Electrode potential and ECell Questions in English

(A) reaction is spontaneous if the standard cell potential $E^{\circ}_{cell}$ is positive,which corresponds to a negative Gibbs free energy change $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.
For system $(A)$: $Cu(s) + 2Ag^{+}(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$.
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = E^{\circ}_{Ag^{+}/Ag} - E^{\circ}_{Cu^{2+}/Cu} = 0.80 \ V - 0.34 \ V = 0.46 \ V$.
Since $E^{\circ}_{cell} > 0$,the reaction is spontaneous.
For system $(B)$: $Ag(s) + Cu^{2+}(aq) \rightarrow Ag^{+}(aq) + Cu(s)$.
$E^{\circ}_{cell} = E^{\circ}_{Cu^{2+}/Cu} - E^{\circ}_{Ag^{+}/Ag} = 0.34 \ V - 0.80 \ V = -0.46 \ V$.
Since $E^{\circ}_{cell} < 0$,the reaction is non-spontaneous.
Therefore,the copper-silver nitrate interface shows a spontaneous reaction.

(A) The cell reaction is: $Ag(s) + AgCl(s) \rightarrow Ag^{+}(aq) + Cl^{-}(aq) + Ag(s)$
Simplified net cell reaction: $AgCl(s) \rightarrow Ag^{+}(aq) + Cl^{-}(aq)$
Calculate $\Delta G^{\circ}_{reaction}$:
$\Delta G^{\circ}_{reaction} = [\Delta G_f^{\circ}(Ag^{+}) + \Delta G_f^{\circ}(Cl^{-})] - [\Delta G_f^{\circ}(AgCl)]$
$\Delta G^{\circ}_{reaction} = [78 + (-129)] - (-109) \ kJ/mol$
$\Delta G^{\circ}_{reaction} = -51 + 109 = 58 \ kJ/mol = 58000 \ J/mol$
Using the relation $\Delta G^{\circ} = -nFE^{\circ}_{cell}$:
Here,$n = 1$ (as $Ag \rightarrow Ag^{+} + e^{-}$ and $AgCl + e^{-} \rightarrow Ag + Cl^{-}$).
$58000 = -1 \times 96500 \times E^{\circ}_{cell}$
$E^{\circ}_{cell} = -\frac{58000}{96500} \approx -0.60 \ V$

(A) The cell reaction is: $Cu_{(s)} + 2Ag^+_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$.
For the given cell,the cathode is $Ag^+/Ag$ and the anode is $Cu^{2+}/Cu$.
The standard cell potential is calculated as:
$E^0_{cell} = E^0_{cathode} - E^0_{anode}$
$E^0_{cell} = E^0_{Ag^+/Ag} - E^0_{Cu^{2+}/Cu}$
$E^0_{cell} = 0.80 \ V - 0.34 \ V = +0.46 \ V$.

(B) Given standard reduction potentials:
$E^{\circ}_{Zn^{2+}|Zn} = -0.76 \ V$
$E^{\circ}_{Cu^{2+}|Cu} = 0.34 \ V$
$E^{\circ}_{Ag^{+}|Ag} = 0.80 \ V$
For cell $(1)$: $Zn|Zn^{2+}||Cu^{2+}|Cu$
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.34 - (-0.76) = 1.10 \ V$
For cell $(2)$: $Zn|Zn^{2+}||Ag^{+}|Ag$
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.80 - (-0.76) = 1.56 \ V$
For cell $(3)$: $Cu|Cu^{2+}||Ag^{+}|Ag$
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.80 - 0.34 = 0.46 \ V$
Comparing the values: $1.56 \ V (2) > 1.10 \ V (1) > 0.46 \ V (3)$.
Therefore,the correct order is $2 > 1 > 3$.

(B) The given cell reaction is $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$.
Oxidation half-reaction: $Zn \rightarrow Zn^{2+} + 2e^{-}$, $E^{\circ}_{ox} = +0.76 \ V$.
Reduction half-reaction: $Fe^{2+} + 2e^{-} \rightarrow Fe$, $E^{\circ}_{red} = -0.41 \ V$ (since $E^{\circ}_{ox}$ for $Fe$ is $+0.41 \ V$, the reduction potential $E^{\circ}_{red} = -E^{\circ}_{ox}$).
The standard e.m.f. of the cell is:
$E^{\circ}_{cell} = E^{\circ}_{ox} + E^{\circ}_{red}$
$E^{\circ}_{cell} = 0.76 \ V + (-0.41 \ V) = +0.35 \ V$.

