Electrochemistry MCQs for NEET — Chemistry Questions with Answers

When NaCl solution is electrolysed then NaOH is formed at cathode. Hence, pH of solution increase.

Nernst equation at 25$°$, where as E_cell = E$°$_cell 0.059 |nlog(A_ox/A_red)

As Cu²⁺ is reductant and its conc. is getting reduced by 100 times, so the log value will increase by the factor of 2. Also the value of n is 2.

${\mathrm{E}}_{\mathrm{cell}} = {\mathrm{E}}^{\circ } - \frac{0.0591}{2}\log \frac{\left[{\mathrm{Cu}}^{+2}\right]}{[{\mathrm{Ag}}^{+}{]}^2}$

By increasing the $\left[\stackrel{\oplus }{\mathrm{Ag}}\right]$ concentration twice E_cell can be increased the most.

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}^{\circ } ={\mathrm{E}}_{\raisebox{1ex}{${\mathrm{Cu}}^{+2}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{Cu}$}\right.}^{\circ } - {\mathrm{E}}_{\raisebox{1ex}{${\mathrm{Fe}}^{+2}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{Fe}$}\right.}^{\circ } \\ =0.34 - (-0.44) = 0.78 \mathrm{V} \\ ∆\mathrm{G}° = -\mathrm{n} {\mathrm{FE}}_{\mathrm{cell}}^{°} \\ =-2 \mathrm{X} 96500 \mathrm{X} 0.78 \\ = -150540 \mathrm{J} \\ = -150.54 \mathrm{KJ} \end{gathered}$$

${\mathrm{H}}_2 \rightleftharpoons 2 {\mathrm{H}}^{+}$+ 2 e^-

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}} = {\mathrm{E}}_{\mathrm{cell}}^{\circ }-\left(\frac{0.0591}{\mathrm{n}}\log \frac{{\left[{\mathrm{H}}^{\oplus }\right]}_{\mathrm{anode}}^2}{{\left[{\mathrm{H}}^{\oplus }\right]}_{\mathrm{cathod}}^2}\right) \\ {\mathrm{E}}_{\mathrm{cell}} =0-\left(\frac{0.0591}{2}\log \frac{{\left(0.02\right)}^2}{{\left(0.01\right)}^2}\right) \\ =-\frac{0.0591}{2}\times \left(0.6\right) \\ =- 0.017 \mathrm{V} \end{gathered}$$

(b) 

$$\begin{gathered} {\mathrm{Zn}}^{+2} + 2{\mathrm{e}}^{-} \stackrel{ }{\to }\mathrm{Zn} \\ {\mathrm{E}}_{\raisebox{1ex}{${\mathrm{Zn}}^{+2}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{Zn}$}\right.} = {\mathrm{E}}_{{}_{\raisebox{1ex}{${\mathrm{Zn}}^{+2}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{Zn}$}\right.}}^{\circ } - \frac{0.0591}{2}\log \frac{1}{\left[{\mathrm{Zn}}^{+2}\right]} \\ \because \left[{\mathrm{Zn}}^{+2}\right] = 0.1 \times \frac{25}{100} = 0.025 \mathrm{M} \\ \therefore {\mathrm{E}}_{\raisebox{1ex}{${\mathrm{Zn}}^{+2}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{Zn}$}\right.} = -0.76-\frac{0.0591}{2}\log \frac{1}{0.025} \\ = -0.76-\frac{0.0591}{2}\times 1.602 \\ = -0.807 \mathrm{V} \\ = -0.81 \mathrm{V} \end{gathered}$$

$$\begin{gathered} 2{\mathrm{H}}^{+} + 2{\mathrm{e}}^{-} \stackrel{ }{\to }{\mathrm{H}}_2 \\ {\mathrm{E}}_{\mathrm{Red}} = -\frac{0.0591}{2}\log \frac{1}{[{\mathrm{H}}^{\oplus }{]}^2} \\ \Rightarrow -0.118 = \frac{0.0591}{2}\log [{\mathrm{H}}^{\oplus }{]}^2 \\ \Rightarrow \log [{\mathrm{H}}^{\oplus }] = \frac{-0.118}{0.0591}=-2 \\ \Rightarrow [{\mathrm{H}}^{\oplus }] = 0.01 \end{gathered}$$

$$\begin{gathered} {\mathrm{W}}_{\max } = -∆\mathrm{G}° = \mathrm{nFE}° \\ \mathrm{E}° = {\mathrm{E}}_{\mathrm{cathode}}^{°} - {\mathrm{E}}_{\mathrm{anode}}^{} \\ = 0.34 - (-0.76) = 1.1 \mathrm{V} \\ \therefore {\mathrm{W}}_{\max } =2 \times 96500 \times 1.1 \\ =212300 \mathrm{J} \\ =212.3 \mathrm{KJ} \end{gathered}$$

${\mathrm{H}}_2{\mathrm{O}}_2 \stackrel{ }{\to }{\mathrm{O}}_2 + 2{\mathrm{H}}^{\oplus } +\hspace{0.17em}2{\mathrm{e}}^{-}$

For 1 mol H₂O₂ required charge = 2 mol e^-

                                                = 2 F

                                                = 2 X 96500

                                                = 193000 C