}^o = -2.71 \ V$
$E_{Mg^{2+}|Mg}^o = -2.37 \ V$
Since $-2.37 \ V > -2.71 \ V > -2.73 \ V > -3.03 \ V$,the species with the highest reduction potential is $Mg^{2+}$.
Therefore,$Mg^{2+}$ is the strongest oxidizing agent among the given options.

(A) The $EMF$ of the cell is calculated using the formula:
$E^o_{cell} = E^o_{cathode} - E^o_{anode}$
Here,$Cu^{2+} | Cu$ acts as the cathode and $Zn^{2+} | Zn$ acts as the anode.
$E^o_{cell} = 0.34 \, V - (-0.76 \, V)$
$E^o_{cell} = 0.34 + 0.76 = 1.10 \, V$

(D) The strength of an oxidizing agent is directly proportional to its standard reduction potential $(E^o_{red})$.
Comparing the given values:
$E^o_{Cr^{3+}/Cr} = -0.74 \ V$
$E^o_{MnO_4^-/Mn^{2+}} = 1.51 \ V$
$E^o_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \ V$
$E^o_{Cl_2/Cl^-} = 1.36 \ V$
Since $MnO_4^-$ has the highest positive standard reduction potential $(1.51 \ V)$,it is the strongest oxidizing agent among the given species.

(D) The given half-cell reactions are:
$(1) \, Fe^{2+} + 2e^{-} \rightarrow Fe, \, E^{0}_{1} = -0.441 \, V$
$(2) \, Fe^{3+} + e^{-} \rightarrow Fe^{2+}, \, E^{0}_{2} = 0.771 \, V$
To obtain the reaction $Fe + 2Fe^{3+} \rightarrow 3Fe^{2+}$,we use the formula $\Delta G^{0} = -nFE^{0}$.
For the overall reaction,the number of electrons transferred $n = 2$.
$\Delta G^{0}_{total} = \Delta G^{0}_{reaction} = \Delta G^{0}_{2} \times 2 - \Delta G^{0}_{1}$
$-nFE^{0}_{cell} = -(n_{2}FE^{0}_{2} \times 2) - (-n_{1}FE^{0}_{1})$
$-2FE^{0}_{cell} = -(1 \times F \times 0.771 \times 2) - (-2 \times F \times -0.441)$
$-2E^{0}_{cell} = -1.542 - 0.882$
$-2E^{0}_{cell} = -2.424$
$E^{0}_{cell} = 1.212 \, V$.

(C) metal can displace another metal from its salt solution if it has a more negative standard reduction potential (i.e.,it is a stronger reducing agent).
Given values are $E^o_A = -0.250 \ V$,$E^o_B = -0.140 \ V$,$E^o_C = -0.126 \ V$,and $E^o_D = -0.402 \ V$.
Comparing these,only $D$ has a more negative reduction potential than $A$ $(E^o_D < E^o_A)$.
Therefore,$D$ can displace $A$ from its aqueous salt solution.

(D) The standard reduction potential is defined as the potential associated with the reduction half-reaction where a species in its standard state ($1$ $M$ concentration for aqueous ions) gains electrons to form the element in its standard state.
For zinc,the reduction half-reaction is $Zn^{2+}_{(aq)} + 2e^{-} \rightarrow Zn_{(s)}$.
Therefore,option $D$ is the correct representation.

(B) For a galvanic cell,the standard $emf$ $(E^o_{cell})$ is calculated as: $E^o_{cell} = E^o_{cathode} - E^o_{anode}$.
Since $Zn$ has a more negative reduction potential than $Ni$,$Zn$ acts as the anode and $Ni$ acts as the cathode.
$E^o_{cathode} = E^o_{Ni^{2+}/Ni} = -0.25 \ V$.
$E^o_{anode} = E^o_{Zn^{2+}/Zn} = -0.76 \ V$.
$E^o_{cell} = (-0.25 \ V) - (-0.76 \ V) = -0.25 \ V + 0.76 \ V = +0.51 \ V$.

