question_answer 36)

Using standard electrode; predict if the reaction between the following is feasible: (i) \[F{{e}^{3+}}(aq)and\,{{I}^{-}}(aq)\] (ii) \[A{{g}^{+}}(aq)and\,Cu\] (iii) \[F{{e}^{3+}}and\,B{{r}^{-}}\] (iv) \[Ag\,and\,F{{e}^{3+}}(aq)\] (v) \[B{{r}_{2}}(aq)and\,F{{e}^{2+}}(aq)\] (Given, \[E_{{{I}_{2}}/{{I}^{-}}}^{{}^\circ }=0.541V,\,E_{C{{u}^{2+}}/Cu}^{{}^\circ }=+0.34V,\] \[E_{B{{r}_{2}}/B{{r}^{-}}}^{{}^\circ }=1.09V,E_{A{{g}^{+}}/Ag}^{{}^\circ }=+0.80V,\] \[E_{F{{e}^{3+}}/F{{e}^{2+}}}^{{}^\circ }=+0.77V)\]  

Answer:

A redox reaction is feasible if its E is positive. (i) \[F{{e}^{3+}}(aq)+{{I}^{-}}(aq)\rightleftharpoons F{{e}^{2+}}(aq)+\frac{1}{2}{{I}_{2}}\] \[E_{\text{Redox}}^{{}^\circ }=E_{\text{Reduced}}^{{}^\circ }-E_{Oxidised\,species}^{{}^\circ }\] \[=E_{F{{e}^{3+}}/F{{e}^{2+}}}^{{}^\circ }-E_{{{I}_{2}}/{{I}^{-}}}^{{}^\circ }\] \[=0.77-0.54=-+0.23V\](Feasible) (ii) \[2A{{g}^{+}}(aq)+Cu\rightleftharpoons 2Ag(s)+C{{u}^{2+}}(aq)\] \[E_{\operatorname{Re}dox}^{{}^\circ }=E_{A{{g}^{+}}/Ag}^{{}^\circ }-E_{C{{u}^{2+}}/Cu}^{{}^\circ }\] \[+0.80-0.34=+0.46V\] (Feasible) (iii) \[F{{e}^{3+}}(aq)+B{{r}^{-}}(aq)\rightleftharpoons F{{e}^{2+}}(aq)+\frac{1}{2}B{{r}_{2}}\] \[E_{\operatorname{Re}dox}^{{}^\circ }=E_{F{{e}^{3+}}/F{{e}^{2+}}}^{{}^\circ }-E_{B{{r}_{2}}^{2+}/Cu}^{{}^\circ }\] \[=+0.77-1.09=-0.32V\] (Not feasible) (iv) \[Ag(s)+F{{e}^{3+}}(aq)\rightleftharpoons A{{g}^{+}}(aq)+F{{e}^{2+}}(aq)\] \[E_{\operatorname{Re}dox}^{{}^\circ }=E_{F{{e}^{3}}/F{{e}^{2+}}}^{{}^\circ }-E_{A{{g}^{+}}/Ag}^{{}^\circ }\] (Not feasible) (v) \[\frac{1}{2}B{{r}_{2}}(aq)+F{{e}^{2+}}(aq)\rightleftharpoons B{{r}^{-}}(aq)+F{{e}^{3+}}(aq)\] \[E_{\operatorname{Re}dox}^{{}^\circ }=E_{\operatorname{Re}duced\,speices}^{{}^\circ }-E_{Oxidised\,species}^{{}^\circ }\] \[=E_{B{{r}_{2}}/B{{r}^{-}}}^{{}^\circ }-E_{F{{e}^{3+}}/F{{e}^{2+}}}^{{}^\circ }\] \[=+1.09-0.77=+0.32V\](Feasible)