Electrode potential and ECell Questions in English

(D) For a spontaneous process,the Gibbs free energy change $(\Delta G)$ must be negative $(\Delta G < 0)$.
For an electrochemical cell,the relationship between Gibbs free energy and cell potential $(E_{cell})$ is given by $\Delta G = -nFE_{cell}$.
For $\Delta G$ to be negative,$E_{cell}$ must be positive $(E_{cell} > 0)$.
Therefore,the correct condition for the spontaneity of a cell is $\Delta G = -ve$ and $E_{cell} = +ve$.

(D) For a spontaneous reaction,the Gibbs free energy change $\Delta G$ must be negative.
Given the relation $\Delta G = -nFE$,where $n$ is the number of moles of electrons transferred,$F$ is the Faraday constant,and $E$ is the cell potential.
For $\Delta G$ to be negative,the value of $E$ must be positive,because $-nFE$ will result in a negative value when $E > 0$.
Therefore,the cell reaction is spontaneous if $E$ is positive.

(C) The strength of an oxidising agent is directly proportional to its standard reduction potential $(E^o)$.
Comparing the given values:
$MnO_4^-/Mn^{2+} (E^o = +1.52 \ V)$
$BrO_3^-/Br_2 (E^o = +1.50 \ V)$
$Cr_2O_7^{2-}/Cr^{3+} (E^o = +1.33 \ V)$
$Fe^{3+}/Fe^{2+} (E^o = +0.76 \ V)$
Since $MnO_4^-/Mn^{2+}$ has the highest reduction potential,it is the strongest oxidising agent.

(B) Fluorine has the highest $E^{o}_{red}$ (equal to $+2.9 \, V$) due to which it can easily accept an electron,and hence it is the best oxidising agent.

(D) Metals that are placed below hydrogen in the electrochemical series cannot displace hydrogen from dilute acids.
$Au$ (Gold) is a noble metal and has a positive standard reduction potential,meaning it is less reactive than hydrogen.
Therefore,$Au$ does not react with dilute $HCl$ to liberate $H_2$ gas.

(B) The standard reduction potentials $(E^o)$ for the given metals are:
$Li^+ + e^- \to Li, E^o = -3.05 \ V$
$K^+ + e^- \to K, E^o = -2.93 \ V$
$Ca^{2+} + 2e^- \to Ca, E^o = -2.87 \ V$
Comparing these values,the increasing order is $-3.05 \ V < -2.93 \ V < -2.87 \ V$.
Therefore,the order is $Li < K < Ca$.

(D) The reduction potential $E^o_{A/A^-}$ is a measure of the tendency of the species $A$ to gain an electron and be reduced to $A^-$.
Since the reduction potential is a large negative value,the tendency for $A$ to be reduced is very low.
Conversely,this implies that the reverse reaction,$A^- \to A + e^-$,which is an oxidation reaction,has a large positive potential.
Therefore,$A^-$ is a strong reducing agent and is readily oxidized to $A$.

(C)
The magnitude of the electrode potential of a metal is a measure of its relative tendency to lose or gain electrons.
Specifically,the tendency of an electrode to lose electrons is defined as its oxidation potential.
$M \to M^{n+} + ne^-$ (Oxidation potential)
$M^{n+} + ne^- \to M$ (Reduction potential)

(A) The standard cell potential is calculated using the formula:
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$
Here,the $Ag^{+}/Ag$ electrode acts as the cathode and the $Zn^{2+}/Zn$ electrode acts as the anode.
Substituting the given values:
$E^{\circ}_{cell} = 0.799 \ V - (-0.763 \ V)$
$E^{\circ}_{cell} = 0.799 + 0.763 = 1.562 \ V$

(A) The strength of a reducing agent is inversely proportional to its standard reduction potential $(E^\circ)$.
Comparing the given values:
$E^\circ (Zn^{2+}/Zn) = -0.762 \ V$
$E^\circ (Cr^{3+}/Cr) = -0.740 \ V$
$E^\circ (H^+/H_2) = 0.00 \ V$
$E^\circ (Fe^{3+}/Fe^{2+}) = 0.770 \ V$
Since $Zn_{(s)}$ has the most negative standard reduction potential $(-0.762 \ V)$,it has the highest tendency to lose electrons and act as the strongest reducing agent.

