Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
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$
Hence (a) is the correct answer.
answered Feb 17, 2014 by sreemathi.v

Related questions

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