Electrode potential and ECell Questions in English

(A) The reducing power of a metal is inversely proportional to its Standard Reduction Potential $(SRP)$ value.
Lower $SRP$ value indicates a stronger reducing agent.
The given $SRP$ values are:
$Ni^{2+}/Ni = -0.25 \, V$
$Zn^{2+}/Zn = -0.76 \, V$
$Mg^{2+}/Mg = -2.36 \, V$
$Ca^{2+}/Ca = -2.87 \, V$
Arranging these in increasing order of $SRP$ values: $Ni < Zn < Mg < Ca$.
Therefore,the order of increasing reducing power is: $Ni < Zn < Mg < Ca$.

(A) The standard cell potential is given by $E^o_{cell} = E^o_{cathode} - E^o_{anode}$.
Given $E^o_{Zn^{2+}/Zn} = -0.76 \ V$,the anode is $Zn$.
We calculate $E^o_{cell}$ for each cathode:
$1$. For $Au^{3+}/Au$ $(n=3)$: $E^o_{cell} = 1.40 - (-0.76) = 2.16 \ V$. Value per electron = $2.16 / 3 = 0.72 \ V$.
$2$. For $Ag^{+}/Ag$ $(n=1)$: $E^o_{cell} = 0.80 - (-0.76) = 1.56 \ V$. Value per electron = $1.56 / 1 = 1.56 \ V$.
$3$. For $Fe^{3+}/Fe^{2+}$ $(n=1)$: $E^o_{cell} = 0.77 - (-0.76) = 1.53 \ V$. Value per electron = $1.53 / 1 = 1.53 \ V$.
$4$. For $Fe^{2+}/Fe$ $(n=2)$: $E^o_{cell} = -0.44 - (-0.76) = 0.32 \ V$. Value per electron = $0.32 / 2 = 0.16 \ V$.
Comparing the values,$Ag^{+}/Ag$ gives the maximum value of $E^o_{cell}$ per electron transferred.

(B) The strength of an oxidizing agent is directly proportional to its standard reduction potential $(E^o)$.
Comparing the given values:
$E_{O_2/H_2O}^o = +1.23 \ V$
$E_{S_2O_8^{2-}/SO_4^{2-}}^o = +2.05 \ V$
$E_{Br_2/Br^-}^o = +1.09 \ V$
$E_{Au^{3+}/Au}^o = +1.40 \ V$
Since $S_2O_8^{2-}$ has the highest standard reduction potential $(+2.05 \ V)$,it is the strongest oxidizing agent.

(D) The cell reaction is: $Fe^{2+}_{(aq)} + Ag^{+}_{(aq)} \to Fe^{3+}_{(aq)} + Ag_{(s)}$
Given standard reduction potentials:
$(1)$ $Ag^{+} + e^{-} \to Ag$,$E^o = x \ V$,$\Delta G^o_1 = -nFE^o = -1 \cdot F \cdot x = -Fx$
$(2)$ $Fe^{2+} + 2e^{-} \to Fe$,$E^o = y \ V$,$\Delta G^o_2 = -2Fy$
$(3)$ $Fe^{3+} + 3e^{-} \to Fe$,$E^o = z \ V$,$\Delta G^o_3 = -3Fz$
We need to find the $\Delta G^o$ for the reaction: $Fe^{2+} \to Fe^{3+} + e^{-}$
This can be obtained by: (Reaction $2$) - (Reaction $3$)
$\Delta G^o_{oxidation} = \Delta G^o_2 - \Delta G^o_3 = -2Fy - (-3Fz) = 3Fz - 2Fy$
For the total cell reaction:
$Fe^{2+} + Ag^{+} \to Fe^{3+} + Ag$
$\Delta G^o_{cell} = \Delta G^o_{oxidation} + \Delta G^o_1 = (3Fz - 2Fy) + (-Fx) = -F(x + 2y - 3z)$
Since $\Delta G^o_{cell} = -nFE^o_{cell}$ and $n = 1$:
$-FE^o_{cell} = -F(x + 2y - 3z)$
$E^o_{cell} = x + 2y - 3z \ V$

(C) The formula for standard Gibbs energy is $\Delta G^o = -nFE^o$.
Here,$n = 2$ (number of electrons transferred in the redox reaction).
$F = 96500 \, C \, mol^{-1}$.
$E^o = 2 \, V$.
Substituting the values: $\Delta G^o = -2 \times 96500 \times 2 = -386000 \, J \, mol^{-1} = -386 \, kJ \, mol^{-1}$.
Given the options,the closest value is $-384 \, kJ \, mol^{-1}$,which is obtained if $F$ is approximated as $96000 \, C \, mol^{-1}$.