(B) The reaction occurring is $Zn(s) + CuSO_4(aq) \rightarrow ZnSO_4(aq) + Cu(s)$.
This is a displacement reaction where $Zn$ displaces $Cu$ from its salt solution.
This occurs because the standard reduction potential of $Zn$ $(-0.76 \ V)$ is less than that of $Cu$ $(+0.34 \ V)$.
Therefore,$Zn$ acts as a stronger reducing agent and gets oxidized while $Cu^{2+}$ ions get reduced to $Cu$ metal.

(C) An element can displace another element from its compound if it has a lower (more negative) standard reduction potential.
The standard reduction potentials are:
$E^o_A = -0.250 \ V$
$E^o_B = -0.136 \ V$
$E^o_C = -0.126 \ V$
$E^o_D = -0.402 \ V$
Comparing these values,$D$ has the lowest reduction potential $(-0.402 \ V < -0.250 \ V)$.
Therefore,$D$ is a stronger reducing agent and will displace $A$ from its compounds.

(A) For a reaction to be spontaneous,the standard cell potential $E^o_{cell}$ must be positive.
$E^o_{cell} = E^o_{ox}(\text{anode}) + E^o_{red}(\text{cathode})$
Alternatively,$E^o_{cell} = E^o_{ox}(\text{anode}) - E^o_{ox}(\text{cathode})$.
For the reaction $Zn_{(s)} + 2Ag^{+}_{(aq)} \to Zn^{2+}_{(aq)} + 2Ag_{(s)}$:
$Zn$ is oxidized $(E^o_{ox} = 0.76 \ V)$ and $Ag^+$ is reduced $(E^o_{red} = -E^o_{ox} = 0.80 \ V)$.
$E^o_{cell} = 0.76 \ V + 0.80 \ V = 1.56 \ V$.
Since $E^o_{cell} > 0$,the reaction is spontaneous. In this reaction,$Zn$ acts as the anode and $Ag$ acts as the cathode.

(C) The standard electrode potential of the Normal Hydrogen Electrode $(NHE)$ is defined as $0.00\,V$ at all temperatures,including $298\,K$.

(D) 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: $E^{\circ}(M_1) = -0.34 \ V$,$E^{\circ}(M_2) = -3.05 \ V$,$E^{\circ}(M_3) = -1.66 \ V$.
Comparing the values: $-3.05 \ V < -1.66 \ V < -0.34 \ V$.
Therefore,the order of reducing power is $M_2 > M_3 > M_1$.

(A) The reactivity of a metal is directly related to its tendency to lose electrons,which is measured by its standard oxidation potential.
Since the standard reduction potential is given,the metal with the most negative standard reduction potential has the highest standard oxidation potential.
Comparing the values: $A = -3.05 \ V$,$B = -1.66 \ V$,$C = -0.40 \ V$,and $D = 0.80 \ V$.
Metal $A$ has the most negative reduction potential $(-3.05 \ V)$,indicating it is the most easily oxidized and therefore the most reactive.

(A) The ability of a metal to displace $H_2$ from acids or $H_2O$ depends on its standard reduction potential $(E^{\circ}_{red})$.
Metals with a negative reduction potential (placed above $H_2$ in the Electrochemical Series) can displace $H_2$.
$Hg$ has a positive standard reduction potential $(E^{\circ}_{Hg^{2+}/Hg} = +0.85 \ V)$,which is greater than that of $H^+/H_2$ $(0.00 \ V)$.
Therefore,$Hg$ cannot displace hydrogen from acids or water.
Thus,the correct option is $(A)$.

(A) The standard electrode potential of $A$ $(E^o_{A^{2+}/A} = -0.76 \ V)$ is lower than that of $B$ $(E^o_{B^{2+}/B} = +0.34 \ V)$.
This indicates that metal $A$ is a stronger reducing agent and has a greater tendency to undergo oxidation compared to metal $B$.
Therefore,the spontaneous reaction is $A(s) + B^{2+}(aq) \to A^{2+}(aq) + B(s)$.
As a result,metal $A$ will dissolve into the solution as $A^{2+}$ ions,and metal $B$ will be deposited on the surface of the rod $A$.

(B) The standard reduction potential $E^{o}_{Zn^{2+}/Zn}$ is $-0.76 \ V$.
Since the value is negative,$Zn$ has a greater tendency to lose electrons (undergo oxidation) compared to hydrogen.
Therefore,$Zn$ acts as a strong reducing agent.

(A) The standard hydrogen electrode $(SHE)$ is used as a reference electrode.
By convention,the standard electrode potential of the hydrogen electrode reaction,$2H^{+} \, (aq) + 2e^- \to H_2 \, (g)$,is defined as $0.00 \ V$ at all temperatures.

(B) The standard electrode potentials $(E^\circ)$ for the given metals are:
$Li^+ + e^- \rightarrow Li$ $(E^\circ = -3.04 \ V)$
$K^+ + e^- \rightarrow K$ $(E^\circ = -2.93 \ V)$
$Ca^{2+} + 2e^- \rightarrow Ca$ $(E^\circ = -2.87 \ V)$
Comparing these values,the order of decreasing standard electrode potential is $Ca > K > Li$.

(C) The cell reaction is: $3Ni(s) + 2Au^{3+}(aq) \rightarrow 3Ni^{2+}(aq) + 2Au(s)$.
For a standard cell,the $EMF$ is calculated as: $E^o_{cell} = E^o_{cathode} - E^o_{anode}$.
Here,the cathode is $Au^{3+}/Au$ and the anode is $Ni^{2+}/Ni$.
$E^o_{cell} = E^o_{Au^{3+}/Au} - E^o_{Ni^{2+}/Ni} = 1.50 \ V - (-0.25 \ V) = 1.75 \ V$.

(A) For a spontaneous electrochemical reaction to occur in a cell,the Gibbs free energy change $(\Delta G)$ must be negative.
The relationship between Gibbs free energy and electromotive force ($EMF$ or $E_{cell}$) is given by the equation: $\Delta G = -nFE_{cell}$.
Since $\Delta G < 0$ and $n, F$ are positive constants,$E_{cell}$ must be positive for the reaction to be spontaneous.
Therefore,when oxidation and reduction take place in a cell,its electromotive force is positive.

(D) The given cell is $A | A^{+}(a = 1) || B^{+}(a = 1) | B$.
In this cell,$A$ acts as the anode and $B$ acts as the cathode.
$EMF = E_{cathode} - E_{anode}$
Given $E^{\circ}_{B^{+}/B} = 0.75 \ V$ and $E^{\circ}_{A^{+}/A} = 0.5 \ V$.
$EMF = 0.75 \ V - 0.5 \ V = 0.25 \ V$.

(D) The reducing capacity is inversely proportional to the standard reduction potential.
The standard reduction potential for $Li^{+}/Li$ is $-3.05 \ V$,which is the most negative value among the given options.
$A$ more negative reduction potential indicates a greater tendency to undergo oxidation (loss of electrons).
Therefore,$Li$ is the strongest reducing agent among the given substances.

(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 $E^o_{Fe^{2+}/Fe} = -0.44 \ V$.
$A$ metal with a lower (more negative) reduction potential can reduce the ion of a metal with a higher (less negative) reduction potential.
Since $E^o_{Zn^{2+}/Zn} (-0.76 \ V) < E^o_{Fe^{2+}/Fe} (-0.44 \ V)$,$Zn$ can reduce $Fe^{2+}$ to $Fe$.
$Zn$ cannot reduce $Mg^{2+}$ because $E^o_{Zn^{2+}/Zn} > E^o_{Mg^{2+}/Mg}$.
Similarly,$Mg$ and $Zn$ act as reducing agents,not oxidizing agents for $Fe$.
Therefore,the correct statement is $A$.

(D) The cell reaction is $Fe^{2+} + Sn \to Fe + Sn^{2+}$.
Here,$Fe^{2+}$ is reduced to $Fe$ (cathode) and $Sn$ is oxidized to $Sn^{2+}$ (anode).
The standard cell potential is calculated as $E^o_{cell} = E^o_{cathode} - E^o_{anode}$.
Given $E^o_{Fe^{2+}/Fe} = -0.44 \ V$ and $E^o_{Sn^{2+}/Sn} = -0.14 \ V$.
$E^o_{cell} = (-0.44 \ V) - (-0.14 \ V) = -0.44 + 0.14 = -0.30 \ V$.

