Electrode potential and ECell Questions in English

(A) The cell reaction is: $Fe^{2+} + Zn \to Zn^{2+} + Fe$.
Here,$Zn$ is oxidized to $Zn^{2+}$ (anode) and $Fe^{2+}$ is reduced to $Fe$ (cathode).
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$
$E^{\circ}_{cell} = E^{\circ}_{Fe^{2+}/Fe} - E^{\circ}_{Zn^{2+}/Zn}$
$E^{\circ}_{cell} = (-0.44 \, V) - (-0.76 \, V)$
$E^{\circ}_{cell} = -0.44 + 0.76 = +0.32 \, V$.

(B) The reducing power of a substance is inversely proportional to its standard reduction potential $(E^o)$.
Lower (more negative) $E^o$ values indicate a stronger reducing agent.
Comparing the given values:
$Cr^{3+} + 3e^- \to Cr$; $E^o = -0.74 \ V$
$Zn^{2+} + 2e^- \to Zn$; $E^o = -0.76 \ V$
$H^{+} + e^- \to 1/2 H_2$; $E^o = 0.00 \ V$
$Fe^{3+} + e^- \to Fe^{2+}$; $E^o = +0.77 \ V$
Since $Zn$ has the most negative reduction potential $(E^o = -0.76 \ V)$,it is the strongest reducing agent among the given options.

(A) The reducing power of a metal is inversely proportional to its standard reduction potential.
Lower reduction potential indicates a greater tendency to lose electrons,hence higher reducing power.
Given reduction potentials: $E^{\circ}_{A} = +0.5 \ V$,$E^{\circ}_{B} = -3.0 \ V$,$E^{\circ}_{C} = -1.2 \ V$.
Comparing the values: $-3.0 \ V < -1.2 \ V < +0.5 \ V$.
Therefore,the order of reducing power is $B > C > A$.

(C) The standard reduction potential of $Mg^{2+}/Mg$ is $-2.37\,V$ and that of $Zn^{2+}/Zn$ is $-0.762\,V$.
$A$ metal with a more negative reduction potential is a stronger reducing agent and can displace a metal with a less negative (or more positive) reduction potential from its salt solution.
Since $E^o_{Mg^{2+}/Mg} < E^o_{Zn^{2+}/Zn}$,$Mg$ is a stronger reducing agent than $Zn$.
Therefore,$Zn$ cannot reduce $Mg^{2+}$ to $Mg$. Consequently,no reaction occurs when zinc dust is added to a solution of $MgCl_2$.

(A) The oxidation reaction is $Fe_{(s)} \rightarrow Fe^{2+}_{(aq)} + 2e^-$,with $E^o_{ox} = +0.44 \ V$.
The reduction reaction is $2H^+_{(aq)} + 2e^- \rightarrow H_{2(g)}$,with $E^o_{red} = 0.00 \ V$.
The standard cell potential is $E^o_{cell} = E^o_{ox} + E^o_{red} = 0.44 \ V + 0.00 \ V = +0.44 \ V$.
Since $E^o_{cell} > 0$,the reaction is spontaneous under standard conditions.
Therefore,$Fe_{(s)}$ will be oxidised to $Fe^{2+}$ by $1 \ M$ $HCl$.

(C) The $EMF$ of a cell is calculated as the difference between the reduction potential of the cathode (right electrode) and the reduction potential of the anode (left electrode).
Mathematically, $E_{cell} = E_{cathode} - E_{anode}$.
Since the cathode is on the right and the anode is on the left, the formula is $E = E_{right} - E_{left}$.

(A) The standard electrode potential $(E^\circ)$ values for the given metals are as follows:
$K^+/K = -2.93 \ V$
$Ba^{2+}/Ba = -2.90 \ V$
$Ca^{2+}/Ca = -2.87 \ V$
$Mg^{2+}/Mg = -2.36 \ V$
Since the values are negative,a less negative value indicates a higher (more positive) potential.
Therefore,the order of decreasing electrode potential is: $Mg > Ca > Ba > K$.

(C) The standard electrode potential $(E^\circ)$ values for the given metals are as follows:
$Li^+ + e^- \rightarrow Li$ $(E^\circ = -3.04 \ V)$
$Al^{3+} + 3e^- \rightarrow Al$ $(E^\circ = -1.66 \ V)$
$Cu^{2+} + 2e^- \rightarrow Cu$ $(E^\circ = +0.34 \ V)$
$Au^{3+} + 3e^- \rightarrow Au$ $(E^\circ = +1.50 \ V)$
Comparing these values,$Au$ has the highest positive standard electrode potential among the given options.

