Chemistry - Electrochemistry Question with Solution | TestHub

The number of faraday's passed$=\frac{2\times 10^{-3}\times 16\times 60}{96500}$

$=\text{1.99}\times 10^{-5}$

$\Rightarrow$number of gram equivalent of${\text{Cu}}^{2+}$deposited${\text{as Cu(s)}}^{\mathrm{}}$

$=\text{1.99}\times 10^{-5}$

$\Rightarrow$number of moles of${\text{Cu}}^{2+}$deposited

$=\frac{\text{1.99}}{2}\times {\text{10}}^{-5}\approx {\text{10}}^{-5}$

Absorbance is directly proportional to$\left[{\text{Cu}}^{2+}\right]$.

No. of moles of Cu²⁺initially present = 2 x 10^-5moles

${\left[{\text{Cu}}^{2+}\right]}_{\text{Initial}}=2\times 10^{-5}\times \frac{1000}{250}$

= 8 x10^-5M