(A) The strength of a reducing agent is inversely proportional to its standard reduction potential $(E^o)$.
The lower (more negative) the standard reduction potential,the easier it is for the species to undergo oxidation,making it a stronger reducing agent.
Comparing the given $E^o$ values:
$Zn^{2+}/Zn = -0.76\,V$
$Cr^{3+}/Cr = -0.74\,V$
$H^+/H_2 = 0.00\,V$
$Fe^{3+}/Fe^{2+} = +0.77\,V$
Since $-0.76\,V$ is the most negative value,$Zn_{(s)}$ has the highest tendency to lose electrons and act as the strongest reducing agent among the given options.

(A) The given cell reaction is: $Sn_{(s)} + 2Fe^{3+}_{(aq)} \rightarrow 2Fe^{2+}_{(aq)} + Sn^{2+}_{(aq)}$
In this reaction,$Sn$ is oxidized to $Sn^{2+}$ (Anode) and $Fe^{3+}$ is reduced to $Fe^{2+}$ (Cathode).
The standard cell potential is calculated as: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$
Here,$E^{\circ}_{cathode} = E^{\circ}_{Fe^{3+}/Fe^{2+}} = +0.77 \ V$ and $E^{\circ}_{anode} = E^{\circ}_{Sn^{2+}/Sn} = -0.14 \ V$.
Substituting the values: $E^{\circ}_{cell} = 0.77 \ V - (-0.14 \ V) = 0.77 \ V + 0.14 \ V = 0.91 \ V$.

(C) metal is oxidized by $NO_3^-$ if its standard reduction potential is lower than that of the $NO_3^-$ reduction reaction $(E^o = +0.96 \, V)$.
Comparing the given reduction potentials:
$V: -1.19 \, V < +0.96 \, V$ (Oxidized)
$Fe: -0.04 \, V < +0.96 \, V$ (Oxidized)
$Hg: +0.86 \, V < +0.96 \, V$ (Oxidized)
$Au: +1.40 \, V > +0.96 \, V$ (Not oxidized)
Since $Au$ has a higher reduction potential than $NO_3^-$,it will not be oxidized. However,the question asks for a pair that is not oxidized. Based on the options provided and the standard chemical behavior,$Au$ is the only metal that resists oxidation by $NO_3^-$. Among the choices,$Fe$ and $Au$ is the pair where $Au$ is not oxidized,but since $Fe$ is,the question implies identifying the pair where the metal is not oxidized. Given the standard interpretation,$Au$ is the correct choice.

(C) The standard cell potential is calculated using the formula: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
Here,the copper electrode acts as the cathode and the magnesium electrode acts as the anode.
$E^{\circ}_{cell} = 0.33 \ V - (-2.37 \ V) = 0.33 + 2.37 = 2.70 \ V$.

(A) For $Cu^{2+} + 2e^{-} \rightarrow Cu$,$\Delta G_1^{0} = -nFE^{0}_1 = -2 \times 0.337 \times F = -0.674 \ F$ $(1)$
For $Cu^{2+} + e^{-} \rightarrow Cu^{+}$,$\Delta G_2^{0} = -nFE^{0}_2 = -1 \times 0.153 \times F = -0.153 \ F$ $(2)$
For the reaction $Cu^{+} + e^{-} \rightarrow Cu$,we subtract equation $(2)$ from equation $(1)$:
$\Delta G_3^{0} = \Delta G_1^{0} - \Delta G_2^{0} = -0.674 \ F - (-0.153 \ F) = -0.521 \ F$
Since $\Delta G_3^{0} = -nFE^{0}_3$ and $n=1$ for this reaction:
$-1 \times F \times E^{0}_3 = -0.521 \ F$
$E^{0}_3 = 0.521 \ V \approx 0.52 \ V$.