(C) The oxidizing power of a species is directly proportional to its standard reduction potential $(E^o)$.
Comparing the given values:
$Bi^{3+} (E^o = +0.20 \ V) < Ce^{4+} (E^o = +1.61 \ V) < Pb^{4+} (E^o = +1.67 \ V) < Co^{3+} (E^o = +1.81 \ V)$.
Therefore,the order of increasing oxidizing power is $Bi^{3+} < Ce^{4+} < Pb^{4+} < Co^{3+}$.

(D) The standard cell potential,$E_{cell}^o$,is a constant value determined by the standard reduction potentials of the half-cells involved in the reaction.
It is defined as $E_{cell}^o = E_{cathode}^o - E_{anode}^o$.
Since $E_{cell}^o$ is independent of the concentration of the reactants or products,changing the concentration of $Cu^{2+}$ ions will not affect the value of $E_{cell}^o$.
Therefore,the value remains unchanged.

(C) Given:
$Ti^{4+} + e^- \to Ti^{3+}, \, E^o = -x \text{ V}$
$Fe^{3+} + e^- \to Fe^{2+}, \, E^o = -y \text{ V}$
Since $x < y$,then $-x > -y$. Therefore,$E^o(Ti^{4+}/Ti^{3+}) > E^o(Fe^{3+}/Fe^{2+})$.
$1$. $Ti^{4+}$ has a higher reduction potential than $Fe^{3+}$,so $Ti^{4+}$ is a stronger oxidant than $Fe^{3+}$.
$2$. $Ti^{3+}$ has a lower reduction potential than $Fe^{2+}$,so $Ti^{3+}$ is a stronger reductant than $Fe^{2+}$.
$3$. For the reaction $Ti^{3+} + Fe^{3+} \to Ti^{4+} + Fe^{2+}$,the cell potential is $E^o_{cell} = E^o_{cathode} - E^o_{anode} = E^o(Fe^{3+}/Fe^{2+}) - E^o(Ti^{4+}/Ti^{3+}) = -y - (-x) = x - y$. Since $x < y$,$E^o_{cell} < 0$,which means the reaction is non-spontaneous.
Thus,the statement in option $C$ is wrong.

(B) The reducing power of a metal is inversely proportional to its standard reduction potential $(E^{\circ}_{red})$.
Lower (more negative) $E^{\circ}_{red}$ values indicate a stronger reducing agent.
Given values:
$E^{\circ}(A^{+}/A) = -2.93 \ V$
$E^{\circ}(C^{2+}/C) = -2.37 \ V$
$E^{\circ}(D^{3+}/D) = -0.74 \ V$
$E^{\circ}(B^{+}/B) = 0.80 \ V$
Comparing the values: $-2.93 < -2.37 < -0.74 < 0.80$.
The reducing power order is the reverse of the reduction potential order: $B < D < C < A$.

(C) The given half-reactions are:
$(1) Au^{+} + e^- \rightarrow Au$,$E_1^o = 1.69 \text{ V}$,$\Delta G_1^o = -1 \times F \times 1.69 = -1.69F$
$(2) Au^{3+} + 3e^- \rightarrow Au$,$E_2^o = 1.40 \text{ V}$,$\Delta G_2^o = -3 \times F \times 1.40 = -4.20F$
For the target reaction $Au^{+} \rightarrow Au^{3+} + 2e^-$,we subtract reaction $(2)$ from reaction $(1)$:
$\Delta G_3^o = \Delta G_1^o - \Delta G_2^o$
$-2FE_3^o = (-1.69F) - (-4.20F)$
$-2FE_3^o = 2.51F$
$E_3^o = -1.255 \text{ V}$

(A) For the given galvanic cell: $Ca_{(s)} | Ca^{2+}_{(aq)} || Fe^{2+}_{(aq)} | Fe_{(s)}$
The anode reaction is: $Ca_{(s)} \rightarrow Ca^{2+}_{(aq)} + 2e^-$
The cathode reaction is: $Fe^{2+}_{(aq)} + 2e^- \rightarrow Fe_{(s)}$
The standard cell potential is given by: $E_{cell}^o = E_{cathode}^o - E_{anode}^o$
Here,$E_{cathode}^o = E_{Fe^{2+}/Fe}^o$. Since $E_{Fe/Fe^{2+}}^o = 0.41 \ V$,then $E_{Fe^{2+}/Fe}^o = -0.41 \ V$.
$E_{anode}^o = E_{Ca^{2+}/Ca}^o = -2.87 \ V$.
Substituting the values: $E_{cell}^o = (-0.41 \ V) - (-2.87 \ V) = -0.41 \ V + 2.87 \ V = 2.46 \ V$.

