Electrode potential and ECell Questions in English

(A) The reducing power of a species is inversely proportional to its standard reduction potential $(E^\circ_{red})$.
Lower the standard reduction potential,higher is the tendency to lose electrons and act as a strong reducing agent.
Given values:
$E^\circ(Al^{3+}/Al) = -1.66 \ V$
$E^\circ(Fe^{2+}/Fe) = -0.45 \ V$
$E^\circ(Br_2/Br^-) = 1.09 \ V$
Comparing the values: $-1.66 \ V < -0.45 \ V < 1.09 \ V$.
Therefore,the order of reducing power is $Al > Fe > Br^-$.

(A) The standard reduction potential of $Hg^{2+}/Hg$ $(+0.85 \ V)$ is higher than that of $2H^+/H_2$ $(0.00 \ V)$.
Metals with a positive reduction potential (below $H_2$ in the electrochemical series) cannot displace hydrogen from acids or water.
Therefore,$Hg$ cannot displace $H_2$.

(A) The strength of a reducing agent is inversely proportional to its standard reduction potential $(E^o_{RP})$.
Lower (more negative) $E^o_{RP}$ values indicate a greater tendency to undergo oxidation,making the species a stronger reducing agent.
Comparing the given values:
$Zn^{2+}/Zn: -0.762 \, V$
$Cr^{3+}/Cr: -0.740 \, V$
$H^{+}/H_2: 0.00 \, V$
$Fe^{3+}/Fe^{2+}: 0.77 \, V$
Since $-0.762 \, V$ is the most negative value,$Zn$ is the strongest reducing agent among the given options.

(C) Metals placed below hydrogen in the electrochemical series cannot displace $H_2$ gas from water.
Mercury $(Hg)$ is placed below hydrogen in the electrochemical series,therefore it does not react with water to release $H_2$ gas.

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

(A) The standard cell potential is given by the formula: $E_{cell}^{o} = E_{cathode}^{o} - E_{anode}^{o}$.
Since the aluminium electrode is coupled with a standard hydrogen electrode $(SHE)$ and the $emf$ is $1.66 \ V$,aluminium acts as the anode because it is more reactive than hydrogen.
Therefore,$E_{cell}^{o} = E_{H^{+}/H_2}^{o} - E_{Al^{3+}/Al}^{o}$.
Substituting the values: $1.66 \ V = 0.00 \ V - E_{Al^{3+}/Al}^{o}$.
Thus,$E_{Al^{3+}/Al}^{o} = -1.66 \ V$.

(B) The strength of a reducing agent is inversely proportional to the standard reduction potential $(E^\circ)$.
The given reduction potentials are:
$E^\circ(Fe^{2+}/Fe) = -0.44 \ V$
$E^\circ(Ni^{2+}/Ni) = -0.25 \ V$
$E^\circ(Sn^{2+}/Sn) = -0.14 \ V$
$E^\circ(Fe^{3+}/Fe^{2+}) = +0.77 \ V$
Since $Fe$ has the lowest (most negative) standard reduction potential of $-0.44 \ V$,it is the strongest reducing agent among the given options.

(C) The tendency to undergo oxidation is determined by the standard oxidation potential of the metal.
Metals with more negative standard reduction potentials (or more positive standard oxidation potentials) are more easily oxidized.
Comparing the standard reduction potentials $(E^\circ_{red})$:
$Mg^{2+} + 2e^- \rightarrow Mg$ $(E^\circ = -2.37 \ V)$
$Al^{3+} + 3e^- \rightarrow Al$ $(E^\circ = -1.66 \ V)$
$Zn^{2+} + 2e^- \rightarrow Zn$ $(E^\circ = -0.76 \ V)$
$Cu^{2+} + 2e^- \rightarrow Cu$ $(E^\circ = +0.34 \ V)$
Since $Mg$ has the most negative reduction potential,it has the highest tendency to lose electrons and undergo oxidation.
Therefore,the correct option is $(C)$.

(D) $E^{0}_{Zn^{2+}/Zn} = -0.76 \ V$ and $E^{0}_{Fe^{2+}/Fe} = -0.44 \ V$.
Since $Zn$ has a more negative standard reduction potential $(SRP)$ than $Fe$,$Zn$ acts as a sacrificial anode and protects iron from corrosion by becoming the anode itself.