(D) We are given:
$(1) \ Fe^{3+} + 3e^{-} \rightarrow Fe \ ; \ E^o_1 = -0.036 \ V \ ; \ n_1 = 3$
$(2) \ Fe^{2+} + 2e^{-} \rightarrow Fe \ ; \ E^o_2 = -0.440 \ V \ ; \ n_2 = 2$
We want to find $E^o_3$ for $Fe^{3+} + e^{-} \rightarrow Fe^{2+}$.
This can be obtained by $(1) - (2)$:
$Fe^{3+} + 3e^{-} - (Fe^{2+} + 2e^{-}) \rightarrow Fe - Fe$
$Fe^{3+} + e^{-} \rightarrow Fe^{2+}$
The Gibbs free energy change is $\Delta G^o = -nFE^o$.
$\Delta G^o_3 = \Delta G^o_1 - \Delta G^o_2$
$-n_3 F E^o_3 = -n_1 F E^o_1 - (-n_2 F E^o_2)$
$-1 \times E^o_3 = -(3 \times -0.036) - (2 \times -0.440)$
$-E^o_3 = 0.108 + 0.880$
$E^o_3 = 0.108 - 0.880 = -0.772 \ V$
Wait,checking the calculation: $\Delta G^o_3 = \Delta G^o_1 - \Delta G^o_2 = (-3 \times F \times -0.036) - (-2 \times F \times -0.440) = 0.108F - 0.880F = -0.772F$.
Since $\Delta G^o_3 = -n_3 F E^o_3$ and $n_3 = 1$,we have $-1 \times F \times E^o_3 = -0.772F$,so $E^o_3 = +0.772 \ V$.

(B) The standard electrode potential for the reaction $Cu^{ } e^{-} \rightarrow Cu$ can be calculated using the Gibbs free energy change $(\Delta G^{\circ} = -nFE^{\circ})$.
For $Cu^{2 } 2e^{-} \rightarrow Cu$,$\Delta G_{1}^{\circ} = -2 \times F \times 0.34 = -0.68F$.
For $Cu^{2 } e^{-} \rightarrow Cu^{ }$,$\Delta G_{2}^{\circ} = -1 \times F \times 0.15 = -0.15F$.
For $Cu^{ } e^{-} \rightarrow Cu$,the reaction is $(Cu^{2 } 2e^{-}$ $\rightarrow Cu) - (Cu^{2 } e^{-}$ $\rightarrow Cu^{ })$.
Therefore,$\Delta G_{3}^{\circ} = \Delta G_{1}^{\circ} - \Delta G_{2}^{\circ} = -0.68F - (-0.15F) = -0.53F$.
Since $\Delta G_{3}^{\circ} = -nFE_{Cu^{ }/Cu}^{\circ}$ and $n=1$,we have $-0.53F = -1 \times F \times E_{Cu^{ }/Cu}^{\circ}$.
Thus,$E_{Cu^{ }/Cu}^{\circ} = 0.53 \ V$.

(A) The reactivity of a metal is directly related to its ability to lose electrons,which is measured by its standard oxidation potential.
Since $E^{0}_{Oxidation} = -E^{0}_{Reduction}$,a more negative $E^{0}_{Reduction}$ value indicates a higher tendency to lose electrons (higher oxidation potential).
Comparing the given values: $A = -3.05 \ V$,$B = -1.66 \ V$,$C = -0.40 \ V$,and $D = 0.80 \ V$.
Element $A$ has the most negative reduction potential,meaning it is the most easily oxidized and therefore the most reactive.

(C) The cell reaction is $2Ag^{+}_{(aq)} + Cd_{(s)} \rightarrow 2Ag_{(s)} + Cd^{2+}_{(aq)}$.
The standard cell potential is $E^o_{cell} = E^o_{cathode} - E^o_{anode} = 0.80 \ V - (-0.40 \ V) = 1.20 \ V$.
The number of electrons transferred in the balanced equation is $n = 2$.
The standard free energy change is given by $\Delta G^o = -nFE^o_{cell}$.
Substituting the values: $\Delta G^o = -2 \times 96500 \ C/mol \times 1.20 \ V = -231600 \ J/mol = -231.6 \ kJ/mol$.

(B) The standard hydrogen electrode $(SHE)$ is assigned a potential of $0.00 \ V$ by convention. This value is used as a reference to determine the electrode potentials of other half-cells. Therefore,the electrode potential is assumed to be zero.