(A) The given half-reactions are:
$(1) MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$; $\Delta G_1^o = -n_1 F E_1^o = -5 \times F \times 1.51 = -7.55 F$
$(2) MnO_2 + 4H^+ + 2e^- \to Mn^{2+} + 2H_2O$; $\Delta G_2^o = -n_2 F E_2^o = -2 \times F \times 1.21 = -2.42 F$
We need the reaction: $MnO_4^- + 4H^+ + 3e^- \to MnO_2 + 2H_2O$
This can be obtained by subtracting reaction $(2)$ from reaction $(1)$:
$(1) - (2): (MnO_4^- + 8H^+ + 5e^-) - (MnO_2 + 4H^+ + 2e^-) \to (Mn^{2+} + 4H_2O) - (Mn^{2+} + 2H_2O)$
$MnO_4^- + 4H^+ + 3e^- \to MnO_2 + 2H_2O$
The change in Gibbs free energy is $\Delta G_3^o = \Delta G_1^o - \Delta G_2^o = -7.55 F - (-2.42 F) = -5.13 F$
Since $\Delta G_3^o = -n_3 F E_3^o$ and $n_3 = 3$,we have:
$-3 F E_3^o = -5.13 F$
$E_3^o = \frac{5.13}{3} = 1.71 \ V$

(A) reaction is spontaneous if the standard cell potential $E^\circ_{cell}$ is positive.
According to the Electrochemical Series,a 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.
For option $A$: $Cu_{(s)} + 2Ag^+_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$. Here,$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = E^\circ_{Ag^+/Ag} - E^\circ_{Cu^{2+}/Cu} = 0.80 \ V - 0.34 \ V = +0.46 \ V$. Since $E^\circ_{cell} > 0$,the reaction is spontaneous.
For options $B, C,$ and $D$,the metal being oxidized has a higher reduction potential than the metal being reduced,resulting in a negative $E^\circ_{cell}$,making them non-spontaneous.

(D) By international convention,the standard electrode potential of the Standard Hydrogen Electrode $(SHE)$ is arbitrarily assigned a value of $0.00 \, V$ at all temperatures.

(A) Given reactions:
$(1) Cu^{+} + e^- \longrightarrow Cu$ ; $\Delta G_1^o = -1 \times F \times X_1 = -X_1 F$
$(2) Cu^{2+} + 2e^- \longrightarrow Cu$ ; $\Delta G_2^o = -2 \times F \times X_2 = -2X_2 F$
We need the value of $E^o$ for the reaction: $Cu^{2+} + e^- \longrightarrow Cu^{+}$
This reaction can be obtained by subtracting reaction $(1)$ from reaction $(2)$:
$(Cu^{2+} + 2e^-$ $\longrightarrow Cu) - (Cu^{+} + e^-$ $\longrightarrow Cu) \implies Cu^{2+} + e^-$ $\longrightarrow Cu^{+}$
Therefore,$\Delta G_3^o = \Delta G_2^o - \Delta G_1^o$
$-1 \times F \times E^o = (-2X_2 F) - (-X_1 F)$
$-FE^o = F(X_1 - 2X_2)$
$E^o = 2X_2 - X_1$

(A) The standard Gibbs free energy change is given by $\Delta G^o = -nFE^o$.
For the reaction $Cu^{2+} + 2e^- \to Cu$,$\Delta G_1^o = -2FX_2$.
For the reaction $Cu^{+} + e^- \to Cu$,$\Delta G_2^o = -FX_1$.
We want to find $E^o$ for $Cu^{2+} + e^- \to Cu^{+}$. This reaction can be obtained by subtracting the second reaction from the first:
$(Cu^{2+} + 2e^- \to Cu) - (Cu^{+} + e^- \to Cu) \implies Cu^{2+} + e^- \to Cu^{+}$.
Therefore,$\Delta G_3^o = \Delta G_1^o - \Delta G_2^o$.
$-FE^o = -2FX_2 - (-FX_1)$.
$-FE^o = -2FX_2 + FX_1$.
Dividing by $-F$,we get $E^o = 2X_2 - X_1$.

(D) The reducing power of a metal is directly proportional to its standard oxidation potential $(E^{\circ}_{OP})$.
Given standard reduction potentials $(E^{\circ}_{RP})$:
$E^{\circ}_{RP}(x) = 0.52 \ V$
$E^{\circ}_{RP}(y) = -3.03 \ V$
$E^{\circ}_{RP}(z) = -1.18 \ V$
Since $E^{\circ}_{OP} = -E^{\circ}_{RP}$,the oxidation potentials are:
$E^{\circ}_{OP}(x) = -0.52 \ V$
$E^{\circ}_{OP}(y) = 3.03 \ V$
$E^{\circ}_{OP}(z) = 1.18 \ V$
Comparing the values,the order of oxidation potential is $y > z > x$.
Therefore,the order of reducing power is $y > z > x$.