(D) The given half-cell reactions are:
$(I) \ Mn^{2+} + 2e^{-} \rightarrow Mn$,$E^{\circ}_{1} = -1.18 \ V$
$(II) \ Mn^{3+} + e^{-} \rightarrow Mn^{2+}$,$E^{\circ}_{2} = +1.51 \ V$
To obtain the reaction $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$,we perform the operation: $(I) - 2 \times (II)$.
The standard Gibbs free energy change is given by $\Delta G^{\circ} = -nFE^{\circ}$.
For reaction $(I)$,$\Delta G^{\circ}_{1} = -2 \times F \times (-1.18) = +2.36F$.
For reaction $(II)$,$\Delta G^{\circ}_{2} = -1 \times F \times (+1.51) = -1.51F$.
For the target reaction,$\Delta G^{\circ}_{net} = \Delta G^{\circ}_{1} - 2 \times \Delta G^{\circ}_{2} = +2.36F - 2 \times (-1.51F) = +2.36F + 3.02F = +5.38F$.
Since $\Delta G^{\circ}_{net} = -nFE^{\circ}_{cell}$,where $n = 2$ electrons are transferred in the balanced equation:
$5.38F = -2 \times F \times E^{\circ}_{cell}$
$E^{\circ}_{cell} = -5.38 / 2 = -2.69 \ V$.
Since $E^{\circ}_{cell}$ is negative,the reaction is non-spontaneous.

(A) The strength of an oxidising agent is directly proportional to the standard reduction potential $(E^o)$. $A$ higher positive $E^o$ value indicates a stronger oxidising agent.
The strength of a reducing agent is inversely proportional to the standard reduction potential $(E^o)$. $A$ lower (or more negative) $E^o$ value indicates a stronger reducing agent.
Comparing the given values:
$F_2$ has the highest $E^o$ $(+2.85 \ V)$,so it is the strongest oxidising agent.
$I^-$ has the lowest $E^o$ ($+0.53 \ V$ for the corresponding $I_2/I^-$ couple),making it the strongest reducing agent among the species listed.
Therefore,the strongest oxidising and reducing agents are $F_2$ and $I^-$ respectively.

(C) The standard reduction potentials are given as:
$E_{X}^{0} = -1.2 \, V$
$E_{Y}^{0} = +0.5 \, V$
$E_{Z}^{0} = -3.0 \, V$
Reducing power is inversely proportional to the standard reduction potential.
Therefore,the metal with the lowest (most negative) reduction potential has the highest reducing power.
Comparing the values: $-3.0 \, V < -1.2 \, V < +0.5 \, V$.
Thus,the order of reducing power is $Z > X > Y$.

(B) For the reaction $Cu^{2+} + e^- \rightarrow Cu^{+}$,$E^0_1 = 0.15 \ V$. The Gibbs free energy change is $\Delta G^0_1 = -n_1 F E^0_1 = -1 \times F \times 0.15 = -0.15 F$.
For the reaction $Cu^{+} + e^- \rightarrow Cu$,$E^0_2 = 0.50 \ V$. The Gibbs free energy change is $\Delta G^0_2 = -n_2 F E^0_2 = -1 \times F \times 0.50 = -0.50 F$.
For the overall reaction $Cu^{2+} + 2e^- \rightarrow Cu$,the total Gibbs free energy change is $\Delta G^0 = \Delta G^0_1 + \Delta G^0_2 = -0.15 F - 0.50 F = -0.65 F$.
Using the relation $\Delta G^0 = -n F E^0_{Cu^{2+}/Cu}$,where $n = 2$:
$-2 F E^0_{Cu^{2+}/Cu} = -0.65 F$.
Therefore,$E^0_{Cu^{2+}/Cu} = \frac{0.65}{2} = 0.325 \ V$.

(B) The standard reduction potential for the cathode $(Sn^{4+}/Sn^{2+})$ is $E_{cathode}^{0} = +0.15 \ V$.
The standard reduction potential for the anode $(Cr^{3+}/Cr)$ is $E_{anode}^{0} = -0.74 \ V$.
The standard cell potential is calculated using the formula: $E_{cell}^{0} = E_{cathode}^{0} - E_{anode}^{0}$.
Substituting the values: $E_{cell}^{0} = 0.15 \ V - (-0.74 \ V) = 0.15 + 0.74 = +0.89 \ V$.

(C) The standard reduction potential $(E^{\circ})$ for $Fe^{3+}/Fe^{2+}$ is $+0.77 \ V$ and for $I_2/2I^{-}$ is $+0.536 \ V$.
Since the reduction potential of $Fe^{3+}/Fe^{2+}$ is higher than that of $I_2/2I^{-}$,$Fe^{3+}$ has a greater tendency to get reduced to $Fe^{2+}$.
Conversely,$I^{-}$ has a greater tendency to get oxidised to $I_2$.
Therefore,the spontaneous reaction is $2Fe^{3+} + 2I^{-} \rightarrow 2Fe^{2+} + I_2$.
Thus,$I^{-}$ will be oxidised to $I_2$.

(D) The $EMF$ of a cell is defined as:
$EMF_{cell} = E_{cathode} (reduction) - E_{anode} (reduction)$
Since $E_{anode} (reduction) = -E_{anode} (oxidation)$,we can write:
$EMF_{cell} = E_{cathode} (reduction) + E_{anode} (oxidation)$
Also,since $E_{cathode} (reduction) = -E_{cathode} (oxidation)$,we can write:
$EMF_{cell} = E_{anode} (oxidation) - E_{cathode} (oxidation)$
Comparing these with the given relations:
$(ii)$ $EMF$ of cell = (Oxidation potential of anode) $+$ (Reduction potential of cathode) is correct.
$(iv)$ $EMF$ of cell = (Oxidation potential of anode) $-$ (Oxidation potential of cathode) is correct.
Therefore,the correct option is $D$.