(A) reaction is spontaneous if the cell potential $E^0_{cell}$ is positive.
For the reaction $X + Y^{2+} \rightarrow X^{2+} + Y$,the cell potential is given by $E^0_{cell} = E^0_{cathode} - E^0_{anode} = E^0_{Y^{2+}/Y} - E^0_{X^{2+}/X}$.
For the reaction to be spontaneous,$E^0_{Y^{2+}/Y} > E^0_{X^{2+}/X}$.
Given values: $E^0_{Zn^{2+}/Zn} = -0.76 \, V$,$E^0_{Fe^{2+}/Fe} = -0.44 \, V$,$E^0_{Ni^{2+}/Ni} = -0.23 \, V$.
Checking option $A$: $X = Zn, Y = Ni$. $E^0_{cell} = E^0_{Ni^{2+}/Ni} - E^0_{Zn^{2+}/Zn} = -0.23 - (-0.76) = +0.53 \, V$. Since $E^0_{cell} > 0$,the reaction is spontaneous.

(C) metal with a lower (more negative) standard reduction potential can displace a metal with a higher (more positive) standard reduction potential from its salt solution.
Comparing the $E^o$ values:
$E^o(Fe^{2+}/Fe) = -0.44 \, V$
$E^o(Cu^{2+}/Cu) = +0.34 \, V$
$E^o(Ag^{+}/Ag) = +0.80 \, V$
$I$. Copper $(+0.34 \, V)$ cannot displace Iron $(-0.44 \, V)$. (Incorrect)
$II$. Iron $(-0.44 \, V)$ can displace Copper $(+0.34 \, V)$. (Correct)
$III$. Silver $(+0.80 \, V)$ cannot displace Copper $(+0.34 \, V)$. (Incorrect)
$IV$. Iron $(-0.44 \, V)$ can displace Silver $(+0.80 \, V)$. (Correct)
Therefore,statements $II$ and $IV$ are correct.

(D) The reducing strength of a metal is directly proportional to its standard oxidation potential $(E^0_{oxi})$.
The relationship between standard reduction potential $(E^0_{red})$ and standard oxidation potential $(E^0_{oxi})$ is: $E^0_{oxi} = -E^0_{red}$.
Given values:
For $M_1$: $E^0_{red} = -0.34 \ V \implies E^0_{oxi} = +0.34 \ V$
For $M_2$: $E^0_{red} = -3.05 \ V \implies E^0_{oxi} = +3.05 \ V$
For $M_3$: $E^0_{red} = -1.66 \ V \implies E^0_{oxi} = +1.66 \ V$
Comparing the oxidation potentials: $3.05 \ V (M_2) > 1.66 \ V (M_3) > 0.34 \ V (M_1)$.
Therefore,the order of reducing strength is $M_2 > M_3 > M_1$.

(A) The standard electrode potential $(E^{\circ})$ is an intensive property,meaning it does not depend on the amount of substance or the stoichiometric coefficients in the balanced chemical equation.
Given reaction: $F_2 + 2e^{-} \to 2F^{-}$,$E^{\circ} = 2.8 \, V$.
For the reaction: $\frac{1}{2} F_2 + e^{-} \to F^{-}$,the value of $E^{\circ}$ remains the same because the potential is independent of the stoichiometric coefficients.
Therefore,$E^{\circ} = 2.8 \, V$.

(D) The standard reduction potentials are given as: $E^{0}_{Mg^{+2}/Mg} = -2.37 \ V$,$E^{0}_{Zn^{+2}/Zn} = -0.76 \ V$,and $E^{0}_{Fe^{+2}/Fe} = -0.44 \ V$.
$A$ metal with a more negative reduction potential acts as a stronger reducing agent.
Comparing the values,$Mg$ is the strongest reducing agent,followed by $Zn$,and then $Fe$.
$A$ stronger reducing agent can reduce the ion of a metal with a higher (more positive) reduction potential.
Since $E^{0}_{Zn^{+2}/Zn} (-0.76 \ V) < E^{0}_{Fe^{+2}/Fe} (-0.44 \ V)$,$Zn$ can reduce $Fe^{+2}$ to $Fe$.

(B) The reducing strength of a substance is inversely proportional to its standard reduction potential.
Since $Mg^{2+}/Mg$ has the most negative reduction potential $(-2.34 \, V)$,it acts as the strongest reducing agent among the given options.

(C) The cell potential $(E_{cell})$ of an electrochemical cell is defined as the difference between the reduction potential of the cathode and the reduction potential of the anode.
Mathematically,it is expressed as: $E_{cell} = E_{cathode} - E_{anode}$.