(A) The standard $EMF$ 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 $(E^o = +0.34\,V)$ and $Zn^{2+}/Zn$ acts as the anode $(E^o = -0.76\,V)$.
Substituting the values: $E^o_{cell} = 0.34\,V - (-0.76\,V) = 0.34\,V + 0.76\,V = 1.10\,V$.

(C) The standard $EMF$ of a cell is calculated using the formula: $E_{cell}^o = E_{cathode}^o - E_{anode}^o$.
In this cell,$Cu^{2+}/Cu$ acts as the cathode (higher reduction potential) and $Mg^{2+}/Mg$ acts as the anode (lower reduction potential).
Substituting the given values: $E_{cell}^o = 0.34 \ V - (-2.37 \ V)$.
$E_{cell}^o = 0.34 + 2.37 = 2.71 \ V$.

(D) The cell reaction involves $Mg$ acting as the anode and $Cu$ acting as the cathode because $Mg$ is a stronger reducing agent than $Cu$.
$E_{cell}^o = E_{cathode}^o - E_{anode}^o$
$E_{cell}^o = E_{Cu^{+2}/Cu}^o - E_{Mg^{+2}/Mg}^o$
Given $E_{cell}^o = 2.7 \ V$ and $E_{Cu^{+2}/Cu}^o = 0.34 \ V$.
$2.7 \ V = 0.34 \ V - E_{Mg^{+2}/Mg}^o$
$E_{Mg^{+2}/Mg}^o = 0.34 \ V - 2.7 \ V = -2.36 \ V$.

(A) The standard reduction potential of $H^{+}/H_2$ is $0.00 \ V$.
Since $E^o_{Ag^{+}/Ag} (0.8 \ V) > E^o_{H^{+}/H_2} (0.00 \ V)$,$Ag^{+}$ is a stronger oxidizing agent than $H^{+}$,meaning $H_2$ can reduce $Ag^{+}$ to $Ag$.
Since $E^o_{Zn^{2+}/Zn} (-0.76 \ V) < E^o_{H^{+}/H_2} (0.00 \ V)$,$Zn$ is a stronger reducing agent than $H_2$,meaning $Zn$ can reduce $H^{+}$ to $H_2$.
Therefore,the correct statement is that $Ag^{+}$ can be reduced by $H_2$.

(D) We use the relation $\Delta G^o = -nFE^o$.
For reaction $(i): Fe^{2+} + 2e^- \to Fe$,$\Delta G_1^o = -2 \times F \times (-0.440 \ V) = 0.880 \ F$.
For reaction $(ii): Fe^{3+} + 3e^- \to Fe$,$\Delta G_2^o = -3 \times F \times (-0.036 \ V) = 0.108 \ F$.
We want the reaction: $Fe^{3+} + e^- \to Fe^{2+}$.
This can be obtained by $(ii) - (i)$:
$\Delta G_3^o = \Delta G_2^o - \Delta G_1^o = 0.108 \ F - 0.880 \ F = -0.772 \ F$.
Since $\Delta G_3^o = -nFE^o$ where $n=1$:
$-0.772 \ F = -1 \times F \times E^o$.
$E^o = +0.772 \ V \approx +0.771 \ V$.

(C) The standard electrode potentials are $E^o_{Ag^+/Ag} = +0.80 \ V$ and $E^o_{Cu^{2+}/Cu} = +0.34 \ V$.
For a spontaneous cell reaction,the electrode with the higher reduction potential acts as the cathode.
Since $0.80 \ V > 0.34 \ V$,the silver electrode acts as the cathode and the copper electrode acts as the anode.
The cell reaction is $Cu_{(s)} + 2Ag^+_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$.
The standard cell potential is calculated as:
$E_{cell}^o = E_{cathode}^o - E_{anode}^o = 0.80 \ V - 0.34 \ V = +0.46 \ V$.
Thus,when the copper electrode works as the anode,the $E_{cell}^o$ is $+0.46 \ V$.