(D) The standard Gibbs free energy change is given by $\Delta G^{o} = -nFE^{o}$.
For reaction $(i): Cu^{2+} + 2e^{-} \rightarrow Cu$,$\Delta G_{1}^{o} = -2 \times F \times 0.337 = -0.674F$.
For reaction $(ii): Cu^{2+} + e^{-} \rightarrow Cu^{+}$,$E^{o} = 0.153 \, V$. We need the reverse reaction: $Cu^{+} \rightarrow Cu^{2+} + e^{-}$,where $E^{o} = -0.153 \, V$.
Thus,for $Cu^{+} \rightarrow Cu^{2+} + e^{-}$,$\Delta G_{2}^{o} = -1 \times F \times (-0.153) = 0.153F$.
Adding the two reactions:
$(Cu^{2+} + 2e^{-}$ $\rightarrow Cu) + (Cu^{+}$ $\rightarrow Cu^{2+} + e^{-})$ $\rightarrow (Cu^{+} + e^{-}$ $\rightarrow Cu)$.
$\Delta G_{total}^{o} = \Delta G_{1}^{o} + \Delta G_{2}^{o} = -0.674F + 0.153F = -0.521F$.
For the target reaction $Cu^{+} + e^{-} \rightarrow Cu$,$n=1$,so $\Delta G^{o} = -1 \times F \times E^{o}$.
$-0.521F = -1 \times F \times E^{o} \implies E^{o} = 0.52 \, V$.

(D) The given half-cell reactions are:
$Fe^{2+} + 2e^{-} \rightarrow Fe, E^{\circ} = -0.441 \ V$ (Anode reaction: $Fe \rightarrow Fe^{2+} + 2e^{-}, E^{\circ} = +0.441 \ V$)
$Fe^{3+} + e^{-} \rightarrow Fe^{2+}, E^{\circ} = +0.771 \ V$ (Cathode reaction: $2Fe^{3+} + 2e^{-} \rightarrow 2Fe^{2+}, E^{\circ} = +0.771 \ V$)
For the overall reaction $Fe + 2Fe^{3+} \rightarrow 3Fe^{2+}$,the standard cell potential is calculated as:
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$
$E^{\circ}_{cell} = E^{\circ}_{Fe^{3+}/Fe^{2+}} - E^{\circ}_{Fe^{2+}/Fe}$
$E^{\circ}_{cell} = 0.771 \ V - (-0.441 \ V)$
$E^{\circ}_{cell} = 1.212 \ V$

(B) The given half-cell reactions are:
$(i) Fe^{3+} + 3e^- \rightarrow Fe, E^{\circ}_1 = -0.036 \ V, n_1 = 3$
$(ii) Fe^{2+} + 2e^- \rightarrow Fe, E^{\circ}_2 = -0.439 \ V, n_2 = 2$
We need to find $E^{\circ}$ for the reaction:
$(iii) Fe^{3+} + e^- \rightarrow Fe^{2+}, n_3 = 1$
This reaction can be obtained by $(i) - (ii)$.
Using the relation $\Delta G^{\circ} = -nFE^{\circ}$,we have:
$n_3 E^{\circ}_3 = n_1 E^{\circ}_1 - n_2 E^{\circ}_2$
$1 \times E^{\circ} = 3 \times (-0.036) - 2 \times (-0.439)$
$E^{\circ} = -0.108 + 0.878$
$E^{\circ} = 0.770 \ V$

(D) For a spontaneous reaction,the standard cell potential $E^{\circ}_{cell}$ must be positive.
The reaction is $X + Y^{2+} \rightarrow X^{2+} + Y$.
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = E^{\circ}_{Y^{2+}/Y} - E^{\circ}_{X^{2+}/X}$.
For $X = Zn$ and $Y = Ni$:
$E^{\circ}_{cell} = E^{\circ}_{Ni^{2+}/Ni} - E^{\circ}_{Zn^{2+}/Zn} = -0.23 \ V - (-0.76 \ V) = +0.53 \ V$.
Since $E^{\circ}_{cell} > 0$,the reaction is spontaneous.

(D) The strength of an oxidising agent is directly proportional to its standard reduction potential $(E^o)$.
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 standard reduction potential $(1.51 \ V)$,it is the strongest oxidising agent.