(D) The relationship between Gibbs free energy change $(\Delta G)$ and cell potential $(E_{cell})$ is given by the equation: $\Delta G = -nFE_{cell}$.
Here,$\Delta G = +966 \ kJ \ mol^{-1} = 966000 \ J \ mol^{-1}$.
For the reaction $\frac{2}{3} Al_2O_3 \to \frac{4}{3} Al + O_2$,the oxidation state of $Al$ changes from $+3$ to $0$. The total number of electrons transferred $(n)$ is calculated as: $n = \frac{4}{3} \times 3 = 4$.
Using $F = 96500 \ C \ mol^{-1}$,we have:
$E_{cell} = -\frac{\Delta G}{nF} = -\frac{966000}{4 \times 96500}$.
Since the decomposition is non-spontaneous,the required external potential is $E_{ext} = -E_{cell} = \frac{966000}{386000} \approx 2.5 \ V$.

(B) The standard cell potential $E^{0}_{cell}$ is calculated using the formula: $E^{0}_{cell} = E^{0}_{cathode} - E^{0}_{anode}$.
Here,$Ag^{+} | Ag$ acts as the cathode $(E^{0}_{red} = 0.80\, V)$ and $Cu^{2+} | Cu$ acts as the anode $(E^{0}_{red} = 0.34\, V)$.
$E^{0}_{cell} = 0.80\, V - 0.34\, V = 0.46\, V$.

(A) chemical reaction is spontaneous when the change in Gibbs free energy,$\Delta G$,is negative.
For a cell reaction,the relationship is given by $\Delta G = -nFE^{\circ}_{cell}$.
Therefore,for a spontaneous reaction,$\Delta G < 0$.

(B) For a spontaneous electrochemical process,the cell potential $(E_{cell})$ is positive.
The relationship between Gibbs free energy change $(\Delta G)$ and cell potential is given by the equation: $\Delta G = -nFE_{cell}$.
Since $E_{cell} > 0$,the value of $\Delta G$ must be negative.
Therefore,the Gibbs free energy decreases during a spontaneous electrochemical process.

(D) In the given reaction,$I^-$ is oxidized and $Cr_2O_7^{2-}$ is reduced.
$E_{cell}^o = E_{cathode}^o - E_{anode}^o$
Here,$Cr_2O_7^{2-}$ acts as the cathode and $I^-$ acts as the anode.
$E_{cell}^o = E_{Cr_2O_7^{2-}}^o - E_{I_2}^o$
$0.79 \ V = 1.33 \ V - E_{I_2}^o$
$E_{I_2}^o = 1.33 \ V - 0.79 \ V$
$E_{I_2}^o = 0.54 \ V$

(B) The given cell reaction is $Mg_{(s)} + Zn^{2+}_{(aq)} \rightarrow Zn_{(s)} + Mg^{2+}_{(aq)}$.
Since the concentrations of both $Mg^{2+}$ and $Zn^{2+}$ ions are $1 \ M$,the cell potential $E_{cell}$ is equal to the standard cell potential $E^{\circ}_{cell}$.
Given that the $emf$ of the cell is $1.60 \ V$,the potential $E$ of the cell under these standard conditions is $1.60 \ V$.

(B) The cell reaction is: $Zn_{(s)} + 2H^{+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + H_{2(g)}$.
The standard cell potential is given by $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
Here,the cathode is the Standard Hydrogen Electrode $(SHE)$,so $E^{\circ}_{cathode} = 0.00 \ V$.
The anode is the $Zn^{2+}/Zn$ electrode,so $E^{\circ}_{anode} = -0.76 \ V$.
Therefore,$E^{\circ}_{cell} = 0.00 \ V - (-0.76 \ V) = +0.76 \ V$.

(A) The reducing power of a metal is inversely proportional to its standard reduction potential.
Lower (more negative) standard reduction potential indicates a stronger reducing agent.
Given values: $E^\circ(X) = 0.52 \, V$,$E^\circ(Y) = -3.03 \, V$,$E^\circ(Z) = -1.18 \, V$.
Comparing the values: $-3.03 < -1.18 < 0.52$.
Therefore,the order of reducing power is $Y > Z > X$.