(C) The cell reaction is $Mg_{(s)} + Cu^{2+}_{(aq)} \to Cu_{(s)} + Mg^{2+}_{(aq)}$.
Here,$Mg$ is oxidized at the anode and $Cu^{2+}$ is reduced at the cathode.
The standard $EMF$ of the cell is calculated as:
$E_{cell}^o = E_{cathode}^o - E_{anode}^o$
$E_{cell}^o = E^o_{(Cu^{2+}/Cu)} - E^o_{(Mg^{2+}/Mg)}$
$E_{cell}^o = 0.34 \ V - (-2.37 \ V) = +2.71 \ V$.

(D) The $Standard \ Hydrogen \ Electrode$ $(SHE)$ is universally accepted as the primary reference electrode.
By convention,the potential of the $SHE$ is assigned a value of $0.00 \ V$ at all temperatures.
The half-cell reaction is represented as:
$H_2(g) \rightarrow 2H^+(aq) + 2e^-$ (as an anode)
$2H^+(aq) + 2e^- \rightarrow H_2(g)$ (as a cathode)

(A) The reactivity of an element is directly proportional to its oxidation potential.
Standard electrode potential $(E^0)$ values given are reduction potentials.
Oxidation potential = $-E^0$ (reduction potential).
Values of oxidation potentials are:
$A: -(-1.36) = 1.36 \ V$
$B: -(-0.32) = 0.32 \ V$
$C: -(0) = 0 \ V$
$D: -(-1.26) = 1.26 \ V$
$E: -(-0.42) = 0.42 \ V$
Comparing these values: $1.36 (A) > 1.26 (D) > 0.42 (E) > 0.32 (B) > 0 (C)$.
Thus,the reactivity order is $A > D > E > B > C$.

(C) For a galvanic cell,the cathode is the electrode with the higher reduction potential and the anode is the electrode with the lower reduction potential.
Here,$E^o_{Ag^+/Ag} = +0.80 \ V$ (cathode) and $E^o_{Cu^{2+}/Cu} = +0.34 \ V$ (anode).
The $emf$ of the cell is calculated as:
$E^o_{cell} = E^o_{cathode} - E^o_{anode}$
$E^o_{cell} = 0.80 \ V - 0.34 \ V = +0.46 \ V$.

(A) The cell reaction is represented as: $\frac{1}{2} H_2 | H^{+} || Ag^{+} | Ag$.
The standard cell potential is calculated using the formula: $E_{Cell}^{0} = E_{Cathode}^{0} - E_{Anode}^{0}$.
Here,the cathode is the $Ag^{+}/Ag$ electrode $(E^{0} = 0.80 \ V)$ and the anode is the standard hydrogen electrode $(E^{0} = 0.00 \ V)$.
Therefore,$E_{Cell}^{0} = 0.80 \ V - 0.00 \ V = 0.80 \ V$.

(A) In a galvanic cell,the electrode with the higher reduction potential acts as the cathode,and the electrode with the lower reduction potential acts as the anode.
Given $E_A = +2.23 \ V$ and $E_B = -1.43 \ V$.
Since $E_A > E_B$,$A$ acts as the cathode and $B$ acts as the anode.
The standard cell potential is calculated as:
$E^o_{cell} = E^o_{cathode} - E^o_{anode}$
$E^o_{cell} = E_A - E_B$
$E^o_{cell} = 2.23 \ V - (-1.43 \ V)$
$E^o_{cell} = 2.23 + 1.43 = 3.66 \ V$.

(B) Since $E^o_{Cu^{2+}/Cu} > E^o_{Mg^{2+}/Mg}$,the copper electrode acts as the cathode and the magnesium electrode acts as the anode.
The standard cell potential is calculated as:
$E^o_{cell} = E^o_{cathode} - E^o_{anode}$
$E^o_{cell} = E^o_{Cu^{2+}/Cu} - E^o_{Mg^{2+}/Mg}$
$E^o_{cell} = (+ 0.34 \ V) - (- 2.37 \ V)$
$E^o_{cell} = + 2.71 \ V$.