(A) The reducing strength of a species is inversely proportional to its standard reduction potential. $A$ lower (more negative) standard reduction potential indicates a higher tendency to undergo oxidation,making it a stronger reducing agent.
Comparing the given values:
$E^o_{Cl_2/Cl^-} = 1.36 \ V$
$E^o_{Cr^{3+}/Cr} = -0.74 \ V$
$E^o_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \ V$
$E^o_{MnO_4^-/Mn^{2+}} = 1.51 \ V$
Among the species listed,$Cr$ has the lowest reduction potential $(-0.74 \ V)$.
Therefore,$Cr$ is the strongest reducing agent.

(B) The cell reaction is: $CH_3COOH + HCOO^- \rightleftharpoons CH_3COO^- + HCOOH$.
The equilibrium constant for this reaction is $K_{eq} = \frac{K_a(CH_3COOH)}{K_a(HCOOH)} = \frac{1.8 \times 10^{-5}}{2.4 \times 10^{-4}} = 0.075$.
Using the relation $E^o_{cell} = \frac{0.059}{n} \log K_{eq}$ at $25^oC$ (where $n=1$):
$E^o_{cell} = 0.06 \times \log(0.075) = 0.06 \times \log(\frac{3}{40}) = 0.06 \times (\log 3 - \log 40) = 0.06 \times (0.477 - 1.602) = 0.06 \times (-1.125) = -0.0675 \ V$.
Given the approximation $\frac{2.303RT}{F} = 0.06$,the value is approximately $-0.0672 \ V$.

(C) The cell reaction is: $2Cr + 3Co^{2+} \rightarrow 2Cr^{3+} + 3Co$.
Since the concentrations are $1.0 \ M$,the reaction quotient $Q = 1$,and $\ln(Q) = 0$.
Using the formula: $E_{cell} = E_{cell}^{o} - \frac{0.0591}{n} \log(Q)$.
Since $Q = 1$,$E_{cell} = E_{cell}^{o}$.
$E_{cell}^{o} = (E_{RP}^{o})_{cathode} - (E_{RP}^{o})_{anode}$.
Here,the cathode is $Co^{2+} | Co$ and the anode is $Cr | Cr^{3+}$.
$E_{cell}^{o} = (-0.28 \ V) - (-0.74 \ V) = +0.46 \ V$.

(A) The given reactions are:
$(1)$ $Cu^{2+} + e^- \rightarrow Cu^+$,$E_1^{\circ} = 0.15 \ V$
$(2)$ $Cu^+ + e^- \rightarrow Cu$,$E_2^{\circ} = 0.5 \ V$
$(3)$ $Cu^{2+} + 2e^- \rightarrow Cu$,$E_3^{\circ} = x \ V$
Using the relationship $\Delta G_3^{\circ} = \Delta G_1^{\circ} + \Delta G_2^{\circ}$:
$-n_3 F E_3^{\circ} = -n_1 F E_1^{\circ} - n_2 F E_2^{\circ}$
$n_3 E_3^{\circ} = n_1 E_1^{\circ} + n_2 E_2^{\circ}$
$2 \times x = 1 \times 0.15 + 1 \times 0.5$
$2x = 0.65$
$x = 0.325 \ V$

(D) solution can be stored in a metal vessel if the metal of the vessel does not act as a reducing agent for the metal ions in the solution. This means the standard reduction potential of the metal vessel $(E^{o}_{vessel})$ must be greater than the standard reduction potential of the metal ion in the solution $(E^{o}_{ion})$,so that $E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$ is negative,indicating a non-spontaneous reaction.
Given standard reduction potentials $(E^{o})$:
$E^{o}_{Zn^{2+}/Zn} = -0.76 \ V$
$E^{o}_{Sn^{2+}/Sn} = -0.14 \ V$
$E^{o}_{Cu^{2+}/Cu} = 0.34 \ V$
$E^{o}_{Ag^{+}/Ag} = 0.80 \ V$
Evaluating the options:
$A$. $CuSO_4$ in $Zn$: $E^{o}_{Zn} < E^{o}_{Cu}$,reaction occurs $(Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu)$.
$B$. $AgNO_3$ in $Zn$: $E^{o}_{Zn} < E^{o}_{Ag}$,reaction occurs $(Zn + 2Ag^{+} \rightarrow Zn^{2+} + 2Ag)$.
$C$. $AgNO_3$ in $Sn$: $E^{o}_{Sn} < E^{o}_{Ag}$,reaction occurs $(Sn + 2Ag^{+} \rightarrow Sn^{2+} + 2Ag)$.
$D$. $CuSO_4$ in $Ag$: $E^{o}_{Ag} > E^{o}_{Cu}$,no reaction occurs ($Ag + Cu^{2+} \rightarrow$ No reaction).
Therefore,$CuSO_4$ solution can be stored in a silver vessel.

