Chemistry MCQs for NEET — Practice Questions with Answers

(d)

$$\begin{gathered} \mathrm{Since}, 22400\mathrm{mL} \mathrm{volume} \mathrm{is} \mathrm{occupied} \mathrm{by} 1 \mathrm{mole} \mathrm{of} {\mathrm{O}}_2 \mathrm{a} \mathrm{STP}. \\ \mathrm{Thus}, 5600 \mathrm{mL} {\mathrm{O}}_2 \mathrm{means} = \frac{5600}{2400}\mathrm{mol} {\mathrm{O}}_2 \\ = \frac{1}{4} \mathrm{mol} {\mathrm{O}}_2 \\ \therefore \mathrm{Weight} \mathrm{of} {\mathrm{O}}_2=\frac{1}{4}\times 32 = 8\mathrm{g} \\ \mathrm{According} \mathrm{to} \mathrm{problem}, \\ \mathrm{Equivalents} \mathrm{of} \mathrm{Ag}=\mathrm{Equivalents} \mathrm{of} {\mathrm{O}}_2 \\ = \frac{\mathrm{Weight} \mathrm{Ag}}{\mathrm{Equivalent} \mathrm{weight} \mathrm{of} \mathrm{Ag}} \\ =\frac{{\mathrm{W}}_{{\mathrm{O}}_2}}{\mathrm{Equivalent} \mathrm{weight} \mathrm{of} {\mathrm{O}}_2} \\ \frac{{\mathrm{W}}_{\mathrm{Ag}}}{{\mathrm{M}}_{\mathrm{Ag}}}=\frac{{\mathrm{W}}_{{\mathrm{O}}_2}}{{\mathrm{M}}_{{\mathrm{O}}_2}} \\ \mathrm{VF} \mathrm{VF} \\ \therefore \frac{{\mathrm{W}}_{\mathrm{Ag}}}{108}\times 1=\frac{8}{32}\times 4 \left[\because 2{\mathrm{H}}_2\mathrm{O} \stackrel{ }{\to }{\mathrm{O}}_2+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}\right] \\ \Rightarrow {\mathrm{W}}_{\mathrm{Ag}}=108\mathrm{g} \end{gathered}$$

According to Kohlrausch's law limiting molar conductivity of NH₄OH:-
Å_m(NH₄OH)=Å_m(NH₄Cl)+Å_m(NaOH)-Å_m(NaCl)

Given $∆$${\mathrm{G}}_{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}^{°}$ = -237.2

$∆{\mathrm{G}}_{{\mathrm{CO}}_2\left(\mathrm{g}\right)}^{°}$ = -394.4

$∆{\mathrm{G}}_{{\mathrm{C}}_5{\mathrm{H}}_{12\left(\mathrm{g}\right)}}^{°}$ = -8.2

C5H12 + 16O2 $\to$5CO2 + 6H2O

$∆$G⁰= 5 x $∆{\mathrm{G}}_{{\mathrm{CO}}_2\left(\mathrm{g}\right)}^{°}$+ 6 x ${\mathrm{G}}_{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}^{°}$-($∆{\mathrm{G}}_{{\mathrm{C}}_5{\mathrm{H}}_{12\left(\mathrm{g}\right)}}^{°}$+$∆{\mathrm{G}}_{{\mathrm{O}}_2}^{°}$)

=5x(-394.4)+6x(-237.2)-(-8.2)+0

=-3387 KJ/mol

In pentane oxygen fuel cell 32 electrons are involved.

-3387 x 10³ = 32 x 96500 x E°_cell 

E°_cell = -3387 x 10³ / 32x96500 = 1.0968V

Electrochemical cell ÅlA⁺ (xM)l l B⁺ (yM)l B⁺
The emf of cell is +0.20V.So,cell reaction is possible.The half cell reaction are given as follows:-

(i) At negative pole:- 
    A^→A⁺+e^-++e^- (Oxidation)

(ii) At positive pole:- 
    B⁺+e^-→B (Reduction)

Hence, cell reaction is
A+B⁺→A⁺+B^-,E°_cell=+0.20V

Cell reaction

Cathode: H₂O(l) + $\frac{1}{2}$O₂(g) + 2e^– $\underset{}{\overset{}{\to }}$ 2OH^–(aq.)

Anode: H₂(g) $\underset{}{\overset{}{\to }}$ 2H⁺ (aq.) + 2e^–

H₂O(l) + $\frac{1}{2}$O₂(g) $\underset{}{\overset{}{\to }}$ 2H⁺ (aq.) + 2OH^–(aq.)

Also we have

H₂(g)(l) + $\frac{1}{2}$O₂(g) $\underset{}{\overset{}{\to }}$ H₂O(l)ΔG°_f = –237.2 kj/mol

H₂O(l) $\underset{}{\overset{}{\to }}$ H⁺ (aq.) + OH^–(aq.)ΔG° = 80 kj/mol

Hence for cell reaction

ΔG° = –77.20 kj/mole

So, E = – $\frac{77200}{2\hspace{0.17em}\times \hspace{0.17em}96500}$ = 0.40 V

E_cell = – $\frac{0.0591}{1}\mathrm{\hspace{0.17em}}\log \mathrm{\hspace{0.17em}}\frac{[{\mathrm{H}}^{\mathrm{+}}{\mathrm{]}}_{\mathrm{a}}}{[{\mathrm{H}}^{\mathrm{+}}{\mathrm{]}}_{\mathrm{c}}}$

