logo

Ask Questions, Get Answers

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

In an electrolysis experiment electric current was passed for 5hr through two cells connected in series.The first cell contains gold salt and second contains $CuSO_4$ solution.9.83g of gold was deposited in first cell.If the oxidation number of Au is +3,find the amount of copper deposited in second cell

$\begin{array}{1 1}(a)\;4g&(b)\;5g\\(c)\;4.7g&(d)\;5.7g\end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
At weight of gold is 197.
Eq.wt of gold =$\large\frac{At.wt}{\text{Oxidation number}}=\frac{197}{3}$
$\Rightarrow 65.66$
As we know that
$\large\frac{W_{Au}}{W_{Cu}}=\frac{E_{Au}}{E_{Cu}}$
$\large\frac{9.83}{W_{Cu}}=\frac{65.66}{63.5/2}$
$W_{Cu}=4.7g$
Hence (c) is the correct answer.
answered Feb 17, 2014 by sreemathi.v
 

Related questions

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