(B) For a reaction to be spontaneous in an electrochemical cell,the cell potential $E_{cell}^0$ must be positive.$E_{cell}^0 = E_{cathode}^0 - E_{anod

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$2Br^- + Cl_2 \rightarrow Br_2 + 2Cl^-$

(B) For a reaction to be spontaneous in an electrochemical cell,the cell potential $E_{cell}^0$ must be positive.$E_{cell}^0 = E_{cathode}^0 - E_{anode}^0$.Given $E_{Cl_2 \mid 2Cl^-}^0 = 1.36 \ V$ and $E_{Br_2 \mid 2Br^-}^0 = 1.09 \ V$.For option $B$: $2Br^- + Cl_2 \rightarrow Br_2 + 2Cl^-$.Here,$Cl_2$ is reduced to $Cl^-$ (cathode) and $Br^-$ is oxidized to $Br_2$ (anode).$E_{cell}^0 = E_{Cl_2 \mid 2Cl^-}^0 - E_{Br_2 \mid 2Br^-}^0 = 1.36 \ V - 1.09 \ V = 0.27 \ V$.Since $E_{cell}^0 > 0$,the reaction is spontaneous.

Class 12 Chemistry Electrochemistry Electrode potential and ECell

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Which of the following reactions will be spontaneous in an electrochemical cell constructed from the given standard electrode potentials: $E_{Cl_2 \mid 2Cl^-}^0 = 1.36 \ V$ and $E_{Br_2 \mid 2Br^-}^0 = 1.09 \ V$?

GSEB 2019 All GSEB

A
$2Cl^- + 2Br^- \rightarrow Cl_2 + Br_2$
B
$2Br^- + Cl_2 \rightarrow Br_2 + 2Cl^-$
C
$Br_2 + 2Cl^- \rightarrow 2Br^- + Cl_2$
D
$Cl_2 + Br_2 \rightarrow 2Cl^- + 2Br^-$

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Given the standard half-cell potentials $(E^{\circ})$ as: $Zn \rightarrow Zn^{2+} + 2e^{-}$; $E^{\circ} = +0.76 \ V$ and $Fe \rightarrow Fe^{2+} + 2e^{-}$; $E^{\circ} = +0.41 \ V$. Then the standard e.m.f. of the cell with the reaction $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$ is:

MediumWBJEE 2018

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If the cell potential of the cell at $298 \ K$ is $2.36 \ V$,write the cell reaction and calculate the standard electrode potential of the $Mg^{2+} \mid Mg$ half-cell.
$Mg_{(s)} \mid Mg^{2+}_{(1 \ M)} \parallel H^{+}_{(1 \ M)} \mid H_{2(g)} (1 \ bar) \mid Pt_{(s)}$

Medium

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Consider the half-cell reduction reactions:
$Mn^{2+} + 2e^{-} \rightarrow Mn$,$E^{\circ} = -1.18 \ V$
$Mn^{3+} + e^{-} \rightarrow Mn^{2+}$,$E^{\circ} = +1.51 \ V$
The $E^{\circ}$ for the reaction $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$,and the possibility of the forward reaction are respectively:

NEET 2013

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The standard $E^{\circ}_{red}$ values of $A, B$ and $C$ are $+0.68 \ V, -2.54 \ V$ and $-0.50 \ V$ respectively. The order of their reducing power is:

EasyMHT CET 2007

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The formal potential of $Fe^{3+}/Fe^{2+}$ in a sulphuric acid and phosphoric acid mixture $(E^{\circ}=+0.61 \ V)$ is much lower than the standard potential $(E^{\circ}=+0.77 \ V)$. This is due to

MediumWBJEE 2014

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Which of the following reactions will be spontaneous in an electrochemical cell constructed from the given standard electrode potentials: $E_{Cl_2 \mid 2Cl^-}^0 = 1.36 \ V$ and $E_{Br_2 \mid 2Br^-}^0 = 1.09 \ V$?

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Given the standard half-cell potentials $(E^{\circ})$ as: $Zn \rightarrow Zn^{2+} + 2e^{-}$; $E^{\circ} = +0.76 \ V$ and $Fe \rightarrow Fe^{2+} + 2e^{-}$; $E^{\circ} = +0.41 \ V$. Then the standard e.m.f. of the cell with the reaction $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$ is:

If the cell potential of the cell at $298 \ K$ is $2.36 \ V$,write the cell reaction and calculate the standard electrode potential of the $Mg^{2+} \mid Mg$ half-cell.$Mg_{(s)} \mid Mg^{2+}_{(1 \ M)} \parallel H^{+}_{(1 \ M)} \mid H_{2(g)} (1 \ bar) \mid Pt_{(s)}$

Consider the half-cell reduction reactions:$Mn^{2+} + 2e^{-} \rightarrow Mn$,$E^{\circ} = -1.18 \ V$$Mn^{3+} + e^{-} \rightarrow Mn^{2+}$,$E^{\circ} = +1.51 \ V$The $E^{\circ}$ for the reaction $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$,and the possibility of the forward reaction are respectively:

The standard $E^{\circ}_{red}$ values of $A, B$ and $C$ are $+0.68 \ V, -2.54 \ V$ and $-0.50 \ V$ respectively. The order of their reducing power is:

The formal potential of $Fe^{3+}/Fe^{2+}$ in a sulphuric acid and phosphoric acid mixture $(E^{\circ}=+0.61 \ V)$ is much lower than the standard potential $(E^{\circ}=+0.77 \ V)$. This is due to

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If the cell potential of the cell at $298 \ K$ is $2.36 \ V$,write the cell reaction and calculate the standard electrode potential of the $Mg^{2+} \mid Mg$ half-cell.
$Mg_{(s)} \mid Mg^{2+}_{(1 \ M)} \parallel H^{+}_{(1 \ M)} \mid H_{2(g)} (1 \ bar) \mid Pt_{(s)}$

Consider the half-cell reduction reactions:
$Mn^{2+} + 2e^{-} \rightarrow Mn$,$E^{\circ} = -1.18 \ V$
$Mn^{3+} + e^{-} \rightarrow Mn^{2+}$,$E^{\circ} = +1.51 \ V$
The $E^{\circ}$ for the reaction $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$,and the possibility of the forward reaction are respectively: