• # 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, $E_{{{I}_{2}}/{{I}^{-}}}^{{}^\circ }=0.541V,\,E_{C{{u}^{2+}}/Cu}^{{}^\circ }=+0.34V,$ $E_{B{{r}_{2}}/B{{r}^{-}}}^{{}^\circ }=1.09V,E_{A{{g}^{+}}/Ag}^{{}^\circ }=+0.80V,$ $E_{F{{e}^{3+}}/F{{e}^{2+}}}^{{}^\circ }=+0.77V)$

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)