Question

Which of the following transformations can be carried out by using HI as a reducing agent, under acidic conditions?[Given: $I_2\left(s\right)\to 2I^{-}$; $E^{\ominus }=0.54V]$.

(i) $C{u}^{+}\to Cu\left(s\right);E^{\ominus }=0.52V$
(ii) $C{r}^{3+}\to C{r}^{2+};E^{\ominus }=-0.41V$
(iii) $F{e}^{3+}\to F{e}^{2+};E^{\ominus }=0.77V$
(iv) $F{e}^{2+}\to Fe\left(s\right);E^{\ominus }=-0.44V$.

• A. (i) and (iii)
• B. (ii) and (iv)
• C. Only (iii)
• D. Only (ii)

Answer 1

Answer: (C)

Only (iii)

When $HI$ is used as reducing agent, $I^{-}$ must be oxidised as $2I^{-}\to I_2$ for which $E^o=-0.54V$.

The reaction will take place spontaneously only when net electrode potential is greater than zero and it becomes greater than zero only in case of $F{e}^{3+}\to F{e}^{2+}+e^{-}$.

i.e. $E_{cell}=E_{cath}-E_{anode}=0.77-0.54=0.23V$