(B) The cell reaction is $2Ce^{4+} + Co \to 2Ce^{3+} + Co^{2+}$.
In this reaction,$Co$ is oxidized to $Co^{2+}$ (anode) and $Ce^{4+}$ is reduced to $Ce^{3+}$ (cathode).
$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$
$E^\circ_{cell} = E^\circ_{Ce^{4+}/Ce^{3+}} - E^\circ_{Co^{2+}/Co}$
Given $E^\circ_{cell} = 1.89 \ V$ and $E^\circ_{Co^{2+}/Co} = -0.28 \ V$.
$1.89 = E^\circ_{Ce^{4+}/Ce^{3+}} - (-0.28)$
$E^\circ_{Ce^{4+}/Ce^{3+}} = 1.89 - 0.28 = 1.61 \ V$.
The closest option provided is $1.64 \ V$.

(B) The relationship between Gibbs free energy change and cell potential is given by the equation: $\Delta G = -nFE^\circ$.
Here,$\Delta G = -21.20 \ kJ = -21200 \ J$.
The number of electrons transferred $(n)$ is $1$.
The Faraday constant $(F)$ is approximately $96500 \ C \ mol^{-1}$.
Rearranging the formula to solve for $E^\circ$: $E^\circ = -\frac{\Delta G}{nF}$.
Substituting the values: $E^\circ = -\frac{-21200 \ J}{1 \times 96500 \ C \ mol^{-1}} = \frac{21200}{96500} \ V \approx 0.21968 \ V$.
Rounding to two decimal places,we get $E^\circ \approx 0.220 \ V$.

(D) $Al$ displaces $H$ from $HCl$ but silver cannot; this means $Al$ is situated above $Ag$ in the $ECS$,hence $Al$ acts as the anode and $Ag$ acts as the cathode.
$E_{Cell}^0 = E_{Cathode}^0 - E_{Anode}^0$
$E_{Cell}^0 = E_{Ag^{+}/Ag}^0 - E_{Al^{3+}/Al}^0$
$2.46 \ V = 0.80 \ V - E_{Al^{3+}/Al}^0$
$E_{Al^{3+}/Al}^0 = 0.80 \ V - 2.46 \ V = -1.66 \ V$.

(A) The given reaction is: $Sn_{(s)} + 2Fe^{3+}_{(aq)} \to 2Fe^{2+}_{(aq)} + Sn^{2+}_{(aq)}$
Here,$Sn$ is oxidized to $Sn^{2+}$ (anode) and $Fe^{3+}$ is reduced to $Fe^{2+}$ (cathode).
$E^0_{cell} = E^0_{cathode} - E^0_{anode}$
$E^0_{cell} = E^0_{Fe^{3+}/Fe^{2+}} - E^0_{Sn^{2+}/Sn}$
$E^0_{cell} = (+0.77 \ V) - (-0.14 \ V)$
$E^0_{cell} = 0.77 \ V + 0.14 \ V = 0.91 \ V$

(D) The given reaction is: $Cr_2O_7^{2-} + I^{-} \to I_2 + Cr^{3+}$
In this reaction,$I^{-}$ is oxidized to $I_2$ (anode) and $Cr_2O_7^{2-}$ is reduced to $Cr^{3+}$ (cathode).
The standard cell potential is given by: $E^0_{cell} = E^0_{cathode} - E^0_{anode}$
Substituting the values: $0.79 \ V = 1.33 \ V - E^0_{I_2}$
Rearranging for $E^0_{I_2}$: $E^0_{I_2} = 1.33 \ V - 0.79 \ V = 0.54 \ V$.
Therefore,the correct option is $(D)$.

(B) The standard electrode potential is a measure of the potential difference between an electrode and a standard hydrogen electrode $(SHE)$ under standard conditions ($1 \ M$ concentration,$298 \ K$).
This potential difference is measured using a high-resistance voltmeter or a potentiometer to ensure that no current flows through the circuit,which would otherwise cause a voltage drop due to internal resistance.

(B) For a galvanic cell,the cell potential is given by $E_{cell} = E_{cathode} - E_{anode}$.
Since $Al$ is more reactive than $Cu$,$Al$ acts as the anode and $Cu$ acts as the cathode.
Given $E_{cell} = 2.0 \ V$ and $E_{Cu^{2+}/Cu} = +0.34 \ V$.
Substituting the values: $2.0 \ V = 0.34 \ V - E_{Al^{3+}/Al}$.
Therefore,$E_{Al^{3+}/Al} = 0.34 \ V - 2.0 \ V = -1.66 \ V$.