(C) The strength of an oxidant is directly proportional to its standard reduction potential $(E^o_{red})$.
Comparing the given $E^o$ values:
$MnO_4^- (1.52 \ V) > Fe^{3+} (0.77 \ V) > I_2 (0.54 \ V) > Sn^{4+} (0.1 \ V)$.
Thus,$MnO_4^-$ has the highest reduction potential and is the strongest oxidant.
The strength of a reductant is inversely proportional to the reduction potential of its corresponding oxidized form.
Looking at the reverse reactions (oxidation potentials):
$Sn^{2+} \to Sn^{4+} + 2e^-; E^o_{ox} = -0.1 \ V$
$2I^- \to I_2 + 2e^-; E^o_{ox} = -0.54 \ V$
$Fe^{2+} \to Fe^{3+} + e^-; E^o_{ox} = -0.77 \ V$
$Mn^{2+} + 4H_2O \to MnO_4^- + 8H^{+} + 5e^-; E^o_{ox} = -1.52 \ V$
$Sn^{2+}$ has the highest oxidation potential (least negative),making it the strongest reductant.
Therefore,the strongest reductant is $Sn^{2+}$ and the strongest oxidant is $MnO_4^-$.

(A) The strength of an oxidant is directly proportional to its standard reduction potential $(E^{\circ}_{RP})$.
The strength of a reductant is inversely proportional to its standard reduction potential $(E^{\circ}_{RP})$.
Comparing the given $E^{\circ}$ values:
$1.52 \, V (MnO_4^{-}) > 0.77 \, V (Fe^{3+}) > 0.54 \, V (I_2) > 0.1 \, V (Sn^{4+})$.
Since $MnO_4^{-}$ has the highest $E^{\circ}_{RP}$,it is the strongest oxidant.
Since $Sn^{2+}$ is the conjugate reductant of the species with the lowest $E^{\circ}_{RP}$ $(Sn^{4+})$,it acts as the strongest reductant among the given species.

(D) The standard reduction potentials $(SRP)$ are given as:
$E^\circ_{X^{+2}/X} = -1.46 \ V$
$E^\circ_{Y^{+2}/Y} = -0.36 \ V$
$E^\circ_{Z^{+2}/Z} = 0.15 \ V$
$E^\circ_{W^{+2}/W} = -1.24 \ V$
$A$ metal with a lower (more negative) reduction potential can displace a metal with a higher (more positive) reduction potential from its salt solution.
Since $E^\circ_{X^{+2}/X} < E^\circ_{Z^{+2}/Z}$,$X$ will displace $Z^{+2}$.
Since $E^\circ_{Y^{+2}/Y} < E^\circ_{Z^{+2}/Z}$,$Y$ will displace $Z^{+2}$.
Since $E^\circ_{W^{+2}/W} < E^\circ_{Z^{+2}/Z}$,$W$ will displace $Z^{+2}$.
Therefore,all the given statements are correct.

(B) The cell potential ${E^o}_{cell}$ is calculated as ${E^o}_{cathode} - {E^o}_{anode}$.
To obtain the largest potential,we must choose the half-cell with the highest ${E^o}$ value as the cathode and the half-cell with the lowest ${E^o}$ value as the anode.
Comparing the given values:
$I: +0.24 \ V$
$II: +1.25 \ V$
$III: +0.15 \ V$
$IV: +0.68 \ V$
The highest value is $II$ $(+1.25 \ V)$ and the lowest value is $III$ $(+0.15 \ V)$.
Therefore,the combination of $II$ and $III$ will result in the largest potential: ${E^o}_{cell} = 1.25 - 0.15 = 1.10 \ V$.

(A) The relationship between standard Gibbs free energy change $(\Delta G^o)$ and standard cell potential $(E^o_{cell})$ is given by the formula: $\Delta G^o = -nFE^o_{cell}$.
Here,the reaction is $H_2 + 2AgCl \rightarrow 2Ag + 2H^{+} + 2Cl^{-}$.
The number of electrons transferred $(n)$ is $2$.
The Faraday constant $(F)$ is approximately $96,500 \, C \, mol^{-1}$.
Given $\Delta G^o = -48,250 \, J \, mol^{-1}$.
Substituting the values: $-48,250 = -(2) \times (96,500) \times E^o_{cell}$.
$E^o_{cell} = \frac{48,250}{2 \times 96,500} = \frac{48,250}{193,000} = 0.25 \, V$.

(B) The cell reaction is $Ti^{2+} + Co^{2+} \rightarrow Ti^{3+} + Co$.
The standard cell potential is given by $E_{cell}^{o} = E_{cathode}^{o} - E_{anode}^{o}$.
Here,the cathode is the $Co^{2+}|Co$ electrode and the anode is the $Ti^{2+}|Ti^{3+}$ electrode.
$E_{cell}^{o} = E_{Co^{2+}|Co}^{o} - E_{Ti^{3+}|Ti^{2+}}^{o}$.
Substituting the given values: $0.09 \ V = -0.28 \ V - E_{Ti^{3+}|Ti^{2+}}^{o}$.
$E_{Ti^{3+}|Ti^{2+}}^{o} = -0.28 \ V - 0.09 \ V = -0.37 \ V$.
The standard oxidation potential is the negative of the standard reduction potential: $E_{ox}^{o} = -E_{red}^{o} = -(-0.37 \ V) = 0.37 \ V$.

