NEET Chemistry Questions: Redox Reactions & Electrochemisty

$$\begin{gathered} \mathrm{For} \mathrm{concentration} \mathrm{cell} -{\mathrm{E}}_{\mathrm{cell}}^{\circ }=0 \\ {\mathrm{E}}_{\mathrm{cell}}=\frac{-0.0591}{2}\log \frac{[{\mathrm{H}}^{\oplus }{]}^2\mathrm{anode}}{[{\mathrm{H}}^{\oplus }{]}^2\mathrm{cathode}} \\ =\frac{-0.0591}{2}\log \frac{({10}^{-6}{)}^2}{({10}^{-4}{)}^2} \\ =(-4) \times \frac{-0.0591}{2}=0.118 \mathrm{V} \end{gathered}$$

 ${\mathrm{MnO}}_2 +\hspace{0.17em}{\mathrm{NH}}_4^{+} +\hspace{0.17em}\mathrm{e} \stackrel{ }{\to }\mathrm{MnO}\left(\mathrm{OH}\right) + {\mathrm{NH}}_3$

NH₃ furthur combines with Zn⁺² and forms $\mathrm{Zn}({\mathrm{NH}}_3{)}_4^{+2}$ 

$$\begin{gathered} \mathrm{W} = \frac{\mathrm{E}}{96500} \times \mathrm{i} \times \mathrm{t} = \frac{31.75}{96500} \times 5 \times 2 \times 3600 \\ \mathrm{W} = 11.84 \mathrm{g} \\ \% \mathrm{of} \mathrm{current} \mathrm{efficiency} = \frac{9.8}{11.84} \times 100 = 82.8\% \end{gathered}$$

(b) Key Idea The relation between molar conductivity (l_m) and electrolytic conductivity (k) is given as ${\mathrm{\lambda }}_{\mathrm{m}}=\frac{\mathrm{k}\times 1000}{\mathrm{M}}$

where, M is molarity of solution.

Given, concentration of solution, M=0.5mol/dm³

Electrolytic conductivity, k=5.76X10^-3Scm^-1 

Temperature, T=298K

$$\begin{gathered} \therefore \mathrm{Molar} \mathrm{conductivity}, \\ {\mathrm{\lambda }}_{\mathrm{m}}=\frac{\mathrm{k}\times 1000}{\mathrm{M}} \\ =\frac{5.76\times {10}^{-3}\times 1000}{0.5} \\ =11.52 \mathrm{S} {\mathrm{cm}}^2/\mathrm{mol} \end{gathered}$$

(b) Key Idea This problem is based on Faraday's first law of electrolysis which states that when an electric current is passed through an electrolytic solution, the amount of substance (w) deposited at the electrode, is proportional to the electric charge (q) passed through the electrolytic solution.

The formula used in the problem is

 w= Eit/96500         ...(i)

where, E = gram -equivalent mass of Cl-

 i= current, t= time, it = q

Given, w=0.10 mol = (0.10 x 71)g,

 i=3A, E=35.5

The following reactions occurred,

At cathode: 2H2O $\stackrel{+2{\mathrm{e}}^{-}}{\to }$H2 + 2OH-

At anode: 2Cl- $\stackrel{-2{\mathrm{e}}^{-}}{\to }$Cl2

              35.5g          71 g

Putting all values in expression (i) we get;

(0.10 x 71) = (35.5/96500)  x 3 x t or t = 6433s

or t = 107.22 min $\approx$ 110 min    [1s = 1/60 min]

(a) Given,  E°_cell = -ve

Now, the relation between $∆$G° and E°_cell is given as

$∆$G° =-nFE°_cell    ...(i)

as E°_cell is negative , so $∆$G° comes out to be positive. Again, relation between $∆$G° and K_eq is given as

$∆$G° = -2.303nRTlogK_eq      ......(ii)

From Eq. (i) we ge; that $∆$G° is positive. Now, if $∆$G° is positive then K_eq comes out to be negative from eq (ii).

i.e. $∆$G° >1 and K_eq< 1

Short trick: As $∆$G° is negative so reaction is non-spontaneous or you can say reaction is moving in backward direction. For non-spontaneous reaction, $∆$G° is always positive and K_eq is always less than 1.

