${\mathrm{E}}^{\Theta }$ values of some redox couples are given below. On the basis of these values choose the correct option.

${\mathrm{Ag}}^{+}/\mathrm{Ag}\left(\mathrm{s}\right)=+0.80$

${\mathrm{I}}_{2}\left(\mathrm{s}\right)/{\mathrm{I}}^{-}=+0.54$

1. Cu will reduce Br-

2. Cu will reduce Ag

3. Cu will reduce I-

4. Cu will reduce Br2

Hint: High reduction potential substance is a good oxidising agent

The given values of $\mathrm{E}°$ are as follows:

${\mathrm{Br}}_{2}/{\mathrm{Br}}^{-}=+1.90\mathrm{V}$
$\mathrm{Ag}/{\mathrm{Ag}}^{+}=-0.80\mathrm{V}$
${\mathrm{Cu}}^{2+}/\mathrm{Cu}\left(\mathrm{s}\right)=+0.34\mathrm{V}$

${\mathrm{Br}}^{-}/{\mathrm{Br}}_{2}=-1.90\mathrm{V}$
The $\mathrm{E}°$ values show that copper will reduce Br2, if the $\mathrm{E}°$ of the following redox reaction is positive.

$\frac{2\mathrm{Cu}+{\mathrm{Br}}_{2}\to {\mathrm{CuBr}}_{2}}{}$

Since, $\mathrm{E}°$ of this reaction is positive, therefore, Cu can reduce Br2.

While other reaction will give negative value.