(D) The standard reduction potentials are $E^{\circ}_{Pb^{2+}/Pb} = -0.13 \, V$ and $E^{\circ}_{Fe^{2+}/Fe} = -0.44 \, V$.
Since $E^{\circ}_{Pb^{2+}/Pb} > E^{\circ}_{Fe^{2+}/Fe}$,$Pb^{2+}$ ions will act as the oxidizing agent and $Fe$ metal will act as the reducing agent.
The spontaneous cell reaction is: $Pb^{2+}(aq) + Fe(s) \rightarrow Fe^{2+}(aq) + Pb(s)$.
In this reaction,$Pb^{2+}$ ions are reduced to $Pb$ metal,so the concentration of $Pb^{2+}$ decreases.
Simultaneously,$Fe$ metal is oxidized to $Fe^{2+}$ ions,so the concentration of $Fe^{2+}$ increases.

(C) The cell reaction is $2Fe^{3+} + 2I^{-} \to 2Fe^{2+} + I_2$.
Here,$Fe^{3+}$ is reduced to $Fe^{2+}$ (cathode) and $I^{-}$ is oxidized to $I_2$ (anode).
$E^o_{\text{cell}} = E^o_{\text{cathode}} - E^o_{\text{anode}}$
$E^o_{\text{cell}} = E^o_{Fe^{3+}/Fe^{2+}} - E^o_{I_2/I^{-}}$
$E^o_{\text{cell}} = 0.771 \ V - 0.536 \ V = 0.235 \ V$

(C) Given half-reactions:
$(I) \ OCl^{-} + H_2O + 2e^{-} \longrightarrow Cl^{-} + 2OH^{-}, \ E^o_1 = 0.94 \ V, \ n_1 = 2$
$\Delta G^o_1 = -n_1 F E^o_1 = -2 \times F \times 0.94 = -1.88F$
$(II) \ \frac{1}{2}Cl_2 + e^{-} \longrightarrow Cl^{-}, \ E^o_2 = 1.36 \ V, \ n_2 = 1$
To get the reaction $OCl^{-} + H_2O + e^{-} \longrightarrow \frac{1}{2}Cl_2 + 2OH^{-}$,we perform $(I) - (II)$:
$\Delta G^o_3 = \Delta G^o_1 - \Delta G^o_2 = -1.88F - (-1.36F) = -0.52F$
For the target reaction,$n_3 = 1$:
$E^o_3 = -\frac{\Delta G^o_3}{n_3 F} = -\frac{-0.52F}{1 \times F} = +0.52 \ V$.

(C) The standard cell potential is calculated using the formula:
$E^o_{cell} = E^o_{cathode} - E^o_{anode}$
Here,the cathode is the silver electrode $(Ag^{+}/Ag)$ and the anode is the zinc electrode $(Zn^{2+}/Zn)$.
$E^o_{cell} = E^o_{Ag^{+}/Ag} - E^o_{Zn^{2+}/Zn}$
$E^o_{cell} = 0.80 \ V - (-0.76 \ V)$
$E^o_{cell} = 0.80 + 0.76 = 1.56 \ V$

(C) The given reaction is $A + 2A^{+3} \to 3A^{+2}$.
This can be split into two half-reactions:
Anode (Oxidation): $A \to A^{+2} + 2e^-$,where $E^o_{ox} = -E^o_{A^{+2}/A} = -(-0.30 \ V) = 0.30 \ V$.
Cathode (Reduction): $2A^{+3} + 2e^- \to 2A^{+2}$,where $E^o_{red} = E^o_{A^{+3}/A^{+2}} = 0.40 \ V$.
The standard $EMF$ of the cell is calculated as $E^o_{cell} = E^o_{ox} + E^o_{red}$.
$E^o_{cell} = 0.30 \ V + 0.40 \ V = 0.70 \ V$.

(C) Given:
$(1) \ Cu^{2+} + e^- \to Cu^{+} ; E_1^o = 0.15 \ V, \Delta G_1^o = -n_1 F E_1^o, n_1 = 1$
$(2) \ Cu^{2+} + 2e^- \to Cu ; E_2^o = 0.34 \ V, \Delta G_2^o = -n_2 F E_2^o, n_2 = 2$
We need to find $E_3^o$ for:
$(3) \ Cu^{+} + e^- \to Cu ; E_3^o = ?, \Delta G_3^o = -n_3 F E_3^o, n_3 = 1$
Reaction $(3)$ can be obtained by subtracting reaction $(1)$ from reaction $(2)$:
$(Cu^{2+} + 2e^- \to Cu) - (Cu^{2+} + e^- \to Cu^{+}) \implies Cu^{+} + e^- \to Cu$
Since $\Delta G^o$ is an extensive property,$\Delta G_3^o = \Delta G_2^o - \Delta G_1^o$
$-n_3 F E_3^o = -n_2 F E_2^o - (-n_1 F E_1^o)$
$-1 \times F \times E_3^o = -(2 \times F \times 0.34) + (1 \times F \times 0.15)$
$-E_3^o = -0.68 + 0.15$
$-E_3^o = -0.53$
$E_3^o = +0.53 \ V$