(A) The standard Gibbs free energy change $(\Delta G^o)$ is related to the standard cell potential $(E_{cell}^o)$ by the equation:
$\Delta G^o = -nFE_{cell}^o$
where $n$ is the number of moles of electrons transferred,$F$ is Faraday's constant,and $E_{cell}^o$ is the standard cell potential.
Therefore,option $(A)$ is correct.

(A) redox reaction is spontaneous if the Gibbs free energy change $(\Delta G)$ is negative.
The relationship between Gibbs free energy and cell e.m.f. $(E^o)$ is given by the equation: $\Delta G^o = -nFE^o$.
Here,$n$ is the number of moles of electrons transferred,$F$ is Faraday's constant,and $E^o$ is the standard cell potential.
For $\Delta G^o$ to be negative,$E^o$ must be positive.
Therefore,for a cell reaction to be feasible (spontaneous),the e.m.f. must be positive.

(B) The given reactions are oxidation half-reactions. The standard reduction potentials $(E^o_{red})$ are the negative of the standard oxidation potentials $(E^o_{ox})$.
$E^o_{red}(Zn^{2+}/Zn) = -0.76 \ V$
$E^o_{red}(Fe^{2+}/Fe) = -0.41 \ V$
In the cell reaction $Fe^{2+} + Zn \to Zn^{2+} + Fe$,$Zn$ is oxidized (anode) and $Fe^{2+}$ is reduced (cathode).
$E^o_{cell} = E^o_{cathode} - E^o_{anode}$
$E^o_{cell} = E^o_{red}(Fe^{2+}/Fe) - E^o_{red}(Zn^{2+}/Zn)$
$E^o_{cell} = -0.41 \ V - (-0.76 \ V)$
$E^o_{cell} = -0.41 \ V + 0.76 \ V = +0.35 \ V$.

(A) The reduction potential represents the tendency to gain electrons. Given the order $Z > Y > X$,$Z$ has the highest tendency to be reduced,followed by $Y$,and then $X$.
An oxidizing agent is a substance that gets reduced itself and oxidizes others. $A$ species with a higher reduction potential can oxidize a species with a lower reduction potential.
Since the reduction potential of $Y > X$,$Y$ can oxidize $X$.
Since the reduction potential of $Z > Y$,$Y$ cannot oxidize $Z$ (in fact,$Z$ would oxidize $Y$).
Therefore,$Y$ will oxidize $X$ but not $Z$.

(B) The standard oxidation potential of $SHE$ (Standard Hydrogen Electrode) is defined as $0.00 \ V$.
Elements that have a standard reduction potential greater than $0.00 \ V$ will have a standard oxidation potential less than $0.00 \ V$.
In the electrochemical series,elements placed below $H$ have positive standard reduction potentials.
$Cu$ is placed below $H$ in the electrochemical series,meaning its standard reduction potential is positive $(+0.34 \ V)$,and thus its standard oxidation potential is negative $(-0.34 \ V)$.
Therefore,$Cu$ has a standard oxidation potential less than $SHE$.

(B) The reducing power depends on the standard oxidation potential. $A$ species with a more positive standard reduction potential is a weaker reducing agent.
In the electrochemical series,the standard reduction potentials $(E^\circ)$ are: $Li^+/Li = -3.04 \ V$,$Zn^{2+}/Zn = -0.76 \ V$,$2H^+/H_2 = 0.00 \ V$,and $Cu^{2+}/Cu = +0.34 \ V$.
Since $Cu$ has the most positive reduction potential,it has the least tendency to lose electrons,making it the weakest reducing agent among the given options.

(D) The standard Gibbs free energy change is given by the formula: $\Delta G^o = -nFE_{cell}^o$.
Here,$n = 2$ (number of electrons transferred),$F = 96500 \ C \ mol^{-1}$ (Faraday's constant),and $E_{cell}^o = 1.20 \ V$.
Substituting the values: $\Delta G^o = -2 \times 96500 \times 1.20 = -231600 \ J \ mol^{-1}$.
Converting to $kJ$: $\Delta G^o = -231.6 \ kJ \ mol^{-1}$.