(A) The standard $EMF$ of the cell is calculated as: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
Given reduction potentials are $E^{\circ}_{Ag^{+}/Ag} = 0.7995 \ V$ and $E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.7710 \ V$.
Since $E^{\circ}_{Ag^{+}/Ag} > E^{\circ}_{Fe^{3+}/Fe^{2+}}$, the silver electrode acts as the cathode and the iron electrode acts as the anode.
$E^{\circ}_{cell} = 0.7995 \ V - 0.7710 \ V = 0.0285 \ V$.

(A) The standard reduction potential of $Fe^{3+}/Fe^{2+}$ is $+0.77 \ V$.
In a mixture of $H_{2}SO_{4}$ and $H_{3}PO_{4}$, $Fe^{3+}$ ions react with phosphate ions to form a stable complex, $[FeHPO_{4}]^{+}$.
According to the Nernst equation, $E = E^{\circ} - (0.059/n) \log(1/[Fe^{3+}])$.
As the concentration of free $Fe^{3+}$ ions decreases due to complex formation, the reduction potential decreases from $+0.77 \ V$ to $+0.61 \ V$.

(A) In the electrochemical series, metals are arranged in the increasing order of their standard reduction potential.
The standard reduction potential $(E^{\circ})$ of $Li^{+} / Li$ is $-3.05 \ V$.
The standard reduction potential $(E^{\circ})$ of $Cu^{2+} / Cu$ is $+0.34 \ V$.
Since $-3.05 \ V < +0.34 \ V$, $Li$ has a lower standard reduction potential than $Cu$.
Therefore, $Li$ is placed higher in the electrochemical series than $Cu$.

(B) The given reactions are oxidation half-reactions, so the given $E^{\circ}$ values are oxidation potentials $(E^{\circ}_{op})$:
$Zn \rightarrow Zn^{2+} + 2e^-, E^{\circ}_{op} = +0.76 \ V$
$Fe \rightarrow Fe^{2+} + 2e^-, E^{\circ}_{op} = +0.41 \ V$
For the cell reaction $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$, $Zn$ is oxidized (anode) and $Fe^{2+}$ is reduced (cathode).
The cell potential is given by: $E^{\circ}_{cell} = E^{\circ}_{op}(\text{anode}) + E^{\circ}_{rp}(\text{cathode})$
Since $E^{\circ}_{rp}(\text{cathode}) = -E^{\circ}_{op}(\text{cathode})$, we have:
$E^{\circ}_{cell} = E^{\circ}_{op}(Zn) - E^{\circ}_{op}(Fe)$
$E^{\circ}_{cell} = 0.76 \ V - 0.41 \ V = +0.35 \ V$.

(A) The electrode reaction for the metal/metal insoluble salt electrode is: $MX_{(s)} + e^- \rightarrow M_{(s)} + X^-_{(aq)}$.
This can be expressed as the sum of two half-reactions:
$M^+_{(aq)} + e^- \rightarrow M_{(s)}$ $(E^{\circ} = 0.79 \ V)$
$MX_{(s)} \rightleftharpoons M^+_{(aq)} + X^-_{(aq)}$ $(K_{sp} = 10^{-10})$
Using the relation $E^{\circ}_{cell} = E^{\circ}_{M^+/M} + \frac{0.059}{n} \log K_{sp}$:
$E^{\circ}_{X^-/MX_{(s)}/M} = 0.79 + 0.059 \log(10^{-10})$
$E^{\circ} = 0.79 + 0.059 \times (-10)$
$E^{\circ} = 0.79 - 0.59 = 0.20 \ V$
Converting to $mV$: $0.20 \ V = 200 \ mV$.

(B) The standard cell potential, denoted as $E_{cell}^0$, is a constant value for a given electrochemical cell at a specific temperature. It depends only on the standard reduction potentials of the half-cells involved and is independent of the concentrations of the reactants or the time the cell has been operating. Therefore, a plot of $E_{cell}^0$ versus time will be a horizontal line, indicating that the value remains constant over time.