(A) The cell reaction is: $2Fe^{3+} + 2I^- \to 2Fe^{2+} + I_2$.
This reaction consists of two half-reactions:
Cathode (Reduction): $2Fe^{3+} + 2e^- \to 2Fe^{2+}$; $E^o_{\text{cathode}} = 0.771 \, V$.
Anode (Oxidation): $2I^- \to I_2 + 2e^-$; $E^o_{\text{anode}} = 0.536 \, V$.
$E^o_{\text{cell}} = E^o_{\text{cathode}} - E^o_{\text{anode}}$.
$E^o_{\text{cell}} = 0.771 \, V - 0.536 \, V = 0.235 \, V$.

(A) The standard reduction potentials are given as: $E^o(Mg^{2+}/Mg) = -2.37 \ V$,$E^o(Zn^{2+}/Zn) = -0.76 \ V$,and $E^o(Fe^{2+}/Fe) = -0.44 \ V$.
According to the electrochemical series,a metal with a lower (more negative) reduction potential can reduce the ion of a metal with a higher (less negative) reduction potential.
Since $E^o(Zn^{2+}/Zn) < E^o(Fe^{2+}/Fe)$ (i.e.,$-0.76 \ V < -0.44 \ V$),$Zn$ can reduce $Fe^{2+}$ to $Fe$.
Therefore,the correct statement is that $Zn$ will reduce $Fe^{2+}$.

(C) The standard cell potential is calculated using the formula: $E^{o}_{cell} = (E^{o}_{red})_{cathode} - (E^{o}_{red})_{anode}$.
In the given cell $Ag_{(s)} | Ag^{\oplus} || Cu^{2+} | Cu_{(s)}$,the reduction occurs at the copper electrode (cathode) and oxidation occurs at the silver electrode (anode).
Therefore,$(E^{o}_{red})_{cathode} = y$ and $(E^{o}_{red})_{anode} = x$.
Substituting these values,we get $E^{o}_{cell} = y - x$.

(D) The reaction $Cu^{2+} + e^- \to Cu^{+}$ can be obtained by subtracting the first reaction from the second reaction:
$(Cu^{2+} + 2e^- \to Cu) - (Cu^{+} + e^- \to Cu) \implies Cu^{2+} + e^- \to Cu^{+}$
The change in Gibbs free energy is given by $\Delta G^o = -nFE^o$.
For the target reaction: $\Delta G_3^o = \Delta G_2^o - \Delta G_1^o$
$-1 \times F \times E^o = (-2 \times F \times X_2) - (-1 \times F \times X_1)$
$-F \times E^o = -2FX_2 + FX_1$
Dividing by $-F$,we get: $E^o = 2X_2 - X_1$

(A) The reducing strength of a metal is inversely proportional to its Standard Reduction Potential $(SRP)$ value.
Stronger reducing agents have more negative $SRP$ values.
Comparing the given values: $A = -3.05 \ V$,$B = -1.66 \ V$,$C = -0.40 \ V$,and $D = 0.80 \ V$.
Since $-3.05 \ V$ is the most negative value,metal $A$ has the highest tendency to lose electrons and acts as the strongest reducing agent.

(A) The standard Gibbs free energy change is given by $\Delta G^o = -nFE^o$.
For reaction $(i)$: $Fe^{2+} + 2e^- \to Fe$,$E^o = -0.47 \ V$,so $\Delta G^o_1 = -2 \times F \times (-0.47) = 0.94 \ F$.
For reaction $(ii)$: $Fe^{3+} + e^- \to Fe^{2+}$,$E^o = +0.77 \ V$,so $\Delta G^o_2 = -1 \times F \times (+0.77) = -0.77 \ F$.
For the overall reaction $(iii)$: $Fe^{3+} + 3e^- \to Fe$,the sum of the reactions is $(i) + (ii)$.
Therefore,$\Delta G^o_3 = \Delta G^o_1 + \Delta G^o_2 = 0.94 \ F - 0.77 \ F = 0.17 \ F$.
Using $\Delta G^o_3 = -nFE^o_{Fe^{3+}/Fe}$,where $n = 3$:
$0.17 \ F = -3 \times F \times E^o_{Fe^{3+}/Fe}$
$E^o_{Fe^{3+}/Fe} = \frac{0.17 \ F}{-3 \ F} = -0.057 \ V$.

