logo

Ask Questions, Get Answers

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

A zinc rod is placed in $0.1\;M$ solution of $ZnSO_4$ at $25^{\large\circ}$C.Assuming that the salt is dissociated to the extent of $95\%$ at this dilusion ,then the potential of electrode at this temperature.Given that $E^0_{Zn^{2+}/Zn}=0.76Volts$.

$\begin{array}{1 1}(a)\;-0.79Volt\\(b)\;0.79Volt\\(c)\;-0.89Volt\\(d)\;0.89Volt\end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
Concentration of $ZnSO_4$ solution =0.1M
$\%$ of dissociation of $ZnSO_4$ solution =95%
Concentration of $Zn^{+2}$ ions in $ZnSO_4$ solution =$\large\frac{0.1\times 95}{100}$
$\Rightarrow 0.095M$
Thus,the electrode can be represented as $Zn(s) /Zn^{+2}(0.095M)$
Reduction reaction taking place at this electrode is
$Zn^{+2}+2e^-\rightarrow Zn(s)$(Here n=2)
According to Nernst equation,the reduction potential of above electrode is given by
$E_{Zn^{+2}/Zn}=E^0_{Zn^{+2}/Zn}-\large\frac{0.059}{n}$$\log \large\frac{1}{0.095}$
$E_{Zn^{+2}/Zn}=-0.76+\large\frac{0.059}{2}$$\log 0.095$
$E_{Zn^{+2}/Zn}$=-0.79Volt
Hence (a) 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
...