(B) The cell voltage $(E_{cell})$ is calculated as $E_{cell} = E_{ox} + E_{red}$.
For a reaction $A + B^{n+} \rightarrow A^{n+} + B$,$E_{cell} = E_{ox}(A) + E_{red}(B^{n+})$.
Given oxidation potentials $(E_{ox})$: $Zn = 0.76 \ V, Cu = -0.34 \ V, Ag = -0.80 \ V, H_2 = 0 \ V, Ni = 0.25 \ V$.
Reduction potentials $(E_{red} = -E_{ox})$: $Cu^{2+} = 0.34 \ V, Ag^{+} = 0.80 \ V, H^{+} = 0 \ V, Ni^{2+} = -0.25 \ V$.
$(A)$ $Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$: $E_{cell} = 0.76 + 0.34 = 1.10 \ V$.
$(B)$ $Zn + 2Ag^{+} \rightarrow Zn^{2+} + 2Ag$: $E_{cell} = 0.76 + 0.80 = 1.56 \ V$.
$(C)$ $H_2 + Cu^{2+} \rightarrow 2H^{+} + Cu$: $E_{cell} = 0 + 0.34 = 0.34 \ V$.
$(D)$ $H_2 + Ni^{2+} \rightarrow 2H^{+} + Ni$: $E_{cell} = 0 + (-0.25) = -0.25 \ V$.
The maximum voltage is $1.56 \ V$ for option $B$.

(D) The reducing power of a metal is inversely proportional to its standard reduction potential $(E^0_{RP})$.
Lower the value of $E^0_{RP}$,higher is the reducing power.
Given values are:
$E^0_A = 0.68 \ V$
$E^0_B = -2.50 \ V$
$E^0_C = 0.5 \ V$
Comparing the values: $-2.50 < 0.5 < 0.68$.
Therefore,the order of reducing power is $B > C > A$.

(B) The given reaction is $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$.
In this reaction,$Zn$ is oxidized to $Zn^{2+}$ (Anode) and $Fe^{2+}$ is reduced to $Fe$ (Cathode).
The standard cell potential is calculated using the formula: $E^{0}_{cell} = E^{0}_{Oxi}(\text{anode}) + E^{0}_{Red}(\text{cathode})$.
Since $E^{0}_{Oxi}(\text{Fe/Fe}^{2+}) = +0.41 \ V$,the reduction potential $E^{0}_{Red}(\text{Fe}^{2+}\text{/Fe}) = -0.41 \ V$.
Thus,$E^{0}_{cell} = E^{0}_{Oxi}(\text{Zn/Zn}^{2+}) + E^{0}_{Red}(\text{Fe}^{2+}\text{/Fe}) = 0.76 \ V + (-0.41 \ V) = +0.35 \ V$.

(C) Given:
$(1)$ $Cu^{2+} + 2e^- \to Cu$,$E_1^o = 0.337 \, V$,$n_1 = 2$
$(2)$ $Cu^{2+} + e^- \to Cu^+$,$E_2^o = 0.153 \, V$,$n_2 = 1$
We need to find $E_3^o$ for $Cu^+ + e^- \to Cu$,$n_3 = 1$.
Using the Gibbs free energy relation $\Delta G^o = -nFE^o$:
$\Delta G_3^o = \Delta G_1^o - \Delta G_2^o$
$-n_3FE_3^o = -n_1FE_1^o - (-n_2FE_2^o)$
$n_3E_3^o = n_1E_1^o - n_2E_2^o$
Substituting the values:
$1 \times E_3^o = (2 \times 0.337) - (1 \times 0.153)$
$E_3^o = 0.674 - 0.153$
$E_3^o = 0.521 \, V$

(D) The given reaction is $Fe^{2+} + Sn \rightarrow Fe + Sn^{2+}$.
In this reaction,$Fe^{2+}$ is reduced to $Fe$ (cathode) and $Sn$ is oxidized to $Sn^{2+}$ (anode).
$E^{0}_{cell} = E^{0}_{cathode} - E^{0}_{anode}$
$E^{0}_{cell} = E^{0}_{Fe^{2+} | Fe} - E^{0}_{Sn^{2+} | Sn}$
$E^{0}_{cell} = (-0.44 \, V) - (-0.14 \, V)$
$E^{0}_{cell} = -0.44 \, V + 0.14 \, V = -0.30 \, V$.