(C) The deposition of metals during electrolysis occurs in the order of their reduction potentials,starting from the highest reduction potential to the lowest reduction potential.
Given standard reduction potentials: $E_A^\circ = -1.2 \ V$,$E_B^\circ = 0.6 \ V$,$E_C^\circ = 0.85 \ V$,and $E_D^\circ = -0.76 \ V$.
Arranging these in decreasing order: $0.85 \ V (C) > 0.6 \ V (B) > -0.76 \ V (D) > -1.2 \ V (A)$.
Thus,the sequence of deposition is $C, B, D, A$.

(B) The reducing power of a species is inversely proportional to its standard reduction potential $(E^o)$.
The given standard reduction potentials are:
$E^o (Al^{3+}/Al) = -1.66 \ V$
$E^o (Fe^{3+}/Fe^{2+}) = +0.77 \ V$
$E^o (Br_2/Br^-) = +1.09 \ V$
$A$ more negative $E^o$ value indicates a stronger reducing agent.
Comparing the values: $-1.66 \ V < +0.77 \ V < +1.09 \ V$.
Therefore,the order of reducing power is $Br^- < Fe^{2+} < Al$.

(C) The strength of an oxidizing agent is directly proportional to its standard reduction potential $(E^o)$.
Comparing the given values:
$E^o(Cu^{+}/Cu) = +0.52 \ V$
$E^o(Fe^{3+}/Fe^{2+}) = +0.77 \ V$
$E^o(\frac{1}{2} I_2/I^{-}) = +0.54 \ V$
$E^o(Ag^{+}/Ag) = +0.88 \ V$
Since $Ag^{+}$ has the highest reduction potential $(+0.88 \ V)$,it is the strongest oxidizing agent.

(A) The reducing power of a species is inversely proportional to its standard reduction potential $(E^o)$.
Comparing the reduction potentials of the corresponding half-reactions:
$1$. For $Cr^{3+} + 3e^- \rightarrow Cr$,$E^o = -0.74 \ V$
$2$. For $Cl_2 + 2e^- \rightarrow 2Cl^-$,$E^o = 1.36 \ V$
$3$. For $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$,$E^o = 1.33 \ V$
$4$. For $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$,$E^o = 1.51 \ V$
To determine the reducing power of the species $(Cr, Cr^{3+}, Mn^{2+}, Cl^-)$,we look at their oxidation potentials (which is $-E^o$ of the reduction reaction):
- $Cr \rightarrow Cr^{3+} + 3e^-$,$E^o_{ox} = +0.74 \ V$
- $Cl^- \rightarrow \frac{1}{2}Cl_2 + e^-$,$E^o_{ox} = -1.36 \ V$
- $Cr^{3+} \rightarrow Cr_2O_7^{2-}$,$E^o_{ox} = -1.33 \ V$
- $Mn^{2+} \rightarrow MnO_4^-$,$E^o_{ox} = -1.51 \ V$
Arranging these in increasing order of oxidation potential (reducing power): $Mn^{2+} < Cl^- < Cr^{3+} < Cr$.

(C) The deposition of metals at the cathode during electrolysis depends on their standard reduction potentials $(E^o)$.
Metals with higher reduction potentials are deposited more easily at the cathode.
The given standard reduction potentials are: $E^o(Ag^+/Ag) = 0.80 \ V$,$E^o(Hg_2^{2+}/2Hg) = 0.79 \ V$,$E^o(Cu^{2+}/Cu) = 0.34 \ V$,and $E^o(Mg^{2+}/Mg) = -2.37 \ V$.
In an aqueous solution,water can also be reduced at the cathode: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$,with $E^o = -0.83 \ V$.
Since $Mg^{2+}$ has a reduction potential of $-2.37 \ V$,which is much lower than that of water $(-0.83 \ V)$,$Mg$ will not deposit from an aqueous solution; instead,$H_2$ gas will be evolved.
Therefore,the metals that will deposit in the order of increasing voltage are $Ag$,$Hg$,and $Cu$.
The correct sequence is $Ag > Hg > Cu$.

(B) For the reaction $Cu^{2+} + e^- \to Cu^{+}$,$\Delta G_1^o = -n_1 E_1^o F = -1 \times 0.15 \times F = -0.15F$.
For the reaction $Cu^{2+} + 2e^- \to Cu$,$\Delta G_2^o = -n_2 E_2^o F = -2 \times 0.34 \times F = -0.68F$.
To find the potential for $Cu^{+} + e^- \to Cu$,we subtract the first reaction from the second:
$(Cu^{2+} + 2e^- \to Cu) - (Cu^{2+} + e^- \to Cu^{+}) \implies Cu^{+} + e^- \to Cu$.
$\Delta G^o = \Delta G_2^o - \Delta G_1^o = -0.68F - (-0.15F) = -0.53F$.
Since $\Delta G^o = -n E^o F$,where $n=1$:
$-0.53F = -1 \times E^o \times F
\implies E^o = 0.53 \, V$.