Chemistry MCQs for NEET — Practice Questions with Answers

(d) Large negative RP or more positive oxidation potential and thus, more is the tendency to get oxidized.

Specific conductance Decreases with dilution since number of ions per unit volume decrease with dilution.

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) 

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

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

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