(C) The reducing power of a metal is inversely proportional to its standard reduction potential $(E_{RP}^\circ)$.
Lower (more negative) $E_{RP}^\circ$ values indicate a stronger tendency to undergo oxidation,meaning higher reducing power.
Given values:
$E_{RP}^\circ (A) = +0.5 \, V$
$E_{RP}^\circ (B) = -3.0 \, V$
$E_{RP}^\circ (C) = -1.2 \, V$
Comparing the values: $-3.0 < -1.2 < +0.5$.
Therefore,the order of reducing power is $B > C > A$.

(B) The cell potential $E^{\circ}_{cell}$ is calculated as $E^{\circ}_{cathode} - E^{\circ}_{anode}$.
To maximize $E^{\circ}_{cell}$,we need the highest possible $E^{\circ}_{cathode}$ (reduction potential) and the lowest possible $E^{\circ}_{anode}$ (oxidation potential).
First,convert all given values to standard reduction potentials $(E^{\circ}_{red})$:
$(i) \, A^{3-}$ $\rightarrow A^{2-} + e^{-}; E^{\circ}_{ox} = 1.5 \, V$ $\Rightarrow E^{\circ}_{red} = -1.5 \, V$
$(ii) \, B^{+} + e^{-} \rightarrow B; E^{\circ}_{red} = 0.5 \, V$
$(iii) \, C^{2+} + e^{-} \rightarrow C^{+}; E^{\circ}_{red} = 0.5 \, V$
$(iv) \, D$ $\rightarrow D^{2+} + 2e^{-}; E^{\circ}_{ox} = -1.15 \, V$ $\Rightarrow E^{\circ}_{red} = 1.15 \, V$
Comparing the reduction potentials,the highest is $E^{\circ}_{red} = 0.5 \, V$ (from $ii$ or $iii$) and the lowest is $E^{\circ}_{red} = -1.5 \, V$ (from $i$).
However,the question asks for the combination of two half-cells. Using $(i)$ as the anode $(E^{\circ}_{ox} = 1.5 \, V)$ and $(ii)$ or $(iii)$ as the cathode $(E^{\circ}_{red} = 0.5 \, V)$,$E^{\circ}_{cell} = 0.5 - (-1.5) = 2.0 \, V$.
Using $(iv)$ as the anode $(E^{\circ}_{ox} = -1.15 \, V)$ and $(ii)$ or $(iii)$ as the cathode $(E^{\circ}_{red} = 0.5 \, V)$,$E^{\circ}_{cell} = 0.5 - (-1.15) = 1.65 \, V$.
Using $(i)$ as the anode $(E^{\circ}_{ox} = 1.5 \, V)$ and $(iv)$ as the cathode $(E^{\circ}_{red} = 1.15 \, V)$,$E^{\circ}_{cell} = 1.15 - (-1.5) = 2.65 \, V$.
Thus,the combination of $(i)$ and $(iv)$ gives the maximum potential.

(B) The cell reaction is $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$.
Here,$Zn$ is oxidized to $Zn^{2+}$ (Anode) and $Fe^{2+}$ is reduced to $Fe$ (Cathode).
Given oxidation potentials are:
$E^{0}_{Zn|Zn^{2+}} = 0.76 \ V$
$E^{0}_{Fe|Fe^{2+}} = 0.41 \ V$
Standard cell potential $E^{0}_{cell} = E^{0}_{oxi}(\text{anode}) + E^{0}_{red}(\text{cathode})$.
Since $E^{0}_{red}(\text{cathode}) = -E^{0}_{oxi}(\text{cathode})$,
$E^{0}_{cell} = E^{0}_{oxi}(Zn|Zn^{2+}) - E^{0}_{oxi}(Fe|Fe^{2+})$
$E^{0}_{cell} = 0.76 \ V - 0.41 \ V = 0.35 \ V$.

