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Which one of the following reactions does not take place?
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The correct answer is:
$\mathrm{Br}_2+2 \mathrm{Cl}^{-} \longrightarrow 2 \mathrm{Br}^{-}+\mathrm{Cl}_2$
The reaction, $\mathrm{Br}_2+2 \mathrm{Cl}^{-} \longrightarrow 2 \mathrm{Br}^{-}+\mathrm{Cl}_2$ does not take place as the $E_{\text {red }}^{\circ}$ of the above reaction is comes out to be negative. Thus, the reaction is non-spontaneous.
Half cell reaction are given as
At cathode, $\mathrm{Br}_2+2 e^{-} \longrightarrow 2 \mathrm{Br}^{-} ; E_{\mathrm{Br}_2 / \mathrm{Br}^{-}}^{\circ}=1.09 \mathrm{~V}$
At anode, $2 \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_2+2 e^{-} ; E_{\mathrm{Cl}_2 / \mathrm{Cl}^{-}}^{\circ}=1.36 \mathrm{~V}$
$\therefore E_{\text {red }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ} \Rightarrow(1.09-1.36) \mathrm{V} \Rightarrow-0.27 \mathrm{~V}$
But the $E_{\text {red }}^{\circ}$ values of other reaction are comes out to be positive. Thus, the all other reaction can take place.
Half cell reaction are given as
At cathode, $\mathrm{Br}_2+2 e^{-} \longrightarrow 2 \mathrm{Br}^{-} ; E_{\mathrm{Br}_2 / \mathrm{Br}^{-}}^{\circ}=1.09 \mathrm{~V}$
At anode, $2 \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_2+2 e^{-} ; E_{\mathrm{Cl}_2 / \mathrm{Cl}^{-}}^{\circ}=1.36 \mathrm{~V}$
$\therefore E_{\text {red }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ} \Rightarrow(1.09-1.36) \mathrm{V} \Rightarrow-0.27 \mathrm{~V}$
But the $E_{\text {red }}^{\circ}$ values of other reaction are comes out to be positive. Thus, the all other reaction can take place.
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