(B) substance with a higher reduction potential acts as a stronger oxidizing agent and can oxidize a substance with a lower reduction potential.
Comparing the reduction potentials:
$E^{o}(Cl_2/Cl^-) = 1.36 \ V > E^{o}(I_2/I^-) = 0.54 \ V$. Therefore,$Cl_2$ can oxidize $I^-$ to $I_2$.
$E^{o}(Mn^{3+}/Mn^{2+}) = 1.50 \ V > E^{o}(Fe^{3+}/Fe^{2+}) = 0.77 \ V$. Therefore,$Mn^{3+}$ can oxidize $Fe^{2+}$.
Option $B$ states that the iodide ion $(I^-)$ is oxidised by chlorine $(Cl_2)$,which is correct because $E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode} = 1.36 \ V - 0.54 \ V = 0.82 \ V > 0$.

(NONE) The oxidation of ferrous ion $(Fe^{2+})$ to ferric ion $(Fe^{3+})$ requires an oxidizing agent with a higher reduction potential than the $Fe^{3+}/Fe^{2+}$ couple $(E^\circ = +0.77 \ V)$.
Among the given options,none of the ions ($Li^{\oplus}$,$Mn^{2+}$,$Pb^{2+}$,$Al^{3+}$) are strong oxidizing agents capable of oxidizing $Fe^{2+}$ to $Fe^{3+}$ under standard conditions.
However,in the context of typical chemistry problems,if we consider the standard reduction potentials: $E^\circ (Li^{\oplus}/Li) = -3.04 \ V$,$E^\circ (Mn^{2+}/Mn) = -1.18 \ V$,$E^\circ (Pb^{2+}/Pb) = -0.13 \ V$,and $E^\circ (Al^{3+}/Al) = -1.66 \ V$.
Since all these species are in their stable oxidation states or are strong reducing agents themselves,they cannot act as oxidizing agents for $Fe^{2+}$.
If the question implies which ion can be reduced by $Fe^{2+}$,none of these are suitable. If the question is flawed and expects an answer from a standard set,please note that $Fe^{2+}$ is typically oxidized by $MnO_4^-$ or $Cr_2O_7^{2-}$. Given the options provided,there is no correct answer.

(A) We use the relationship between Gibbs free energy change $(\Delta G^{o})$ for the given half-reactions.
Let the reactions be:
$(1) Cu^{+2} + 2e^{-} \to Cu$,$\Delta G_{1}^{o} = -2F X_{1}$
$(2) Cu^{+} + e^{-} \to Cu$,$\Delta G_{2}^{o} = -1F X_{2}$
$(3) Cu^{+2} + e^{-} \to Cu^{+}$,$\Delta G_{3}^{o} = -1F E^{o}$
From Hess's Law,reaction $(3) = (1) - (2)$:
$\Delta G_{3}^{o} = \Delta G_{1}^{o} - \Delta G_{2}^{o}$
$-1F E^{o} = -2F X_{1} - (-1F X_{2})$
$-E^{o} = -2X_{1} + X_{2}$
$E^{o} = 2X_{1} - X_{2}$

(B) The cell reaction is $2Ce^{4+} + Co \to 2Ce^{3+} + Co^{2+}$.
Anode reaction: $Co \to Co^{2+} + 2e^-$,so $E_{anode}^o = E_{Co^{2+}/Co}^o = -E_{Co/Co^{2+}}^o = -0.28 \, V$.
Cathode reaction: $2Ce^{4+} + 2e^- \to 2Ce^{3+}$,so $E_{cathode}^o = E_{Ce^{4+}/Ce^{3+}}^o = x$.
The standard cell potential is given by $E_{cell}^o = E_{cathode}^o - E_{anode}^o$.
Substituting the values: $1.89 = x - (-0.28)$.
$1.89 = x + 0.28$.
$x = 1.89 - 0.28 = 1.61 \, V$.