For E_cell to be highest [H⁺]_a should be lower and [H⁺]_c should be higher and that why anode compartment should be more basic and cathodic compartment should be acidic.

anode : Ag(s) $\underset{}{\overset{}{\to }}$ Ag+ (aq) + 1e^–

Cathode : Ag+ (aq) + 1e^– $\underset{}{\overset{}{\to }}$ Ag

Net : ${\mathrm{Ag}}_{\left(\mathrm{AgBr}\right)}^{\mathrm{+}}\mathrm{\hspace{0.17em}}\stackrel{1{\mathrm{e}}^{-}}{\to }\hspace{0.17em}{\mathrm{Ag}}_{({\mathrm{Ag}}_{\mathrm{2}}{\mathrm{Co}}_3\mathrm{)}}^{\mathrm{+}}$

0 = 0 + $\frac{0.059}{1}\mathrm{\hspace{0.17em}}\log \mathrm{\hspace{0.17em}}\frac{\left(\frac{{\mathrm{K}}_{\mathrm{SP}}\mathrm{AgBr}}{\left[{\mathrm{Br}}^{-}\right]}\right)}{\sqrt{\frac{{\mathrm{K}}_{\mathrm{SP}}{\mathrm{Ag}}_2{\mathrm{Co}}_3}{\left[{\mathrm{CO}}_3^{2-}\right]}}}$ ⇒ $\frac{{\mathrm{K}}_{\mathrm{SP}}\mathrm{AgBr}}{\left[{\mathrm{Br}}^{-}\right]}$ = $\sqrt{\frac{{\mathrm{K}}_{\mathrm{SP}}{\mathrm{Ag}}_2{\mathrm{Co}}_3}{\left[{\mathrm{CO}}_3^{2-}\right]}}$

⇒ $\frac{4\mathrm{\hspace{0.17em}}\times \mathrm{\hspace{0.17em}}{10}^{-13}}{\sqrt{8\mathrm{\hspace{0.17em}}\times \mathrm{\hspace{0.17em}}{10}^{-12}}}\mathrm{\hspace{0.17em}}=\mathrm{\hspace{0.17em}}\frac{\left[{\mathrm{Br}}^{-}\right]}{\sqrt{\left[{\mathrm{CO}}_3^{2-}\right]}}$ ⇒ $\frac{\left[{\mathrm{Br}}^{-}\right]}{\sqrt{\left[{\mathrm{CO}}_3^{2-}\right]}}$ = $\sqrt{2}\mathrm{\hspace{0.17em}}\times \mathrm{\hspace{0.17em}}{10}^{-7}$

The linear dependence of voltage on the log of the square root of hydrogen pressure and the cube root of M^(x+) concentration suggests that the balanced reaction involves the transfer of 3 electrons and the formation of 1.5 H2 molecules, which implies x = 3.

Reaction is H_c⁺ $\stackrel{1\mathrm{e}-}{\to }\hspace{0.17em}{{\mathrm{H}}_{\mathrm{A}}}^{\mathrm{+}}$

∴ E = $\frac{0.059}{\mathrm{1}}\mathrm{\hspace{0.17em}}\log \mathrm{\hspace{0.17em}}\frac{2.1\mathrm{\hspace{0.17em}}\times \mathrm{\hspace{0.17em}}{10}^{-4}}{1.8\mathrm{\hspace{0.17em}}\times \mathrm{\hspace{0.17em}}{10}^{-5}}$ = 0.0629 V 

${\mathrm{E}}_{{\mathrm{Br}}^{\mathrm{-}}/\mathrm{AgBr}/\mathrm{Ag}}^{\mathrm{0}}$ = ${\mathrm{E}}_{{\mathrm{Ag}}^{\mathrm{+}}/\mathrm{Ag}/\mathrm{Ag}}^{\mathrm{0}}\mathrm{\hspace{0.17em}}+\mathrm{\hspace{0.17em}}\frac{0.059}{1}$ log K_SPAgBr = ${\mathrm{E}}_{{\mathrm{Ag}}^{\mathrm{+}}/\mathrm{Ag}}^{\mathrm{0}}$ – 0.7257

and ${\mathrm{E}}_{{\mathrm{Cr}}^{\mathrm{-}}/\mathrm{AgCl}/\mathrm{Ag}}^{\mathrm{0}}$ = ${\mathrm{E}}_{{\mathrm{Ag}}^{\mathrm{+}}/\mathrm{Ag}}^{\mathrm{0}}\mathrm{\hspace{0.17em}}+\mathrm{\hspace{0.17em}}\frac{0.059}{1}$ log K_SPAgCl = ${\mathrm{E}}_{{\mathrm{Ag}}^{\mathrm{+}}/\mathrm{Ag}}^{\mathrm{0}}$ – 0.59

Now cell reaction is

Ag + Br^– $\stackrel{}{\to }$ AgBr + 1e^–

AgCl + 1e^– $\stackrel{}{\to }$ AgBr + Cl^–

Br^–AgCl $\stackrel{1e^{-}}{\to }$ Cl^– + AgBr 

0 = (0.7257–0.59) + $\frac{0.059}{1}\mathrm{\hspace{0.17em}}\log \mathrm{\hspace{0.17em}}\frac{[{\mathrm{Br}}^{-}\mathrm{]}}{[{\mathrm{Cl}}^{-}\mathrm{]}}$ 

⇒ $\frac{[{\mathrm{Br}}^{-}\mathrm{]}}{[{\mathrm{Cl}}^{-}\mathrm{]}}$ = 0.005