Question

Using the standard electrode potential, find out the pair between which redox reaction is not feasible.
${E^ \circ }$ values: $F{e^{3 + }}/F{e^{2 + }}$= +0.77; ${I_2}/{I^ - }$= +0.54; $C{u^{ + 2}}/Cu$= +0.34; $A{g^ + }/Ag$= +0.80V
(A) $F{e^{ + 3}}$ and ${I^ - }$
(B) $A{g^ + }$ and Cu
(C) $F{e^{ + 3}}$ and Cu
(D) $A{g^ + }$ and $F{e^{ + 3}}$

Answer 1

Hint: The question gives us the values of reduction potentials. More positive is the reduction potential more is its ability to get reduced. Any two elements having least difference between their reduction potentials will not undergo redox reduction because both will have similar ability to get reduced.

Complete step by step answer:
-Here we have to tell which of the redox reactions is not feasible or is not possible. And for 2 compounds to undergo redox reaction one should have the ability to reduce and other to oxidise. This means that for one element the reduction potential should be highly positive and for the other element it should be less positive.
-The question gives us reduction potential for various elements:
$F{e^{3 + }}/F{e^{2 + }}$= +0.77 V
${I_2}/{I^ - }$= +0.54 V
$C{u^{ + 2}}/Cu$= +0.34 V
$A{g^ + }/Ag$= +0.80 V
We all know that more positive is the reduction potential; more is their ability to get reduced. So from the above given redox potentials we can give the order of getting reduced to be: $A{g^ + } > F{e^{ + 3}} > {I_2} > C{u^{ + 2}}$
So, the possible reaction will be:
${I_2} + Cu \to {I^ - } + C{u^{ + 2}}$
$F{e^{ + 3}} + {I^ - } \to F{e^{ + 2}} + {I_2}$
$F{e^{ + 3}} + Cu \to F{e^{ + 2}} + C{u^{ + 2}}$
$A{g^ + } + F{e^{ + 2}} \to Ag + F{e^{ + 3}}$
$A{g^ + } + {I^ - } \to Ag + {I_2}$
$A{g^ + } + Cu \to Ag + C{u^{ + 2}}$
But since we can see that the reduction potential of $F{e^{3 + }}/F{e^{2 + }}$= +0.77 V and $A{g^ + }/Ag$= +0.80 V are very close by and almost same. So, $A{g^ + }$ and $F{e^{ + 3}}$cannot undergo a redox reaction since both have almost the same ability to get reduced and poor ability to get oxidized.

So, the correct option will be: (D) $A{g^ + }$ and $F{e^{ + 3}}$



Note: In a redox reaction one element gets reduced while the other gets oxidised. Also when we write the ${E^ \circ }$ value for any element ${X^{ + 2}}/X$, we are talking about its reduction potential and for $X/{X^{ + 2}}$ it is oxidation potential.