(C) In the electrochemical series,metals with a negative standard reduction potential can displace $H_2$ from dilute acids.
$Cu$ has a positive standard reduction potential $(E^o = +0.34 \ V)$,which means it is placed below hydrogen in the electrochemical series.
Therefore,$Cu$ cannot displace $H_2$ from dilute $H_2SO_4$ solution.

(A) The reduction potential indicates the tendency of a species to get reduced. $A$ higher reduction potential means a stronger oxidizing agent.
Given the order of reduction potential: $x > y > z$.
Since $z$ has the lowest reduction potential,it acts as the strongest reducing agent and is easily oxidized.
Species with higher reduction potential can oxidize species with lower reduction potential.
Therefore,$y$ (having higher reduction potential than $z$) can oxidize $z$,but $y$ cannot oxidize $x$ (as $x$ has a higher reduction potential than $y$).

(C) The ability of a metal to reduce $H_2O$ depends on its position in the electrochemical series.
Metals that are more reactive than hydrogen (those with negative standard reduction potentials) can displace hydrogen from $H_2O$.
$Ca$,$Fe$,and $Li$ have negative standard reduction potentials and can reduce $H_2O$.
$Cu$ has a positive standard reduction potential $(E^o = +0.34 \ V)$,meaning it is less reactive than hydrogen and cannot reduce $H_2O$ to $H_2$ gas.

(A) $Cu$ is less reactive than hydrogen in the electrochemical series. Therefore,it cannot displace hydrogen from an acid.

(D) The strength of an oxidizing agent is directly proportional to its standard reduction potential.
$A$ higher reduction potential indicates a greater tendency to gain electrons,thus acting as a stronger oxidizing agent.
Comparing the given values:
$Li^+/Li = -3.05 \, V$
$Ba^{2+}/Ba = -2.73 \, V$
$Na^+/Na = -2.71 \, V$
$Mg^{2+}/Mg = -2.36 \, V$
Among the given options,$Mg^{2+}$ has the highest (least negative) reduction potential.
Therefore,$Mg^{2+}$ is the strongest oxidizing agent.

(D) The standard cell potential is given by the formula: ${E^0}_{cell} = {E^0}_{cathode} - {E^0}_{anode}$.
Since $Cu$ has a higher reduction potential than $Mg$,$Cu$ acts as the cathode and $Mg$ acts as the anode.
Thus,${E^0}_{cell} = {E^0}_{Cu^{2+}|Cu} - {E^0}_{Mg^{2+}|Mg}$.
Substituting the given values: $2.7 = 0.34 - {E^0}_{Mg^{2+}|Mg}$.
Rearranging the equation: ${E^0}_{Mg^{2+}|Mg} = 0.34 - 2.7$.
Therefore,${E^0}_{Mg^{2+}|Mg} = -2.36 \ V$.

(A) The reducing power of an element is inversely proportional to its standard reduction potential $(E^{0}_{Red})$.
Lower $E^{0}_{Red}$ value indicates a stronger reducing agent.
Given values are: $E^{0}_{A} = +0.5 \ V$,$E^{0}_{B} = -3.0 \ V$,$E^{0}_{C} = -1.2 \ V$.
Comparing the values: $-3.0 < -1.2 < +0.5$.
Therefore,the order of reducing power is $B > C > A$.

(C) The cell reaction involves the oxidation of $Ni$ to $Ni^{2+}$ (anode) and the reduction of $Cu^{2+}$ to $Cu$ (cathode).
$E^o_{cell} = E^o_{cathode} - E^o_{anode}$
Given $E^o_{Cu^{2+}/Cu} = +0.34\,V$ and $E^o_{Ni^{2+}/Ni} = -0.25\,V$.
$E^o_{cell} = (+0.34\,V) - (-0.25\,V) = 0.59\,V$.