(A) The reducing power of a species is inversely proportional to its standard reduction potential $(E^{\circ})$.
Lower reduction potential indicates a stronger tendency to undergo oxidation, thus acting as a better reducing agent.
The reduction potentials are:
$Al^{3+}/Al = -1.66 \ V$
$Cr^{3+}/Cr = -0.74 \ V$
$Fe^{3+}/Fe^{2+} = +0.77 \ V$
$Co^{3+}/Co^{2+} = +1.81 \ V$
Arranging these in increasing order of reduction potential:
$Al < Cr < Fe^{2+} < Co^{2+}$
Therefore, the order of reducing power (decreasing) is:
$Al > Cr > Fe^{2+} > Co^{2+}$

(C) An extensive property is a property that depends on the amount of matter present in the system. $\Delta G$ (Gibbs free energy) is proportional to the number of moles of reactants, making it an extensive property.
An intensive property is a property that is independent of the amount of matter present. $E_{cell}$ (electromotive force) is a potential difference that does not depend on the size or amount of the cell, making it an intensive property.
Therefore, $\Delta G$ is extensive and $E_{cell}$ is intensive. Thus, option $(C)$ is correct.

(C) The reducing power of a metal is inversely proportional to its standard reduction potential $(E^\circ)$.
$A$ lower (more negative) $E^\circ$ value indicates a stronger reducing agent, while a higher (more positive) $E^\circ$ value indicates a weaker reducing agent.
Comparing the given values:
$E^\circ_{(\text{Li}^+/\text{Li})} = -3.05 \text{ V}$
$E^\circ_{(\text{Mg}^{2+}/\text{Mg})} = -2.36 \text{ V}$
$E^\circ_{(\text{Ag}^+/\text{Ag})} = 0.80 \text{ V}$
$E^\circ_{(\text{Au}^{3+}/\text{Au})} = 1.40 \text{ V}$
Since $\text{Au}^{3+}/\text{Au}$ has the highest positive standard reduction potential $(1.40 \text{ V})$, it is the weakest reducing agent.

(B) Metals that have a negative standard reduction potential relative to hydrogen $(E^\circ < 0 \text{ V})$ can liberate $\text{H}_2$ gas from $\text{HCl}$.
Copper $(Cu)$ has a positive standard reduction potential $(E^\circ_{Cu^{2+}/Cu} \approx +0.34 \text{ V})$.
Since its reduction potential is higher than that of hydrogen $(E^\circ_{H^+/H_2} = 0.00 \text{ V})$, it is less reactive than hydrogen and cannot displace it from acids.

(B) $1$. For the reaction $Fe^{2+} + 2e^- \to Fe$, the standard Gibbs free energy change is $\Delta G^\circ_1 = -n_1 F E^\circ_{Fe^{2+}/Fe} = -2F X$.
$2$. For the reaction $Fe^{3+} + 3e^- \to Fe$, the standard Gibbs free energy change is $\Delta G^\circ_2 = -n_2 F E^\circ_{Fe^{3+}/Fe} = -3F Y$.
$3$. To find the potential for $Fe^{3+} + e^- \to Fe^{2+}$, we subtract the first reaction from the second: $(Fe^{3+} + 3e^- \to Fe) - (Fe^{2+} + 2e^- \to Fe) \implies Fe^{3+} + e^- \to Fe^{2+}$.
$4$. The change in Gibbs free energy for this reaction is $\Delta G^\circ_3 = \Delta G^\circ_2 - \Delta G^\circ_1 = -3FY - (-2FX) = 2FX - 3FY$.
$5$. Since $\Delta G^\circ_3 = -n_3 F E^\circ_{Fe^{3+}/Fe^{2+}}$ where $n_3 = 1$, we have $-1 \cdot F \cdot E^\circ_{Fe^{3+}/Fe^{2+}} = 2FX - 3FY$.
$6$. Therefore, $E^\circ_{Fe^{3+}/Fe^{2+}} = 3Y - 2X$.