(A) The reducing power of an element is inversely proportional to its standard reduction potential $(E^{0}_{Red})$.
Lower the value of $E^{0}_{Red}$,stronger is the reducing agent.
Comparing the given values: $-3.04 \, V < -1.90 \, V < 0.00 \, V < 1.90 \, V$.
Since element $I$ has the lowest reduction potential $(-3.04 \, V)$,it acts as the strongest reducing agent.

(D) Given: $E^{0}_{Mg^{2+}/Mg} = -2.37 \, V$ and $E^{0}_{Cu^{2+}/Cu} = +0.33 \, V$.
Since $E^{0}_{Cu^{2+}/Cu} > E^{0}_{Mg^{2+}/Mg}$,$Cu$ acts as the cathode and $Mg$ acts as the anode.
The standard cell potential is calculated as:
$E^{0}_{cell} = E^{0}_{cathode} - E^{0}_{anode}$
$E^{0}_{cell} = 0.33 \, V - (-2.37 \, V)$
$E^{0}_{cell} = 0.33 \, V + 2.37 \, V = 2.7 \, V$.

(D) Given reactions:
$(i) Fe^{3+} + 3e^{-} \rightarrow Fe ; E_{1}^{0} = -0.036 \, V$
$(ii) Fe^{2+} + 2e^{-} \rightarrow Fe ; E_{2}^{0} = -0.439 \, V$
Using $\Delta G^{0} = -nFE^{0}$:
$\Delta G_{1}^{0} = -3 \times F \times (-0.036) = 0.108F$
$\Delta G_{2}^{0} = -2 \times F \times (-0.439) = 0.878F$
For the reaction $Fe^{3+} + e^{-} \rightarrow Fe^{2+}$,we have $\Delta G_{3}^{0} = \Delta G_{1}^{0} - \Delta G_{2}^{0}$:
$-1 \times F \times E_{3}^{0} = 0.108F - 0.878F$
$-E_{3}^{0} = -0.770$
$E_{3}^{0} = 0.770 \, V$

(A) Chemical reactivity of a metal is directly related to its ability to lose electrons,which is measured by its standard oxidation potential.
Metals with more negative standard reduction potentials are stronger reducing agents and thus more reactive.
The given standard reduction potentials are:
$A = -3.05 \ V$
$B = -1.66 \ V$
$C = -0.40 \ V$
$D = 0.80 \ V$
Since $A$ has the most negative reduction potential $(-3.05 \ V)$,it has the highest tendency to lose electrons and is the most chemically reactive metal among the given options.

(A) We are given the following standard reduction potentials:
$(1)$ $Fe^{+3} + 3e^{-} \rightarrow Fe$ ; $E_1^{\circ} = -0.036 \ V$,$\Delta G_1^{\circ} = -n_1 F E_1^{\circ} = -3 \times F \times (-0.036) = 0.108 F$
$(2)$ $Fe^{+2} + 2e^{-} \rightarrow Fe$ ; $E_2^{\circ} = -0.439 \ V$,$\Delta G_2^{\circ} = -n_2 F E_2^{\circ} = -2 \times F \times (-0.439) = 0.878 F$
We need the potential for: $Fe^{+3} + e^{-} \rightarrow Fe^{+2}$
This reaction is obtained by subtracting equation $(2)$ from equation $(1)$:
$(Fe^{+3} + 3e^{-}) - (Fe^{+2} + 2e^{-}) \rightarrow Fe - Fe$
$Fe^{+3} + e^{-} \rightarrow Fe^{+2}$
For this reaction,$\Delta G_3^{\circ} = \Delta G_1^{\circ} - \Delta G_2^{\circ} = 0.108 F - 0.878 F = -0.770 F$
Since $\Delta G_3^{\circ} = -n_3 F E_3^{\circ}$ and $n_3 = 1$:
$-0.770 F = -1 \times F \times E_3^{\circ}$
$E_3^{\circ} = 0.770 \ V$

(D) The reactivity of a metal is inversely proportional to its standard reduction potential $(E^{0}_{Red})$.
Lower $E^{0}_{Red}$ value indicates a stronger reducing agent and higher reactivity.
The given $E^{0}_{Red}$ values are:
$P: -2.90 \, V$
$S: -0.76 \, V$
$Q: +0.34 \, V$
$R: +1.20 \, V$
Comparing these values,the order of reactivity is $P > S > Q > R$.