Chemistry MCQs for NEET — Practice Questions with Answers

(a) 

$$\begin{gathered} {\mathrm{Cu}}^{2+} + {\mathrm{e}}^{-} \stackrel{ }{\to }\mathrm{Cu}+ \\ E_1^0 = 0.15 \mathrm{V}, ∆{\mathrm{G}}_1^{\circ }={\mathrm{n}}_1{E}_1^{}\mathit{ }F \\ {\mathrm{Cu}}^{+} + {\mathrm{e}}^{-}\stackrel{ }{\to }\mathrm{Cu}, E_2^{\circ }\mathit{ }=0.50 \mathrm{V}, ∆{\mathrm{G}}_2^{\circ }=-{\mathrm{n}}_2{E}_2^{\circ }\mathit{ }F \\ {\mathrm{Cu}}^{2+} + 2{\mathrm{e}}^{-} \stackrel{ }{\to }\mathrm{Cu}, E_{}^{\circ }= ?\mathit{ }∆\mathrm{G}°=-\mathrm{n}E\mathit{°}F \end{gathered}$$

$$\begin{gathered} ∆\mathrm{G}°=∆{\mathrm{G}}_1^{\circ } + ∆{\mathrm{G}}_2^{\circ } \\ -\mathrm{nE}°\mathrm{F} = -{\mathrm{n}}_1{\mathrm{E}}_1^{\circ }\mathrm{F} - {\mathrm{n}}_2{\mathrm{E}}_2^{\circ }\mathrm{F} \\ \mathrm{or} -2\mathrm{E}°\mathrm{F} = -1\mathrm{F} \times 0.15 +(-1\mathrm{F} \times 0.50) \\ \mathrm{or} -2\mathrm{E}°\mathrm{F} = -0.15 \mathrm{F} - 0.50\mathrm{F} \\ \mathrm{or} -2\mathrm{FE}° = -\mathrm{F}(0.15 + 0.50) \\ \therefore \mathrm{E}°=\frac{0.65}{2}=0.325 \mathrm{V} \end{gathered}$$

(b) Zn is oxidise in place of iron and iron acts as cathode if Zn plating or galvanisation of iron is made and thus, also called sacrificial protection.

(c)

$$\begin{gathered} 2{\mathrm{H}}^{+}+2\mathrm{e} \stackrel{ }{\to }{\mathrm{H}}_2 \left(\mathrm{Cathode}\right) \\ 2{\mathrm{OH}}^{-}\stackrel{ }{\to }{\mathrm{H}}_2\mathrm{O} + \left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right){\mathrm{O}}_2 + 2\mathrm{e} \left(\mathrm{Anode}\right) \end{gathered}$$

Concept used: Faradays' first law

wt deposited at Electrode is given by

w= z x Q

w = Z x 2 x 96500                           [1 F=96500 c]

$$\begin{gathered} \mathrm{w}=\frac{\mathrm{F}}{96500}\times 2\times 96500 \\ =\mathrm{w}=\frac{63.5}{{\displaystyle \frac{2}{96500}}}\times 2\times 96500 \\ =1 \\ \mathrm{w}=\frac{63.5\times 2}{2} \\ \mathrm{w}=63.5 \mathrm{g} \end{gathered}$$

(b) Sodium chloride in water dissociates as

NaCl $\rightleftharpoons$  Na⁺ + Cl^-

H₂O $\rightleftharpoons$ H⁺ + OH^-

When electric current is passed through this solution using platinum electrodes, Na⁺ and H⁺ move towards cathode whereas Cl^- and OH^- ions move towards anode.

At cathode

H⁺ + e^- $\to$ H

H+H $\to$ H₂

At anode

Cl^- $\to$ C1 + e^-

Cl + Cl $\to$ Cl₂

If mercury is used as cathode, H⁺ ions are not discharged at mercury cathode because mercury has a high hydrogen over voltage. Na⁺ ions are discharged at cathode in preference of H⁺ ions yielding sodium, which dissolves in mercury to form sodium amalgam.