(A) The cell reaction is represented as $Zn|Zn^{2+}||CuI|Cu$.
At the anode (oxidation): $Zn_{(s)} \to Zn^{2+}_{(aq)} + 2e^{-}$,$E^o_{ox} = -(-0.76 \ V) = +0.76 \ V$.
At the cathode (reduction): $2CuI_{(s)} + 2e^{-} \to 2Cu_{(s)} + 2I^{-}_{(aq)}$,$E^o_{red} = -0.17 \ V$.
The overall cell reaction is $Zn_{(s)} + 2CuI_{(s)} \to Zn^{2+}_{(aq)} + 2Cu_{(s)} + 2I^{-}_{(aq)}$.
The standard cell potential is $E^o_{cell} = E^o_{cathode} - E^o_{anode} = E^o_{red(cathode)} - E^o_{red(anode)}$.
$E^o_{cell} = -0.17 \ V - (-0.76 \ V) = -0.17 \ V + 0.76 \ V = 0.59 \ V$.

(C) The standard hydrogen electrode $(SHE)$ is used as a reference electrode to measure the potential of other electrodes.
By international convention,the potential of the standard hydrogen electrode is arbitrarily assumed to be $0.00 \ V$ at all temperatures.

(B) The cell reaction is: $Ti^{2+}(aq.) + Co^{2+}(aq.) \rightarrow Ti^{3+}(aq.) + Co(s)$.
$E_{cell}^o = E_{cathode}^o - E_{anode}^o$.
Here,the cathode is the $Co^{2+}|Co$ electrode and the anode is the $Ti^{3+}|Ti^{2+}$ electrode.
$E_{cell}^o = E_{Co^{2+}|Co}^o - E_{Ti^{3+}|Ti^{2+}}^o$.
Given $E_{cell}^o = 0.09 \ V$ and $E_{Co^{2+}|Co}^o = -0.28 \ V$.
$0.09 = -0.28 - E_{Ti^{3+}|Ti^{2+}}^o$.
$E_{Ti^{3+}|Ti^{2+}}^o = -0.28 - 0.09 = -0.37 \ V$.
The electrode potential for the oxidation half-reaction $Ti^{2+} \rightarrow Ti^{3+} + e^-$ is the negative of the reduction potential $E_{Ti^{3+}|Ti^{2+}}^o$.
Therefore,$E_{Ti^{2+}|Ti^{3+}}^o = -(-0.37 \ V) = 0.37 \ V$.

(A) Metals that are less reactive than hydrogen in the electrochemical series cannot displace hydrogen from dilute acids.
$Cu$ (Copper) is placed below hydrogen in the reactivity series,so it does not react with dilute acids to release $H_{2(g)}$.
$Fe$,$Mn$,and $Zn$ are more reactive than hydrogen and will release $H_{2(g)}$ when reacting with dilute acids.

(B) reducing agent is a substance that undergoes oxidation (loses electrons). The strength of a reducing agent is determined by its standard oxidation potential,which is the negative of its standard reduction potential $(E^o_{ox} = -E^o_{red})$.
Comparing the standard reduction potentials $(E^o_{red})$:
$E^o_{K^{+}/K} = -2.93 \ V$
$E^o_{Zn^{2+}/Zn} = -0.76 \ V$
$E^o_{Fe^{2+}/Fe} = -0.44 \ V$
$E^o_{Cu^{2+}/Cu} = 0.34 \ V$
The species with the most negative standard reduction potential is the strongest reducing agent because it has the highest tendency to lose electrons.
Since $E^o_{K^{+}/K} = -2.93 \ V$ is the most negative value,$K_{(s)}$ is the strongest reducing agent.

(A) reducing agent is a substance that undergoes oxidation,and its strength is determined by its standard oxidation potential. The standard oxidation potential is the negative of the standard reduction potential $(E^\circ_{ox} = -E^\circ_{red})$.
$1$. For $Zn_{(s)} \rightarrow Zn^{2+} + 2e^-$,$E^\circ_{ox} = -(-0.762 \ V) = +0.762 \ V$
$2$. For $Cr_{(s)} \rightarrow Cr^{3+} + 3e^-$,$E^\circ_{ox} = -(-0.740 \ V) = +0.740 \ V$
$3$. For $H_{2(g)} \rightarrow 2H^+ + 2e^-$,$E^\circ_{ox} = -(0.00 \ V) = 0.00 \ V$
$4$. For $Fe^{2+} \rightarrow Fe^{3+} + e^-$,$E^\circ_{ox} = -(0.770 \ V) = -0.770 \ V$
Since $Zn_{(s)}$ has the highest standard oxidation potential $(+0.762 \ V)$,it is the strongest reducing agent among the given options.

(B) The standard cell potential is calculated using the formula:
$E_{cell}^o = E_{cathode}^o - E_{anode}^o$
Here,the cathode is the $N^{+}/N$ electrode and the anode is the $M^{+}/M$ electrode.
$E_{cell}^o = E_{N^{+}/N}^o - E_{M^{+}/M}^o$
$E_{cell}^o = 0.25 \ V - 0.52 \ V = -0.27 \ V$
Since $E_{cell}^o < 0$,the cell reaction is non-spontaneous.

