Question

The solution of $CuS{O}_4$ in which $Cu$ rod is dipped is diluted to 10 times the reduction electrode potential will be

• A. Reduces by 0.03 V
• B. Increases by 1 V
• C. Reduces by 1 V
• D. Increases by 0.03 V

Answer 1

The correct option is A Reduces by 0.03 V

$E_{\left(Cu/C{u}^{2+}\right)}=E_{\left(Cu/C{u}^{2+}\right)}^0+\frac{0.059}{2}logC{u}^{2+}$

$E_1=E_{\left(Cu/C{u}^{2+}\right)}^0+\frac{0.059}{2}logx$
$E_2=E_{\left(Cu/C{u}^{2+}\right)}^0+\frac{0.059}{2}log\frac{x}{10}$

$E_2-E_1=\frac{0.059}{2}log\frac{1}{10}$
$E_2-E_1=-\frac{0.059}{2}$

= - 0.03 V
Hence it reduces by 0.03 V