(B) To find the standard electrode potential for the reaction $Fe^{3+} + e^- \rightarrow Fe^{2+}$, we use the Gibbs free energy change $(\Delta G^\circ)$.
For reaction $(1): Fe^{3+} + 3e^- \rightarrow Fe$, $\Delta G_1^\circ = -n_1 F E_1^\circ = -3 \times F \times (-0.04) = 0.12 F$.
For reaction $(2): Fe^{2+} + 2e^- \rightarrow Fe$, $\Delta G_2^\circ = -n_2 F E_2^\circ = -2 \times F \times (-0.44) = 0.88 F$.
We want the reaction: $Fe^{3+} + e^- \rightarrow Fe^{2+}$, which is reaction $(1) - (2)$.
Therefore, $\Delta G_3^\circ = \Delta G_1^\circ - \Delta G_2^\circ = 0.12 F - 0.88 F = -0.76 F$.
Since $\Delta G_3^\circ = -n_3 F E_3^\circ$ and $n_3 = 1$, we have $-0.76 F = -1 \times F \times E_3^\circ$.
Thus, $E_3^\circ = +0.77 V$ (approximately $+0.76 V$ based on given values).

(B) The cell reaction is: $2Al(s) + 3Pb^{2+}(aq) \rightarrow 2Al^{3+}(aq) + 3Pb(s)$.
The standard cell potential is given by: $E^0_{cell} = E^0_{cathode} - E^0_{anode}$.
Here, the cathode is $Pb^{2+}/Pb$ and the anode is $Al^{3+}/Al$.
So, $E^0_{cell} = E^0_{Pb^{2+}/Pb} - E^0_{Al^{3+}/Al}$.
Given $E^0_{cell} = 1.5 \text{ V}$ and $E^0_{Pb^{2+}/Pb} = -0.14 \text{ V}$.
Substituting the values: $1.5 \text{ V} = -0.14 \text{ V} - E^0_{Al^{3+}/Al}$.
$E^0_{Al^{3+}/Al} = -0.14 \text{ V} - 1.5 \text{ V} = -1.64 \text{ V}$.

(B) $1$. The standard electrode potential $(E^\circ)$ is defined for the reduction half-reaction.
$2$. The notation $Cu^{+2}/Cu$ represents the reduction of $Cu^{+2}$ ions to metallic $Cu$.
$3$. The corresponding reduction half-reaction is $Cu^{+2} + 2e^- \rightarrow Cu$.
$4$. Therefore, the correct reaction is option $B$.

(C) The cell reaction is: $Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$.
The standard cell potential is given by $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
Since the concentrations are $1M$, the cell potential $E_{cell}$ is equal to the standard cell potential $E^{\circ}_{cell} = 0.463 \text{ V}$.
Here, $Ag$ is the cathode and $Cu$ is the anode.
$E^{\circ}_{cell} = E^{\circ}_{Ag^+/Ag} - E^{\circ}_{Cu^{2+}/Cu}$.
$0.463 \text{ V} = E^{\circ}_{Ag^+/Ag} - 0.337 \text{ V}$.
$E^{\circ}_{Ag^+/Ag} = 0.463 \text{ V} + 0.337 \text{ V} = 0.800 \text{ V}$.

(A) reaction is non-spontaneous if the standard cell potential $E^0_{cell} < 0$.
$E^0_{cell} = E^0_{cathode} - E^0_{anode}$.
For option $(A)$: $2Ag(s) + Fe^{+2}_{(aq)} \rightarrow 2Ag^{+1}_{(aq)} + Fe(s)$.
$E^0_{cathode} = E^0_{(Fe^{+2}|Fe)} = -0.44 \text{ V}$.
$E^0_{anode} = E^0_{(Ag^{+1}|Ag)} = +0.79 \text{ V}$.
$E^0_{cell} = -0.44 - 0.79 = -1.23 \text{ V}$.
Since $E^0_{cell} < 0$, the reaction is non-spontaneous.

