Electrochemistry MCQs for NEET — Chemistry Questions with Answers

(d) 

The Nernst equation is an important relation which is used to determine reaction equilibrium constants and concentration potentials

(b) Specific conductivity, k=0.012${\mathrm{\Omega }}^{-1}{\mathrm{cm}}^{-1}$

              $$\begin{gathered} \mathrm{k}=\frac{1}{\mathrm{resistance}}\times \frac{1}{\mathrm{a}}\left[\mathrm{G}=\frac{1}{\mathrm{R}}\right] \\ \frac{1}{\mathrm{a}}=\mathrm{cell} \mathrm{constant} \\ \frac{1}{\mathrm{a}}=55 \times 0.012 = 0.66{\mathrm{cm}}^{-1} \end{gathered}$$

(a) $2Mn2+ \to \; 2Mn3+ +\; 2e \; \; \; E0OP\; =\; -1.51\; V$

$2e\; +\; Mn2+ \to \; Mn$

$After\; solving\; these\; equations$

$3Mn2+ \to \; 2Mn3+ +\; Mn \; \; \; E0Cell\; =\; -2.69\; V$

$Negative\; value\; of\; E0Cell suggest\; for\; non-spontaneous\; nature.$

(a) ${E^0}_{OP{Z}_n}>{E^0}_{O{P}_{Fe}}$

(c) $E_{\mathrm{cell}} = E_{\mathrm{cell}}^{\circ } + \frac{0.059}{2}\log \frac{{\left[{\mathrm{Ag}}^{+}\right]}^2}{\left[{\mathrm{Sn}}^{2+}\right]}$

Concept used: Faradays first law.

wt of sub deposited is given by :

w = 2 x Q

w = 2 x i x t       (i always in ampere t always in sec)

$$\begin{gathered} \mathrm{w}=\frac{\mathrm{Eq} \mathrm{wt}}{96500}\times \mathrm{i}\times \mathrm{t} \\ =\frac{\mathrm{Mol} \mathrm{wt}}{{\displaystyle \frac{\mathrm{n}-\mathrm{factor} }{86500}}}\times \mathrm{i} \times \mathrm{t} \\ =\frac{65}{{\displaystyle \frac{2}{96500}}}\times 5\times (40\times 60) \\ =\frac{65\times 5\times 4\times 6}{2\times 965} \\ =\frac{65\times 60}{965} \\ = 4 . \left(\mathrm{something}\right) \end{gathered}$$

$\Rightarrow \mathrm{hence}: \left(\mathrm{c}\right) 4.065 \mathrm{g} \mathrm{is} \mathrm{correct}.$

Concept used: Faraday's first law:

w = 2 x Q

$$\begin{gathered} =\mathrm{w}=\frac{\mathrm{E}}{96500}\times \mathrm{i}\times \mathrm{t} \\ =\frac{27}{{\displaystyle \frac{3}{96500}}}\times 4\times 10\times 6\times 60\times 60 [\mathrm{eq} \mathrm{wt} \mathrm{of} \mathrm{AT}: 27 \mathrm{valancy}: 3] \\ =\frac{27\times 4\times 10\times 6\times 6\times 6}{3\times 965} \\ =\frac{108\times 36\times 2\times {10}^4}{965} \\ =\frac{7776}{965} \\ =8\times {10}^4 \mathrm{g} (\mathrm{app}.) \end{gathered}$$

(a) Zn has more +ve E°_OP and thus possesses more tendency to get oxidise and act as reducingt agent.

it prevents rusting because it provides cathdic protection (b) prevent action of water and salt. Several blocks of magnesium are fixed to the bottom of a ship to. Magnesium, on account of its lightness, great affinity for oxygen and toughness is used in ship

(a) 

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

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