Concept used: Faraday's second law:

acc. to second law

$$\begin{gathered} =\frac{{\mathrm{w}}_1}{{\mathrm{w}}_2}=\frac{\mathrm{Eq} {\mathrm{wt}}_1}{\mathrm{Eq} {\mathrm{wt}}_2} \\ =\frac{{\mathrm{w}}_1}{\mathrm{Eq} {\mathrm{wt}}_1}=\frac{{\mathrm{w}}_2}{\mathrm{Eq} {\mathrm{wt}}_2} \\ =\frac{{\mathrm{w}}_{\mathrm{A}1}}{{\mathrm{Eq}}_{\mathrm{A}1}}=\frac{{\mathrm{W}}_{\mathrm{Na}}}{{\mathrm{Eq}}_{\mathrm{Na}}} \\ =\frac{1.8}{{\displaystyle \frac{27}{3}}}=\frac{{\mathrm{w}}_{\mathrm{Na}}}{{\displaystyle \frac{23}{1}}} \\ = {\mathrm{w}}_{\mathrm{Na}}=\frac{1.8}{9}\times 23 \\ ={\mathrm{w}}_{\mathrm{Na}}=\frac{18\times 23}{90}=4.6 \mathrm{g} \end{gathered}$$

Concept used: Faraday's second law  + products of electrolysis

given sol: ${\mathrm{AgNO}}_3, {\mathrm{CuSO}}_4, \mathrm{Al}{\left({\mathrm{NO}}_3\right)}_3$   and   NaCl

Note:

in case of $\mathrm{Al} {(\mathrm{NO}}_{}$₃)3 and NaCl,

Al and Na will not be deposited at their respective electrodes because their discharge potential is less than that of hydrogen hence instead of Al, Na,

${\mathrm{H}}_2$ gas will be evolved at cathode.

$$\begin{gathered} {\mathrm{AgNO}}_3 \to {\mathrm{Ag}}^{+}+{\mathrm{NO}}_3^{-} \\ 2{\mathrm{H}}_2\mathrm{O} \rightleftharpoons 2{\mathrm{H}}^{+}+2{\mathrm{O}}^{2-} \end{gathered}$$

at cathode:

$$\begin{gathered} {\mathrm{Ag}}^{+}+ {\mathrm{e}}^{-} \to \mathrm{Ag} \\ (1 \mathrm{F}) \end{gathered}$$

here:

1 mole ${\mathrm{e}}^{-}$ deposits 1 mole Ag

1F charge deposite $\to$ 1 Mol Ag

3F $\to$x

${\mathrm{x}}_1= 3 \mathrm{mol}$

Similarly for Cu

$$\begin{gathered} {\mathrm{Cu}}^{2+}+ 2{\mathrm{e}}^{-}\to \mathrm{Cu} \\ 2\mathrm{F} \mathrm{deposites} \to 1 \mathrm{mole} \\ 3\mathrm{F} \to \mathrm{x} \\ {\mathrm{x}}_2 =\frac{3}{2}\mathrm{mole} \\ {\mathrm{x}}_1:{\mathrm{x}}_2 \\ =3 \mathrm{mol}:\frac{3}{2}\mathrm{mol} \\ =3\times 2:\frac{3}{2}\times 2 (\mathrm{multiplied} \mathrm{by} 2 \mathrm{to} \mathrm{remove} \mathrm{fraction}) \\ =6:3:0:0 \end{gathered}$$

(b) In presence of Hg electrode preferential discharge of Na⁺ (in comparison to H⁺) occurs.

(b) ${\mathrm{E}}_{\mathrm{X}}^{°}$ = -1.2V, ${\mathrm{E}}_{\mathrm{Y}}^{°}$ = +0.5V, ${\mathrm{E}}_{\mathrm{Z}}^{°}$ = -3V, $\therefore$Z>X>Y [$\because$ higher the reduction potential, lesser the reducing power.]