(C) The standard $emf$ of a cell is given by $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$.
To maximize $E^\circ_{\text{cell}}$, we must choose the cathode with the highest reduction potential and the anode with the lowest reduction potential.
Given potentials: $E^\circ_A = +2.5 \text{ V}$, $E^\circ_B = +3.0 \text{ V}$, $E^\circ_C = -2.0 \text{ V}$, $E^\circ_D = -1.5 \text{ V}$.
Highest potential (cathode) is $B$ $(+3.0 \text{ V})$.
Lowest potential (anode) is $C$ $(-2.0 \text{ V})$.
$E^\circ_{\text{cell}} = E^\circ_B - E^\circ_C = 3.0 - (-2.0) = 5.0 \text{ V}$.
Thus, $C$ is the anode and $B$ is the cathode.

(D) For a reaction to be spontaneous, the standard cell potential $E^\circ_{\text{cell}}$ must be positive.
$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = E^\circ_{Y^{+2}/Y} - E^\circ_{X^{+2}/X} > 0$.
This implies $E^\circ_{Y^{+2}/Y} > E^\circ_{X^{+2}/X}$.
Given potentials: $E^\circ_{Zn^{+2}/Zn} = -0.76 \text{ V}$, $E^\circ_{Fe^{+2}/Fe} = -0.44 \text{ V}$, $E^\circ_{Ni^{+2}/Ni} = -0.23 \text{ V}$.
Checking option $(D)$: $X = Zn$ $(E^\circ = -0.76 \text{ V})$ and $Y = Ni$ $(E^\circ = -0.23 \text{ V})$.
$E^\circ_{\text{cell}} = (-0.23) - (-0.76) = +0.53 \text{ V}$.
Since $E^\circ_{\text{cell}} > 0$, the reaction is spontaneous.

(B) $1$. In a galvanic cell, the electrode with the higher reduction potential acts as the cathode (positive electrode).
$2$. Given $E^0_{(Cu^{2+}|Cu)} = +0.34 \text{ V}$ and $E^0_{(H^+|H_2)} = 0.00 \text{ V}$.
$3$. Since $+0.34 \text{ V} > 0.00 \text{ V}$, the copper electrode is the cathode.
$4$. Reduction occurs at the cathode: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$.

(A) reaction is spontaneous if $E^0_{cell} > 0$. $E^0_{cell} = E^0_{cathode} - E^0_{anode}$.
For $(A)$: $E^0_{cell} = (-0.403) - (-2.866) = +2.463 \text{ V}$. Since $E^0_{cell} > 0$, it is spontaneous.
For $(B)$: $E^0_{cell} = (-0.136) - (1.08) = -1.216 \text{ V}$. Non-spontaneous.
For $(C)$: $E^0_{cell} = (-0.257) - (0.799) = -1.056 \text{ V}$. Non-spontaneous.
For $(D)$: $E^0_{cell} = (-0.763) - (1.68) = -2.443 \text{ V}$. Non-spontaneous.

(B) The standard cell potential is given by the formula: $E^0_{cell} = E^0_{cathode} - E^0_{anode}$.
Here, the cathode is $Pb^{2+}/Pb$ and the anode is $Al^{3+}/Al$.
Given: $E^0_{cell} = 1.5 \text{ V}$ and $E^0_{cathode} = -0.14 \text{ V}$.
Substituting the values: $1.5 \text{ V} = -0.14 \text{ V} - E^0_{anode}$.
Rearranging the equation: $E^0_{anode} = -0.14 \text{ V} - 1.5 \text{ V}$.
Therefore, $E^0_{Al^{3+}/Al} = -1.64 \text{ V}$.

(B) The standard electrode potential $E^\circ$ is defined for the reduction half-reaction.
For the $Cu^{+2}/Cu$ electrode, the reduction reaction is the gain of electrons by the copper ion to form metallic copper.
The reaction is: $Cu^{+2}(aq) + 2e^- \rightarrow Cu(s)$.
Thus, option $B$ represents the correct reduction reaction.

(C) The standard cell potential $E^0_{cell}$ is given by the difference between the standard reduction potentials of the cathode and the anode:
$E^0_{cell} = E^0_{cathode} - E^0_{anode}$
Here, the cathode is $Ag$ and the anode is $Cu$, so:
$E^0_{cell} = E^0_{Ag^+/Ag} - E^0_{Cu^{2+}/Cu}$
Given $E^0_{cell} = 0.463 \text{ V}$ and $E^0_{Cu^{2+}/Cu} = 0.337 \text{ V}$:
$0.463 \text{ V} = E^0_{Ag^+/Ag} - 0.337 \text{ V}$
$E^0_{Ag^+/Ag} = 0.463 \text{ V} + 0.337 \text{ V} = 0.800 \text{ V}$
Thus, the standard potential of the $Ag$ electrode is $0.8 \text{ V}$.

