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)
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