Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

$CuSO_4$ solution(250ml) was electrolysed using a platinum anode and a copper cathode.At constant current of 2mA was passed for 16 minutes.It was found that after electrolysis the absorbance of this solution was reduced to 50% of its original value.Then the concentration of copper sulphate in solution to begin with is

$\begin{array}{1 1}(a)\;8\times 10^{-4}\\(b)\;8\times 10^{-5}\\(c)\;8\times 10^{-6}\\(d)\;8\times 10^{-7}\end{array}$

Can you answer this question?

1 Answer

0 votes
Total no of Faradays passed=$\large\frac{2\times 10^{-3}\times 16\times 60}{96500}$
$\Rightarrow 1.9\times 10^{-5}$
Moles of $Cu^{+2}$ ions deposited $=1.9\times \large\frac{10^{-5}}{2}$
Since absorbance was reduced to 50% of its original value the initial moles of $Cu^{+2}$ would be two times the moles of $Cu^{+2}$ reduced.
$\therefore$ Initial moles of $Cu^{+2}=\large\frac{1.9\times 10^{-5}}{2}$$\times 2$
$\Rightarrow 1.9\times 10^{-5}$
The concentration of $CuSO_4$ in solution =$1.9896\times 10^{-5}\times 4$
$\Rightarrow \approx 8\times 10^{-5}mol/lit$
Hence (b) is the correct answer.
answered Feb 17, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App