(C) The cell reaction is: $2Fe^{3+} + 2I^- \to 2Fe^{2+} + I_2$
Reduction half-reaction: $2Fe^{3+} + 2e^- \to 2Fe^{2+}$; $E^o_{red} = 0.771 \, V$
Oxidation half-reaction: $2I^- \to I_2 + 2e^-$; $E^o_{ox} = -E^o_{red} = -0.536 \, V$
$E^o_{cell} = E^o_{red} + E^o_{ox} = 0.771 \, V + (-0.536 \, V) = 0.235 \, V$

(C) The ability of an element to displace others from their salt solutions depends on its reducing power.
An element with a lower $E^{0}_{Red}$ value has a higher $E^{0}_{Oxi}$ value,making it a stronger reducing agent.
Comparing the given values: $E^{0}_{Red}(A) = 0.8 \, V$,$E^{0}_{Red}(B) = 0.79 \, V$,$E^{0}_{Red}(C) = 0.34 \, V$,and $E^{0}_{Red}(D) = -2.37 \, V$.
Since $D$ has the lowest $E^{0}_{Red}$ value $(-2.37 \, V)$,it has the highest oxidation potential.
Therefore,$D$ acts as the strongest reducing agent and can displace $A, B,$ and $C$ from their respective salt solutions.

(D) The reducing power of a substance is inversely proportional to its standard reduction potential $(E^{0}_{Red})$.
Lower (more negative) $E^{0}_{Red}$ values indicate a greater tendency to lose electrons,thereby acting as a stronger reducing agent.
Comparing the given values: $Li (-3.05 \ V) < Zn (-0.76 \ V) < H_2 (0.00 \ V) < Ag (0.80 \ V)$.
Since $Li$ has the most negative reduction potential,it has the highest reducing power.

(C) The standard cell potential $E^o_{cell}$ is calculated using the formula: $E^o_{cell} = E^o_{cathode} - E^o_{anode}$.
In the given cell $Zn_{(s)} | Zn^{2+}_{(aq)} || Ag^{+}_{(aq)} | Ag_{(s)}$,the oxidation occurs at the $Zn$ electrode (anode) and reduction occurs at the $Ag$ electrode (cathode).
Here,$E^o_{cathode} = E^o_{Ag^{+}/Ag} = +0.80 \ V$ and $E^o_{anode} = E^o_{Zn^{2+}/Zn} = -0.76 \ V$.
Substituting the values: $E^o_{cell} = 0.80 \ V - (-0.76 \ V) = 0.80 \ V + 0.76 \ V = 1.56 \ V$.

(C) For $Fe^{3+} + 3e^{-} \rightarrow Fe$,$\Delta G_1 = -nFE^0 = -3 \times F \times (-0.036) = 0.108F$.
For $Fe^{2+} + 2e^{-} \rightarrow Fe$,$\Delta G_2 = -nFE^0 = -2 \times F \times (-0.439) = 0.878F$.
We need the reaction $Fe^{3+} + e^{-} \rightarrow Fe^{2+}$.
This can be obtained by: $(Fe^{3+} + 3e^{-}$ $\rightarrow Fe) - (Fe^{2+} + 2e^{-}$ $\rightarrow Fe)$.
Therefore,$\Delta G_3 = \Delta G_1 - \Delta G_2 = 0.108F - 0.878F = -0.770F$.
Since $\Delta G_3 = -nFE^0_{cell}$ and $n=1$,we have $-1 \times F \times E^0_{cell} = -0.770F$.
Thus,$E^0_{cell} = 0.770 \ V$.

(A) The cell potential $E^o_{cell}$ is calculated as $E^o_{cell} = E^o_{cathode} - E^o_{anode}$.
To obtain the maximum potential,we must choose the half-cell with the highest reduction potential as the cathode and the half-cell with the lowest reduction potential as the anode.
Comparing the given values:
$E^o_{(i)} = -0.24 \ V$
$E^o_{(ii)} = +1.25 \ V$ (Highest)
$E^o_{(iii)} = -1.25 \ V$ (Lowest)
$E^o_{(iv)} = +0.68 \ V$
Thus,the cathode is $(ii)$ and the anode is $(iii)$.
$E^o_{cell} = 1.25 - (-1.25) = 2.50 \ V$.

(D) The strength of an oxidizing agent is directly proportional to its standard reduction potential $(E^o_{red})$.
Greater the value of $E^o_{red}$,stronger is the oxidizing agent.
The given reduction potentials are:
$E^o(Li^{+}/Li) = -3.05 \ V$
$E^o(Ba^{2+}/Ba) = -2.73 \ V$
$E^o(Na^{+}/Na) = -2.71 \ V$
$E^o(Mg^{2+}/Mg) = -2.37 \ V$
Comparing these values,$-2.37 \ V$ is the highest value.
Therefore,$Mg^{2+}$ is the strongest oxidizing agent among the given species.