Electrochemistry

Λ? = K * 1000 / M = (2 * 10? * 10³) / 10? ³ = 20 Scm²mole? ¹
For weak acid (α) = Λ? / Λ? = 20 / 190 = 2/19
K? = Cα² / (1 - α) = 10? ³ * (2/19)² / (1 - 2/19)
= 10? ³ * (4/361) / (17/19) = 0.01238 * 10? ³
= 12.38 * 10?
So, ans is 12 (the nearest integer).

$E_{cell}={\left(E_{RHS}^o-E_{LHS}^o\right)}_{RP}-\frac{0.0591}{n}lo{g}_{10}Q$

$Q=\frac{\left[Z{n}^{++}\right]}{\left[C{u}^{++}\right]}=\frac{0.04}{0.02}andn=2$

$\therefore E_{cell}=\left(0.34+0.76\right)-\frac{0.0591}{2}lo{g}_{10}\frac{0.04}{0.02}$

$=1.1-\frac{0.0591}{2}*0.3010=1.091V=109*10^{-2}V$

$F{e}^{+3}+3e^{-}\to Fe$  

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

${\wedge }_M=\frac{K*1000}{M}$ $M_2=\frac{20}{\left(80/1000\right)}$

$\frac{{\wedge }_{M_1}}{{\wedge }_{M_2}}=\frac{M_2}{M_1}$ $M_1=\frac{10}{\left(20/1000\right)}$

$\frac{{\wedge }_{M_1}}{{\wedge }_{M_2}}=\frac{1}{2}\Rightarrow {\wedge }_{M_2}=2{\wedge }_{M_1}$

(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 $H_{2}O\left(\mathcal{l}\right)$ ${}_{}$

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

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

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

Using Nernst equation

No. of electron involved = 4

Anode, H₂ (g) -> 2H⁺ + 2e^-

Cathode, Cu²⁺ + 2e -> Cu (s)

Net cell reaction H₂ + Cu²⁺ -> 2H⁺ Cu (s)

$E_{cell}^0=E_{C{u}^{2+}/Cu}^0-E_{H^{+}/H_2}^0$

= 0.34 V – 0

= 0.34 V

$E_{cell}=E_{cell}^0-\frac{0.06}{2}log\frac{{\left[H^{+}\right]}^2}{\left[C{u}^{2+}\right]}$

$\begin{array}{l}\Rightarrow 0.576=0.34-\frac{0.06}{2}log\frac{{\left[H^{+}\right]}^2}{0.01}\\ \Rightarrow -log\left[H^{+}\right]=4.93\\ \Rightarrow pHofsolution=4.93~5\end{array}$

Antiseptic Dettol is mixture of chloroxylenol and terpineol

Brown ring is formed due to nitrosoferrous sulphate formation.