Question

$E^0$ values of some redox couples are given below. On the basis of these values choose the correct option.

$E^0$ values $B{r}_2|B{r}^{-}$ = +1.90V; $A{g}^{+}|A{g}_{(}s)$ = +0.80V; $C{u}^{2+}|C{u}_{(}s)$ = + 0.34V; $I_{2\left(s\right)}|I^{-}$= +0.54V

• A. $Cu$ will reduce $B{r}^{-}$
• B. $Cu$ will reduce Ag
• C. $Cu$ will reduce $I^{-}$
• D. $Cu$ will reduce $B{r}_2$

Answer 1

The correct option is D $Cu$ will reduce $B{r}_2$

Given reduction potential values as:

$B{r}_2|B{r}^{-}$ = +1.90V; $A{g}^{+}|A{g}_{(}s)$ = +0.80V; $C{u}^{2+}|C{u}_{(}s)$ = + 0.34V; $I_{2\left(s\right)}|I^{-}$= +0.54V

Reduction potential determines the tendency of the specie to get reduced. More positive the value of $E^0$, greater is the tendency of the species to get reduced and stronger is the oxidising agent.

On the basis of the given $E^0$ values, the order of getting reduced is:

$B{r}_2>A{g}^{+}>I_2>C{u}^{2+}$

Thus Cu has the least tendency to get reduced and will itself gets oxidise and reduce other species as:

$B{r}_2,A{g}^{+}$ and $I_2$