Chemistry Electrochemistry

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

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

[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

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

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

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

$F{e}^{+3}+3e^{-}\to Fe$

3 Faraday is required to deposit 1 mole Fe

 $?56g$  deposited by   $3*96500$ C charge

$\therefore 0.3482g$ deposited by $\frac{3*96500}{56}*0.3482C=\frac{100803.9}{56}C=1800.7C$

$\therefore Q=It$

1800.07 = 1.5 t

$\therefore t=1200sec=\frac{120\cancel{0}}{6\cancel{0}}$ min = 20 min 

$Pb{\left(N{O}_3\right)}_2\underset{673K}{\overset{\Delta }{\to }}PbO+\underset{\left(A\right)}{N{O}_2}+O_2$

$\underset{\left(A\right)}{N{O}_2}\underset{}{\overset{Dimerise}{\to }}\underset{\left(B\right)}{N_2{O}_4}$

$E_{C{\mathcal{l}}_2/C{\mathcal{l}}^{-}}^0=+1.36V$

$E_{l_2/l^{-}}^0=+0.54V$

$E_{A{g}^{+}/Ag}^0=+0.80V$

$E_{N{a}^{+}/Na}^0=-2.71V$

$E_{L{i}^{+}/Li}^0=-3.05V$

$\therefore E_{C{\mathcal{l}}_2/C{\mathcal{l}}^{-}}^0>E_{A{g}^{+}/Ag}^0>E_{N{a}^{+}/Na}^0>E_{L{i}^{+}/Li}^0$