(c) From Faraday's first law of electrolysis, 

W/E = it/96500   ......... (i)

Given i = 1A; t=60s

Putting these values in eq. (i) we get 

W/E = 1x60/96500

or W/E = 6/9650 = Number of mole of electrons

Number of electrons = 6/9650 x 6.022x 10²³ = 3.75 x 10²⁰

(d) The metal with higher negative standard reduction potential, have higher tendency to get reduced.

Zn²⁺ +2e^- $\to$ Zn;  E° = -0.76V

Fe²⁺ + 2e^- $\to$ Fe;  E° =-0.44 V

Here, In galvanised iron, Zn has higher negative reduction potential means Zn takes electrons given by iron and itself gets reduced.

Thus, Zn works as anode and protects iron from rusting by making iron as cathode.

(d) From the question, we have an equation

                     $2{\mathrm{H}}^{+}+ 2{\mathrm{e}}^{-} \stackrel{ }{\to }{\mathrm{H}}_2\left(\mathrm{g}\right)$

According to Nernst equation, 

$$\begin{gathered} \mathrm{E}=\mathrm{E}°-\frac{0.0591}{2}\log \frac{{\mathrm{P}}_{{\mathrm{H}}_2}}{{\left[{\mathrm{H}}^{+}\right]}^2} \\ =0-\frac{0.0591}{2}\log \frac{{\mathrm{P}}_{{\mathrm{H}}_2}}{{\left({10}^{-7}\right)}^2} \left[\therefore \left[{\mathrm{H}}^{+}\right]={10}^{-7}\right] \\ \therefore \mathrm{For} \mathrm{potential} \mathrm{of} {\mathrm{H}}_2 \mathrm{electrode} \mathrm{to} \mathrm{be} \mathrm{zero}, {{\mathrm{P}}_{\mathrm{H}}}_2 \mathrm{should} \mathrm{be} \mathrm{equal} \mathrm{to} {\left[{\mathrm{H}}^{+}\right]}^2, \mathrm{i}.\mathrm{e}. {10}^{-14}\mathrm{atm}. \\ \therefore \log \frac{{10}^{-14}}{{\left({10}^{-7}\right)}^2}=0 \end{gathered}$$

(a) Fuel cell is a device that converts energy of combustion of fuels like hydrogen and methane directly into electical energy.

Electrolytic cell converts electrical energy into chemical energy. Dynamo is an electrical generator
that produces direct current with the use of a commutator.

Ni-Cd cell is a type of rechargeable battery which consists of a cadmium anode and a metal gid containing NiO₂ acting as a cathode.

Molality of solution X = molality of solution Y
                               = 0.2 mol/kg

We know that, elevation in the boiling point ($∆$T_b) of a solution is proportional to the molal concentration of the solution i.e

$∆{\mathrm{T}}_{\mathrm{b}}\propto \mathrm{m} \mathrm{or} ∆{\mathrm{T}}_{\mathrm{b}}={\mathrm{K}}_{\mathrm{b}}\mathrm{m}$

where, m is the molality of the solution and K, is molal boiling point constant or ebullioscopic constant.

$\therefore$By elevation in boiling point relation $∆{\mathrm{T}}_{\mathrm{b}}={\mathrm{iK}}_{\mathrm{b}}\mathrm{m} \mathrm{or} ∆{\mathrm{T}}_{\mathrm{b}}\propto \mathrm{i}$

where, i is van't Hoff is factor

Since, $∆{\mathrm{T}}_{\mathrm{b}}$ of solution X is greater than ΔΤ_b, of solution Y

(Observed colligative property is greater than normal colligative property)

$\therefore$ i of solution X > i of solution Y

$\therefore$ Solution X undergoing dissociation.

(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