(A) The relationship between the standard Gibbs free energy change $(\Delta G^{\circ})$ and the cell potential $(E^{\circ}_{cell})$ is given by the equation: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.
For a reaction to be spontaneous,the change in Gibbs free energy $(\Delta G^{\circ})$ must be negative $(\Delta G^{\circ} < 0)$.
Since $n$ (number of moles of electrons) and $F$ (Faraday's constant) are always positive,for $\Delta G^{\circ}$ to be negative,the cell potential $(E^{\circ}_{cell})$ must be positive.
Therefore,the reaction is spontaneous if the cell potential is positive.

(B) Electrode potential is a relative term,meaning it is always measured with respect to a reference electrode.
In electrochemistry,the standard hydrogen electrode is arbitrarily chosen as the reference electrode.
To establish a basis for comparison with all other electrode reactions,its standard electrode potential is defined as $0.00 \ V$ at all temperatures.

(C) The cell potential $(E_{cell})$ is defined as the difference between the reduction potential of the cathode and the reduction potential of the anode.
$E_{cell} = E_{cathode} - E_{anode}$.

(B) metal with a lower standard reduction potential acts as a stronger reducing agent and can displace a metal with a higher standard reduction potential from its salt solution.
The standard reduction potential of $Fe^{2+}/Fe$ is $-0.44 \ V$,while that of $Cu^{2+}/Cu$ is $+0.34 \ V$.
Since $E^{\circ}_{Fe^{2+}/Fe} < E^{\circ}_{Cu^{2+}/Cu}$,iron can displace copper from its salt solution according to the reaction: $Fe(s) + Cu^{2+}(aq) \rightarrow Fe^{2+}(aq) + Cu(s)$.

(A) metal with a more negative standard reduction potential acts as a stronger reducing agent.
From $(i)$,$Cu + 2Ag^+ \rightarrow Cu^{2+} + 2Ag$,so $Cu$ is a stronger reducing agent than $Ag$.
From $(ii)$,$Ag$ does not displace $Zn^{2+}$,so $Zn$ is a stronger reducing agent than $Ag$.
From $(iii)$,$Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$,so $Zn$ is a stronger reducing agent than $Cu$.
Combining these,the order of decreasing strength as reducing agents is $Zn > Cu > Ag$.

(B) The standard hydrogen electrode $(SHE)$ is defined as having an electrode potential of $0.00 \ V$ at all temperatures.
Therefore,the correct option is $(B)$.

(A) The standard free energy change $\Delta G^o$ is related to the standard cell potential $E^o$ by the equation: $\Delta G^o = -nFE^o$.
In the given reaction,$\frac{1}{2}Cu_{(s)} + \frac{1}{2}Cl_{2(g)} \rightleftharpoons \frac{1}{2}Cu^{2+} + Cl^-$,the number of electrons transferred $(n)$ is $1$.
The Faraday constant $F$ is $96500 \ C \ mol^{-1}$ and the standard cell potential $E^o$ is $1.02 \ V$.
Substituting these values: $\Delta G^o = -1 \times 96500 \times 1.02 = -98430 \ J$.

(A) The reducing strength of an element is inversely proportional to its standard reduction potential.
Lower (more negative) reduction potential indicates a stronger tendency to undergo oxidation,making it a stronger reducing agent.
Comparing the given values: $-3.04 \ V < -1.90 \ V < 0 \ V < 1.90 \ V$.
Since $I$ has the lowest (most negative) standard reduction potential,it is the most suitable reducing agent.

(A) The reducing power is defined as the tendency of a species to lose electrons,which is inversely proportional to its standard reduction potential $(E^o)$.
Comparing the given standard reduction potentials:
$E^o(Br_2/Br^-) = +1.08 \, V$
$E^o(Fe^{3+}/Fe^{2+}) = +0.77 \, V$
$E^o(Al^{3+}/Al) = -1.66 \, V$
Since the reducing power increases as the reduction potential decreases,the order of reducing power is:
$Br^{-} < Fe^{2+} < Al$
Therefore,the correct option is $A$.

