NEET Chemistry Questions: Redox Reactions & Electrochemisty

B). 2F = 1 mole Fe deposited from Fe²⁺

1 F = $\frac{1}{2}$ mol Fe    deposited from F$e^{2+}=28g$

3 F = 1 mol Fe        deposited from F$e^{3+}$

1 F = $\frac{1}{3}$ mol Fe   deposited from F$e^{3+}=18.6g$  

(A). Specific conductance = Observed conductance $\times \frac{\mathcal{l}}{a}$

 $=oh{m}^{-1}\times \frac{cm}{c{m}^2}=oh{m}^{-1}c{m}^{-1}$                                 

$$\begin{gathered} \mathrm{\alpha } =\frac{\mathrm{\lambda m}}{{\mathrm{\lambda }}^0\mathrm{m}}=\frac{7.8}{390}=0.02 \\ {\mathrm{C\alpha }}^2=\mathrm{Ka}=0.047\times {(0.02)}^2=16\times {10}^{-6} \\ \mathrm{pKa}=-\log \mathrm{Ka}=-\log 16\times {10}^{-6} \\ 6-\log {\left(2\right)}^4=6-4\times 0.3=4.8 \\ \mathrm{pKa}=14-\mathrm{pKa}=14-4.8=9.2 \end{gathered}$$

C). This is based on the application of Faraday’s laws of electrolysis. One mole of water can be obtained by the combination of a mole of $H_2$ and $\frac{1}{2}$ mole of oxygen which may be produced by the passage of 2 Faradays.

(C). 1 mol H₂ gas at STP = $2F = 2 \times 96500 C$

$22400 c{m}^3 of H_2 gas at STP = 193 \times {10}^3 C$

$1 c{m}^{3 }of H_2$ gas at STP per second $= \frac{193\times {10}^3}{22400}= 8.61 amp.$

(B). After electrolysis aqueous NaCl is converted into aqueous NaOH.
The quantity of electricity passed

$= \frac{0.250 \times 35\times 60}{96500} F= 5.44 \times {10}^{-3}F$

The number of equivalents of $O{H}^{-}$ ion formed

                                                  $= 5.44 \times {10}^{-3}$

 $\therefore$ Molarity of NaOH =$\frac{5.44 \times {10}^{-3}}{0.4L}=1.36 \times {10}^{-2}$

 ∴ pOH = – log $(1.36 \times {10}^{-2}) = 1.87$

$\therefore$ pH = 12.13

(B). During the discharge of lead storage cell, sulphuric acid is consumed. Its concentration decreases and therefore, density decreases.

(A). For the given reaction

$E=E_{S{n}^{2+}|S{n}^{4+}}^{°}- E_{F{e}^{2+}, F{e}^{3+}=}^{°}0.15-0.77=-0.62 V$

C). Cell constant 

$=\frac{Specific conduc\tan ce}{Observed conduc\tan ce}=\frac{x}{y}c{m}^{-1}$

Specific conductance of $0.01 M H_{2}S{O}_4$

= Observed conductance $\times$ Cell constant$=z\times \frac{x}{y}oh{m}^{-1}c{m}^{-1}$

(B). Since Fluorine is the strongest oxidising agent so it will destroy the electrode employed.

(A) Cell reaction

Cathode : $H_{2}O\left(\mathcal{l}\right)+\frac{1}{2}{O}_2\left(g\right)+2e^{-}\to 2O{H}^{-}\left(aq\right)$

Anode : $H_2\left(g\right)\to 2H^{+}\left(aq.\right)+2e^{-}$

$H_{2}O\left(\mathcal{l}\right)+\frac{1}{2}{O}_2\left(g\right)+H_2\left(g\right)\to 2H^{+}\left(aq.\right)+2O{H}^{-}\left(aq.\right)$

Also we have

$$\begin{gathered} H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(\mathcal{l}\right) .......\left(1\right) \\ ∆G{°}_f=-237.2 kJ/mol \\ {H}_{2}O\left(\mathcal{l}\right)\to H^{+}\left(aq.\right)+O{H}^{-}\left(aq.\right) .......\left(2\right) \\ ∆G{°}_f=-237.2 kJ/mol \end{gathered}$$

Hence for cell reaction $\left[eq.\left(1\right)+eq\left(2\right)\times 2\right]$

$$\begin{gathered} ∆G°=-77.20kJ/mol \\ So, E°-\frac{∆G°}{nF}=\frac{77200}{2\times 96500}=0.40 V \end{gathered}$$

(C) At anode $Zn\to Z{n}^{2+}+2e^{-}$

 At cathode :$C{u}^{2+}+2e^{-}\to Cu$

Cell reaction : $Zn+C{u}^{2+}\to Z{n}^{2+}+Cu$

As electricity is withdrawn from the cell, the concentration of $C{u}^{2+}$ decreases while that of $Z{n}^{2+}$ increases

$$\begin{gathered} E_{cell }={E^{°}}_{cell }-\frac{0.059}{n}\log \frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]} \\ 1=1.103- \frac{0.059}{2}\log \frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]} \\ \therefore \frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}=3041 \end{gathered}$$

If x is the amount of $C{u}^{2+}$ that is converted to Cu when the cell potential drops to 1V then

$\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}=\frac{1+x}{1-x}=3041 ; x=\frac{3040}{3042}=0.9993$

To change $C{u}^{2+} to Cu and Zn to Z{n}^{2+}, 2$ electrons are requried, hence quantity of charge drawn from the cell 

$= 2 \times 0.9993 \times 96500 = 1.029 \times {10}^5 C.$

(B) Since $$\begin{gathered} B^{-}+e^{-}\to B^{-2} ; E°=1.25 ....\left(i\right) \\ having largest reduction potential and \\ {C}^{-3}\to C^{-}+2e^{-} ; E°=+1.25 ....\left(iii\right) \end{gathered}$$

having largest oxidation potential. So the cell having these two half cell reaction cell would result in maximum potential.

(C). pH changes from 0 to 7.

$\therefore \left[H^{+}\right]$changes from 1 to ${10}^{-7}$ M.

Accordingly $E_{red}$ decreases by 0.059 log ${10}^{-7}$

i.e. 0.059 × (–7) = – 0.41 volt.

 (D). Solubility product reaction is

$AgI\to A{g}^{+}+I^{-}$

By calculating the EMF of this cell reaction from the given data and relating to $K_{sp}$via the $∆G°$ of the reaction, we can obtain  $K_{sp}$.