Chemistry Electrochemistry

Except electron gain enthalpy all the physical properties like sublimation enthalpy, ionization enthalpy and hydration energy affected the value of reduction potential.

Ans. = 3

Molar conductivity measured at infinite dilution is called limiting molar conductivity. Hence at infinite dilution stage, both strong electrolyte (KCl) and weak electrolyte (CH₃COOH) have identical value of conductivity. Therefore statement – I is false.

Statement – II is also false because molar conductivity increases by the decreasing the concentration of electrolyte.

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

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

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

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

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

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

${\lambda }_m^{\infty }$  of Nal = 12.7 mSm² mol^-1

${\lambda }_m^{\infty }$ of NaNO₃ = 12.0 m Sm² mol^-1

${\lambda }_m^{\infty }$ of AgNO₃ = 13.3 mSm² mol^-1

$\therefore {\lambda }_{Agl}^{\infty }=\left[{\lambda }_m^{\infty }\left(AgN{O}_3\right)+{\lambda }_m^{\infty }\left(Nal\right)\right]-{\lambda }_m^{\infty }NaN{O}_3$                

= (13.3 + 12.7) - 12.0 = 14.0 m Sm² mol^-1

Each ion makes a definite contribution irrespective of the other ion

$\begin{array}{rr}& {\left({\mathrm{\Lambda }}_{\mathrm{m}}^0\right)}_{\mathrm{N}\mathrm{a}\mathrm{B}\mathrm{r}}-{\left({\mathrm{\Lambda }}_{\mathrm{m}}^0\right)}_{\mathrm{N}\mathrm{a}\mathrm{l}}={\mathrm{\Lambda }}_{{\mathrm{m}\mathrm{B}\mathrm{r}}^{-}}^0-{\mathrm{\Lambda }}_{{\mathrm{m}}_{{\mathrm{r}}^{-}}}^0\\ & {\left({\mathrm{\Lambda }}_{\mathrm{m}}^0\right)}_{\mathrm{K}\mathrm{B}\mathrm{r}}-{\left({\mathrm{\Lambda }}_{\mathrm{m}}^0\right)}_{\mathrm{N}\mathrm{a}\mathrm{B}\mathrm{r}}={\mathrm{\Lambda }}_{{\mathrm{m}\mathrm{K}}^{+}}^0-{\mathrm{\Lambda }}_{{\mathrm{m}\mathrm{N}\mathrm{a}}^{+}}^0\end{array}$

(a)  $Cd\left(s\right)+2Ni{\left(OH\right)}_3\left(s\right)\to CdO\left(s\right)+2Ni{\left(OH\right)}_2\left(s\right)+H_{2}O\left(\mathcal{l}\right)$ ${}_{}{}_{}{}_{}$

During discharging of secondary battery this reaction takes place.

(b) $Zn\left(Hg\right)+HgO\left(s\right)\to ZnO\left(s\right)+Hg\left(\mathcal{l}\right)$ $$

Primary battery mercury cell reaction

(c) $2PbS{O}_4\left(s\right)+2H_{2}O\left(\mathcal{l}\right)\to Pb\left(s\right)+Pb{O}_2\left(s\right)+2H_{2}S{O}_4\left(aq\right)$ ${}_{}{}_{}{}_{}{}_{}{}_{}$

During charging of secondary battery PbSO4 reacts and $H_{2}S{O}_4$ ${}_{}{}_{}$ generated

(d) $H_2\&O_2$ ${}_{}{}_{}$ reacts in fuel cell to form $H_{2}O\left(\mathcal{l}\right)$ ${}_{}$

Cu_{(s) $\underset{}{\overset{}{\to }}$} Cu⁺⁺ + 2e^- (Oxidⁿ at cathode)

$\frac{2A{g}^{+}+2e\underset{}{\overset{}{\to }}2Ag\left(s\right)}{C{u}_{\left(s\right)}+2A{g}^{+}\underset{}{\overset{}{\to }}2A{g}_{\left(s\right)}+C{u}^{++}}\left(Re{d}^{n}atanode\right)$

$Q=\frac{\left[C{u}^{++}\right]}{{\left[A{g}^{+}\right]}^2}andn=2$

0.43 =   $E_{cell}^o=\frac{0.06}{2}log\frac{\left(0.001\right)}{{\left(0.01\right)}^2}$

$\therefore E_{cell}^o=0.46$

$E_{cell}^o=E_{A{g}^{+}/Ag}^o-E_{C{u}^{++}/Cu}^o$

0.46 = 0.80 - $E_{c{u}^{++}/Cu}^o$

$\therefore E_{c{u}^{++}/Cu}^o=0.34Volt=34*10^{-2}$  

  $Anode:\left(Zn\to Z{n}^{2+}+2e\right)*x$

$Cathode:\underset{\_}{\left(S{n}^{x+}+xe\to Sn\right)*2}$      

Net call reaction :  $xZn+2S{n}^{x+}\to xZ{n}^{2+}+2Sn\left(n=2x\right)$

Using Nernst equation

No. of electron involved = 4

Molar mass of C₇H₅N₃O₆ = 12 * 7 + 1 * 5 + 14 * 3 + 16 * 6 = 84 + 42 + 96 = 227 g/mol

Number of moles = $\frac{681}{227}$  = 3 mol

$\therefore$ number of N-atoms

$=3*6.02*10^{23}*3=9*6.02*10^{23}$

$=5418*10^{21}$

$\therefore x=5418$         

$F{e}^{+3}+l^{-}\to I_2+F{e}^{+2}$

Fe⁺² undergoes reduction & I₂ undergoes oxidation.

$E_{cell}^o=E_{cathode}^o-E_{anode}^o$

= (0.77 – 0.54) V = 0.23 V

= 23 * 10^-2 V