(C) To find the standard electrode potential for the reaction $OCl^{-} \to \frac{1}{2}Cl_2$,we use the Gibbs free energy change relation $\Delta G^o = -nFE^o$.
For reaction $1$: $OCl^{-} + H_2O + 2e^{-} \to Cl^{-} + 2OH^{-}$,$E^o_1 = 0.94 \ V$,$n_1 = 2$.
$\Delta G^o_1 = -2 \times F \times 0.94 = -1.88F$.
For reaction $2$: $Cl^{-} \to \frac{1}{2}Cl_2 + e^{-}$,$E^o_2 = -1.36 \ V$,$n_2 = 1$.
$\Delta G^o_2 = -1 \times F \times (-1.36) = 1.36F$.
Adding the two reactions: $OCl^{-} + H_2O + e^{-} \to \frac{1}{2}Cl_2 + 2OH^{-}$.
Total $\Delta G^o = \Delta G^o_1 + \Delta G^o_2 = -1.88F + 1.36F = -0.52F$.
Since $\Delta G^o = -nFE^o_{total}$ and $n = 1$ for the net reaction:
$-0.52F = -1 \times F \times E^o_{total}$.
$E^o_{total} = 0.52 \ V$.

(C) substance with a higher reduction potential has a greater tendency to gain electrons.
Therefore,it is easily reduced and acts as a strong oxidising agent.

(A) The correct answer is $A$.
$Hg$ (mercury) is a metal that is placed below hydrogen in the electrochemical series.
Therefore,it has a positive standard reduction potential $(E^0_{Hg^{2+}/Hg} = +0.85 \ V)$ and cannot displace hydrogen from $H_2S$ or other acids.
$Hg + H_2S \to \text{No reaction}$.

(D) The strength of an oxidizing agent is directly proportional to its standard reduction potential $(E^\circ_{red})$.
Higher reduction potential indicates a greater tendency to gain electrons,making the species a stronger oxidizing agent.
Comparing the given values: $E^\circ(Li^{+}|Li) = -3.05 \ V$,$E^\circ(Ba^{2+}|Ba) = -2.90 \ V$,$E^\circ(Na^{+}|Na) = -2.71 \ V$,and $E^\circ(Mg^{2+}|Mg) = -2.37 \ V$.
Since $-2.37 \ V$ is the highest value among the given potentials,$Mg^{2+}$ is the strongest oxidizing agent.

(B) The given reactions are oxidation half-reactions:
$Zn \to Zn^{2+} + 2e^-; E^o_{ox} = 0.76 \, V$
$Fe \to Fe^{2+} + 2e^-; E^o_{ox} = 0.44 \, V$
For the cell reaction $Fe^{2+} + Zn \to Zn^{2+} + Fe$,the oxidation occurs at the $Zn$ electrode (anode) and reduction occurs at the $Fe$ electrode (cathode).
$E^o_{cell} = E^o_{cathode(red)} - E^o_{anode(red)}$
Since $E^o_{red} = -E^o_{ox}$:
$E^o_{cathode(red)} = -E^o(Fe \to Fe^{2+}) = -0.44 \, V$
$E^o_{anode(red)} = -E^o(Zn \to Zn^{2+}) = -0.76 \, V$
$E^o_{cell} = -0.44 - (-0.76) = -0.44 + 0.76 = +0.32 \, V$.

(A) The standard electrode potentials are given as $E^o_{Fe^{2+}/Fe} = -0.44 \, V$ and $E^o_{Cu^{2+}/Cu} = +0.32 \, V$.
Since the reduction potential of $Cu^{2+}/Cu$ is higher than that of $Fe^{2+}/Fe$,$Cu^{2+}$ acts as a stronger oxidizing agent.
Therefore,$Cu^{2+}$ can oxidize $Fe$ to $Fe^{2+}$ according to the reaction: $Fe(s) + Cu^{2+}(aq) \rightarrow Fe^{2+}(aq) + Cu(s)$.

(B) The electrode represented by $Pt, O_2(1 \, atm) | 2H^+(1 \, M)$ is the Standard Hydrogen Electrode $(SHE)$ or a related reference electrode system.
By definition,the standard electrode potential of the Standard Hydrogen Electrode is $E^o = 0 \, V$ at all temperatures.
Therefore,the correct option is $B$.

(B) For a cell to be in operation (spontaneous reaction),the Gibbs free energy change $\Delta G$ must be negative.
Since $\Delta G = -nFE_{cell}$,for $\Delta G < 0$,the cell potential $E_{cell}$ must be positive.
Therefore,the correct option is $(B)$.