Chemistry Electrochemistry

Main product of electrolysis of conc. H? SO? is H? S? O? i.e. HO? SO-OSO? H

$Mn{O}_4^{-}+8H+5e^{-}\to Mn+4H_{2}O{E}^o=1.51V$

Quantity of electricity required to reduce 1 mole of $Mn{O}_4^{-}$ is 5F

So, for 5 mole $Mn{O}_4^{-}$ 25F electricity is required.

  ${E^o}_{Cell}={E^o}_{A{g}^{+}/Ag}-{E^o}_{Z{n}^{+2}/Zn}=0.8+0.76=1.56V$

Anode :      $Zn\left(s\right)\to Z{n}^{2+}\left(aq\right)+2e^{-}$

Cathode :  2Ag⁺(aq) + 2e^- ® 2Ag(s)

Zn(s) + 2Ag⁺(aq) ® Zn²⁺ (aq) + 2Ag(s)

$E_{cell}={E^o}_{Cell}-\frac{0.0591}{n}log\frac{\left[Z{n}^{2+}\right]}{{\left[A{g}^{+}\right]}^2}=1.56-\frac{0.0591}{2}log\left(\frac{0.1}{10^{-4}}\right)$    

 x = 147

$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$

$E_1=E^0-\frac{0.059}{5}log\frac{\left[M{n}^{2+}\right]}{\left[Mn{O}_4^{-}\right]}{\left[\frac{1}{H^{+}}\right]}^8$  

[H⁺] = 1M

$E_2=E^0-\frac{0.059}{5}log\frac{\left[M{n}^{2+}\right]}{\left[Mn{O}_4^{-}\right]}+\frac{0.059}{5}log10^{-32}$   

So, difference in E₁ & E₂ is

= 0.3776 V

= 3776 * 10^-4 V

Using ; Nernst equation

$E_{cell}=E_{cell}^o-\frac{0.06}{n}lo{g}_{10}\frac{\left[C{u}^{2+}\right]}{{\left[A{g}^{+}\right]}^2}$              

$E_{cell}=2.97-\frac{0.06}{2}lo{g}_{10}\frac{0.25}{10^{-6}}V$              

$=2.97-0.03\left(log2.5+5\right)V$              

$E_{cell}=2.8V$              

Ans. is 3 (the nearest integer)

(a) Cell constant = $\left(\frac{\mathcal{l}}{A}\right)=m/m^2=m^{-1}$  

(b) molar conductivity $\left({\lambda }_m\right)=\frac{\kappa *1000}{molarity}$

= Scm² mol^-1

(c) Conductivity $\left(\kappa \right)=\frac{1}{\rho }=\frac{\mathcal{l}}{RA}$

$={\Omega }^{-1}{m}^{-1}$                                

(d) degree of dissociation = $\frac{numberofmoledissociated}{Totalmole}$ , it is dimensionless 

Here, C₃H₆ is propene

Second reaction is iodoform reaction.

Change in oxidation number = 2

$\Delta G°=-n{E}_{cell}^{o}F$

$\Delta G°=-n{E}_{cell}^{o}F$

=-2 * 4.315 * 96487 Jmol^-1

^{$\therefore \Delta G°=\Delta H°-T\Delta S°$}

^{$\Delta S°=\frac{\Delta H°-\Delta G°}{T}$}

$=\frac{-825.2*1000-\left(-2*4.315*96487\right)}{298.15}J{K}^{-1}$

$=\frac{7482.81}{298.15}=25.09J{K}^{-1}$

$\approx$ 25.1 JK^-1

Ans. = 25

$\kappa =\frac{1}{R}*\frac{\mathcal{l}}{A}=\left[\left(\frac{1}{1500}\right)*1.14\right]Sc{m}^{-1}=\frac{1.14}{1500}Sc{m}^{-1}$

${\lambda }_m=\frac{\kappa }{M}*1000Sc{m}^{2}mo{l}^{-1}$

${\lambda }_m=1000*\frac{\left(\frac{1.14}{1500}\right)}{0.001}Sc{m}^{2}mo{l}^{-1}=760Sc{m}^{2}mo{l}^{-1}$

Energy of emitted photon in 0.1 sec = 10^-4 J

$n*\frac{hc}{\lambda }=10^{-4}$

$\frac{n*6.63*10^{-34}*3*10^8}{1000*10^{-9}}=10^{-4}$

$n=5.02*10^{14}=50.2*10^{13}\approx 50*10^{13}$