(C) The given reactions are oxidation half-reactions. To determine the oxidizing agent,we look at the reduction potentials $(E^o_{red})$.
For the first reaction: $[Fe(CN)_6]^{3-} + e^- \to [Fe(CN)_6]^{4-}$,$E^o_{red} = +0.35 \ V$.
For the second reaction: $Fe^{3+} + e^- \to Fe^{2+}$,$E^o_{red} = +0.77 \ V$.
$A$ higher reduction potential indicates a stronger tendency to get reduced,making the species a stronger oxidizing agent.
Comparing the two,$Fe^{3+}$ has a higher reduction potential $(+0.77 \ V > +0.35 \ V)$.
Therefore,$Fe^{3+}$ is the strongest oxidizing agent. The correct option is $(C)$.

(D) The Assertion is incorrect because copper $(Cu)$ is a noble metal with a positive standard reduction potential $(E^{\circ}_{Cu^{2+}/Cu} = +0.34 \ V)$,meaning it cannot displace hydrogen from $HCl$.
The Reason is correct because hydrogen is indeed placed above copper in the electrochemical reactivity series,which explains why copper cannot reduce $H^+$ ions to $H_2$ gas.

(A) The given cell reaction is: $2 Fe^{3+}_{(aq)} + 2 I^{-}_{(aq)} \rightarrow 2 Fe^{2+}_{(aq)} + I_{2(aq)}$
Here,the number of electrons involved in the reaction is $n = 2$.
The formula for standard Gibbs energy is: $\Delta_r G^{\ominus} = -n F E^{\ominus}_{cell}$
Substituting the values: $\Delta_r G^{\ominus} = -2 \times 96500 \ C \ mol^{-1} \times 0.24 \ V$
$\Delta_r G^{\ominus} = -46320 \ J \ mol^{-1}$
Converting to $kJ \ mol^{-1}$: $\Delta_r G^{\ominus} = -46.32 \ kJ \ mol^{-1}$

(B) The reducing power of a metal is inversely proportional to its Standard Reduction Potential $(SRP)$ value.
Given $SRP$ values are:
$E^{\circ}_{K^{+}/K} = -2.93 \ V$
$E^{\circ}_{Al^{3+}/Al} = -1.66 \ V$
$E^{\circ}_{Cr^{3+}/Cr} = -0.74 \ V$
$E^{\circ}_{Ag^{+}/Ag} = 0.80 \ V$
Since the reducing power follows the order of decreasing $SRP$ values (more negative $SRP$ means stronger reducing agent),the order is:
$K > Al > Cr > Ag$.

(A) The standard Gibbs free energy change $(\Delta G^{\circ})$ is related to the standard electrode potential $(E^{\circ})$ by the equation $\Delta G^{\circ} = -nFE^{\circ}$.
For the reaction $Cu^{2+} + 2e^{-} \longrightarrow Cu$,$\Delta G^{\circ}_{1} = -2F(0.34) = -0.68F$.
For the reaction $Cu^{+} + e^{-} \longrightarrow Cu$,$\Delta G^{\circ}_{2} = -1F(0.522) = -0.522F$.
We want to find $E^{\circ}$ for the reaction $Cu^{2+} + e^{-} \longrightarrow Cu^{+}$.
This reaction can be obtained by subtracting the second reaction from the first: $(Cu^{2+} + 2e^{-}$ $\longrightarrow Cu) - (Cu^{+} + e^{-}$ $\longrightarrow Cu) \implies Cu^{2+} + e^{-}$ $\longrightarrow Cu^{+}$.
Therefore,$\Delta G^{\circ}_{3} = \Delta G^{\circ}_{1} - \Delta G^{\circ}_{2} = -0.68F - (-0.522F) = -0.158F$.
Since $\Delta G^{\circ}_{3} = -nFE^{\circ}_{3}$ where $n=1$,we have $-0.158F = -1F(E^{\circ}_{3})$.
Thus,$E^{\circ}_{3} = 0.158 \ V$.

(N/A) metal with a stronger reducing power displaces a metal with a weaker reducing power from its salt solution.
The order of increasing reducing power (based on standard electrode potentials) for the given metals is $Cu < Fe < Zn < Al < Mg$.
Therefore,a metal higher in this series can displace any metal lower than it from its salt solution.
Thus,the order in which the metals can displace each other is $Mg > Al > Zn > Fe > Cu$.