(A) reaction is non-spontaneous if the standard cell potential $E^0_{cell} < 0$.
For option $A$: $E^0_{cell} = E^0_{cathode} - E^0_{anode} = E^0_{Fe^{+2}/Fe} - E^0_{Ag^{+}/Ag} = -0.44 - 0.79 = -1.23 \text{ V}$. Since $E^0_{cell} < 0$, this reaction is non-spontaneous.
For option $B$: $E^0_{cell} = 0.34 - (-1.66) = 2.00 \text{ V} > 0$ (Spontaneous).
For option $C$: $E^0_{cell} = 0.34 - (-0.44) = 0.78 \text{ V} > 0$ (Spontaneous).
For option $D$: $E^0_{cell} = -0.44 - (-1.66) = 1.22 \text{ V} > 0$ (Spontaneous).

(C) The standard $emf$ of a cell is given by $E^0_{cell} = E^0_{cathode} - E^0_{anode}$.
To maximize $E^0_{cell}$, we must choose the electrode with the highest reduction potential as the cathode and the electrode with the lowest reduction potential as the anode.
Given potentials: $E^0_A = +2.5 \text{ V}$, $E^0_B = +3.0 \text{ V}$, $E^0_C = -2.0 \text{ V}$, $E^0_D = -1.5 \text{ V}$.
Highest potential is $E^0_B = +3.0 \text{ V}$ (Cathode).
Lowest potential is $E^0_C = -2.0 \text{ V}$ (Anode).
$E^0_{cell} = E^0_B - E^0_C = 3.0 \text{ V} - (-2.0 \text{ V}) = 5.0 \text{ V}$.
Thus, the cell is maximum when $C$ is the anode and $B$ is the cathode.

(D) For a redox reaction to be spontaneous, the standard cell potential $E^0_{cell}$ must be positive.
$E^0_{cell} = E^0_{cathode} - E^0_{anode} = E^0_{Y^{+2}/Y} - E^0_{X^{+2}/X} > 0$.
This implies $E^0_Y > E^0_X$.
Given potentials: $E^0_{Zn} = -0.76 \text{ V}$, $E^0_{Fe} = -0.44 \text{ V}$, $E^0_{Ni} = -0.23 \text{ V}$.
Checking option $D$: $X = Zn$ $(E^0 = -0.76 \text{ V})$ and $Y = Ni$ $(E^0 = -0.23 \text{ V})$.
$E^0_{cell} = -0.23 - (-0.76) = +0.53 \text{ V}$.
Since $E^0_{cell} > 0$, the reaction is spontaneous.

(B) $1$. In a galvanic cell, the electrode with the higher reduction potential acts as the cathode (positive electrode).
$2$. The standard reduction potential of the standard hydrogen electrode $(SHE)$ is $E^\circ (H^+ | H_2) = 0.00 \text{ V}$.
$3$. Since $E^\circ (Cu^{2+} | Cu) = +0.34 \text{ V} > 0.00 \text{ V}$, the copper electrode acts as the cathode.
$4$. Reduction occurs at the cathode: $Cu^{2+}_{(aq)} + 2e^- \longrightarrow Cu_{(s)}$.

(A) Step $1$: Identify the oxidation and reduction half-reactions from the overall reaction $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$.
Step $2$: Oxidation: $Mn^{2+} \rightarrow Mn^{3+} + e^-$ $(E^0_{ox} = -1.51 \text{ V})$.
Step $3$: Reduction: $Mn^{2+} + 2e^- \rightarrow Mn$ $(E^0_{red} = -1.18 \text{ V})$.
Step $4$: Calculate $E^0_{cell} = E^0_{red} + E^0_{ox} = -1.18 \text{ V} + (-1.51 \text{ V}) = -2.69 \text{ V}$.
Step $5$: Since $E^0_{cell} < 0$, the reaction is non-spontaneous.