On the basis of standard electrode potential | vedclass

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Question

(B, D) reaction is spontaneous if the standard cell potential $E^{\circ}_{cell}$ is positive. $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{ano

Vedclass · Accepted Answer

$Br_{2} + 2Cl^{-} \rightarrow Cl_{2} + 2Br^{-}$

Vedclass · Answer

(B, D) reaction is spontaneous if the standard cell potential $E^{\circ}_{cell}$ is positive. $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.$(a)$ $Cu + Zn^{+2} \rightarrow Cu^{+2} + Zn$: $E^{\circ}_{cell} = E^{\circ}_{Zn^{+2}/Zn} - E^{\circ}_{Cu^{+2}/Cu} = -0.76 - 0.34 = -1.10 \ V$. (Non-spontaneous)$(b)$ $Mg + Fe^{+2} \rightarrow Mg^{+2} + Fe$: $E^{\circ}_{cell} = E^{\circ}_{Fe^{+2}/Fe} - E^{\circ}_{Mg^{+2}/Mg} = -0.44 - (-2.37) = +1.63 \ V$. (Spontaneous)$(c)$ $Br_{2} + 2Cl^{-} \rightarrow Cl_{2} + 2Br^{-}$: $E^{\circ}_{cell} = E^{\circ}_{Br_{2}/Br^{-}} - E^{\circ}_{Cl_{2}/Cl^{-}} = 1.08 - 1.36 = -0.28 \ V$. (Non-spontaneous)$(d)$ $Fe + Cd^{+2} \rightarrow Cd + Fe^{+2}$: $E^{\circ}_{cell} = E^{\circ}_{Cd^{+2}/Cd} - E^{\circ}_{Fe^{+2}/Fe} = -0.44 - (-0.74) = +0.30 \ V$. (Spontaneous)Both $(b)$ and $(d)$ are spontaneous reactions.

On the basis of standard electrode potential values,suggest which of the following reactions would take place?

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