(A) Reducing strength is inversely proportional to the standard reduction potential $(E^{\ominus})$.
Lower (more negative) $E^{\ominus}$ values indicate stronger reducing agents.
The given $E^{\ominus}$ values are:
$Li^{+}/Li = -3.04 \ V$
$Na^{+}/Na = -2.71 \ V$
$I_{2}/I^{-} = +0.53 \ V$
$Ag^{+}/Ag = +0.79 \ V$
$Cl_{2}/Cl^{-} = +1.36 \ V$
Arranging these in decreasing order of reducing strength (from most negative $E^{\ominus}$ to most positive $E^{\ominus}$):
$Li > Na > I^{-} > Ag > Cl^{-}$

(N/A) The standard electrode potential $(E^{\Theta})$ for $Al^{3+}/Al$ is $-1.66 \ V$,which indicates a high tendency for $Al$ to lose electrons and form $Al^{3+}$ ions in solution.
Conversely,the $E^{\Theta}$ for $Tl^{3+}/Tl$ is $+1.26 \ V$,which is a high positive value. This indicates that $Tl^{3+}$ has a strong tendency to gain electrons and be reduced to $Tl$.
Therefore,$Al^{3+}$ is stable in solution,whereas $Tl^{3+}$ is unstable and acts as a strong oxidizing agent.
Since $Al$ forms $Al^{3+}$ ions more easily than $Tl$ forms $Tl^{3+}$ ions,$Al$ is more electropositive than $Tl$.

(N/A) The formula for standard Gibbs energy is ${\Delta _r}{G^\Theta } = - nFE_{(cell)}^\Theta $.
In this reaction,the number of electrons transferred $(n)$ is $2$.
The Faraday constant $(F)$ is $96487 \, C \, mol^{-1}$ and the standard cell potential $(E_{(cell)}^\Theta )$ is $1.1 \, V$.
Substituting these values into the equation:
${\Delta _r}{G^\Theta } = - 2 \times 96487 \, C \, mol^{-1} \times 1.1 \, V$
${\Delta _r}{G^\Theta } = - 212271 \, J \, mol^{-1}$
Converting to kilojoules:
${\Delta _r}{G^\Theta } = - 212.27 \, kJ \, mol^{-1}$

(N/A) The standard electrode potential of $Mg^{2+} | Mg$ is determined by constructing an electrochemical cell with the standard hydrogen electrode $(SHE)$.
$1$. Set up a cell with a magnesium electrode ($Mg$ rod dipped in $1 \, M \, MgSO_4$ solution) as the anode and a standard hydrogen electrode $(Pt_{(s)}, H_{2_{(g)}} (1 \, bar) | H^{+}_{(aq)} (1 \, M))$ as the cathode.
$2$. The cell representation is: $Mg | Mg^{2+} (aq, 1 \, M) || H^{+} (aq, 1 \, M) | H_2 (g, 1 \, bar), Pt_{(s)}$.
$3$. Measure the electromotive force $(emf)$ of this cell using a voltmeter.
$4$. The standard cell potential is given by $E^{\Theta}_{cell} = E^{\Theta}_{cathode} - E^{\Theta}_{anode}$.
$5$. Since the standard electrode potential of the $SHE$ is defined as $0.00 \, V$,we have $E^{\Theta}_{cell} = 0 - E^{\Theta}_{Mg^{2+}/Mg}$.
$6$. Therefore,$E^{\Theta}_{Mg^{2+}/Mg} = -E^{\Theta}_{cell}$.

(N/A) Substances that are stronger oxidizing agents than ferrous ions can oxidize ferrous ions.
The oxidation reaction is: $Fe^{2+} \to Fe^{3+} + e^{-}; E^{\Theta} = -0.77 \ V$.
This implies that any substance with a standard reduction potential higher than $+0.77 \ V$ can oxidize ferrous ions $(Fe^{2+})$ to ferric ions $(Fe^{3+})$.
Three substances that satisfy this condition are $F_{2}$ $(E^{\Theta} = +2.87 \ V)$,$Cl_{2}$ $(E^{\Theta} = +1.36 \ V)$,and $O_{2}$ $(E^{\Theta} = +1.23 \ V)$.

