Ncert Exercises & Solutions

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

$E^{Θ} values : {Br}_{2} / {Br}^{-} = + 1.90$
${Ag}^{+} / Ag (s) = + 0.80$
${Cu}^{2 +} / Cu (s) = + 0.34;$
$I_{2} (s) / I^{-} = + 0.54$

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

E °

are as follows:

{Br}_{2} / {Br}^{-} = + 1.90 V

Ag / {Ag}^{+} = - 0.80 V

{Cu}^{2 +} / Cu (s) = + 0.34 V

I^{-} / I_{2} (s) = - 0.54 V

{Br}^{-} / {Br}_{2} = - 1.90 V

The

E °

values show that copper will reduce Br₂, if the

E °

of the following redox reaction is positive.

2 Cu + {Br}_{2} \to {CuBr}_{2}

Cu \to {Cu}^{2 +} + 2 e^{-}; E ° = - 0.34 V

{Br}_{2} + 2 e^{-} \to 2 {Br}^{-}; E = + 1.09 V Cu + {Br}_{2} \to {CuBr}_{2}; E ° = + 0.75 V

Since,

E °

of this reaction is positive, therefore, Cu can reduce Br₂.

While other reaction will give negative value.