(A) The reducing power of a metal is inversely proportional to its standard reduction potential. $A$ lower (more negative) reduction potential indicates a stronger reducing agent.
The given standard reduction potentials are:
$K^{+} / K = -2.93 \, V$
$Mg^{2+} / Mg = -2.37 \, V$
$Cr^{3+} / Cr = -0.74 \, V$
$Hg^{2+} / Hg = 0.79 \, V$
$Ag^{+} / Ag = 0.80 \, V$
Arranging these in increasing order of reduction potential:
$K < Mg < Cr < Hg < Ag$
Therefore,the increasing order of reducing power is:
$Ag < Hg < Cr < Mg < K$

(A) The displacement ability of a metal depends on its standard reduction potential $(E^{\circ})$. $A$ metal with a more negative $E^{\circ}$ value acts as a stronger reducing agent and can displace a metal with a more positive $E^{\circ}$ value from its salt solution.
The standard reduction potentials are:
$E^{\circ} Mg^{+2}/Mg = -2.36 \ V$
$E^{\circ} Al^{+3}/Al = -1.66 \ V$
$E^{\circ} Zn^{+2}/Zn = -0.76 \ V$
$E^{\circ} Fe^{+2}/Fe = -0.44 \ V$
$E^{\circ} Cu^{+2}/Cu = +0.34 \ V$
Since the reduction potential increases in the order $Mg < Al < Zn < Fe < Cu$,the reducing power decreases in the same order. Therefore,the order in which they displace each other is $Mg > Al > Zn > Fe > Cu$.

(A) The reducing power of a metal is inversely proportional to its standard reduction potential. $A$ more negative reduction potential indicates a stronger reducing agent.
The given standard reduction potentials are:
$K^{+}/K = -2.93 \, V$
$Mg^{2+}/Mg = -2.37 \, V$
$Cr^{3+}/Cr = -0.74 \, V$
$Hg^{2+}/Hg = 0.79 \, V$
$Ag^{+}/Ag = 0.80 \, V$
Arranging these in increasing order of reduction potential: $K < Mg < Cr < Hg < Ag$.
Since the reducing power increases as the reduction potential becomes more negative,the increasing order of reducing power is: $Ag < Hg < Cr < Mg < K$.

(C) metal with a more negative standard reduction potential is a stronger reducing agent and can displace a metal with a more positive reduction potential from its salt solution.
The given standard reduction potentials are:
$E^o Zn^{2+}/Zn = -0.76 \ V$
$E^o Co^{2+}/Co = -0.28 \ V$
$E^o Cu^{2+}/Cu = +0.34 \ V$
$E^o Ag^+/Ag = +0.80 \ V$
Comparing these values,$Ag^+$ has the highest positive reduction potential $(+0.80 \ V)$,meaning $Ag$ is the weakest reducing agent and can be displaced by all other metals $(Zn, Co, Cu)$ from their respective solutions.
Therefore,the correct answer is $Ag$.

(D) An oxidizing agent is a species that gets reduced,meaning it has a higher reduction potential.
The given reactions are oxidation half-reactions. We must convert them to reduction potentials:
$1. [Fe(CN)_6]^{3-} + e^- \to [Fe(CN)_6]^{4-}, E^o_{red} = +0.35 \ V$
$2. Fe^{3+} + e^- \to Fe^{2+}, E^o_{red} = +0.77 \ V$
Comparing the reduction potentials,$E^o_{red}(Fe^{3+}/Fe^{2+}) = +0.77 \ V$ is greater than $E^o_{red}([Fe(CN)_6]^{3-}/[Fe(CN)_6]^{4-}) = +0.35 \ V$.
Therefore,$Fe^{3+}$ is a stronger oxidizing agent than $[Fe(CN)_6]^{3-}$.

(A) The reducing power is inversely proportional to the standard reduction potential $(E^{\circ}_{red})$. The lower the value of $E^{\circ}_{red}$,the stronger the reducing agent.
Among the given values,$Li$ has the lowest potential $(-3.05 \ V)$,so it is the strongest reducing agent.
The oxidising power is directly proportional to the standard reduction potential $(E^{\circ}_{red})$. The higher the value of $E^{\circ}_{red}$,the stronger the oxidising agent.
Among the given values,$Ag^{+}$ (derived from $Ag$ with $+0.80 \ V$) has the highest potential,so it is the strongest oxidising agent.

(A) Standard electrode potential is defined as the potential of an electrode measured at $298 \ K$ temperature,when the concentration of all species involved in the half-cell reaction is unity $(1 \ M)$.

(A) By international convention,the standard electrode potential of a standard hydrogen electrode $(SHE)$ is arbitrarily assigned a value of $0.00 \ V$ at all temperatures.

(B) The strength of a reducing agent is inversely proportional to the standard reduction potential value.
Lower reduction potential indicates a stronger reducing agent.
Given values:
$A = 0.34 \ V$
$C = -0.46 \ V$
$B = -0.80 \ V$
Comparing the values: $-0.80 < -0.46 < 0.34$.
Therefore,the order of reducing strength is $B > C > A$.