Electrochemistry MCQs for NEET — Chemistry Questions with Answers

Electrochemical cell ÅlA⁺ (xM)l l B⁺ (yM)l B⁺
The emf of cell is +0.20V.So,cell reaction is possible.The half cell reaction are given as follows:-

(i) At negative pole:- 
    A^→A⁺+e^-++e^- (Oxidation)

(ii) At positive pole:- 
    B⁺+e^-→B (Reduction)

Hence, cell reaction is
A+B⁺→A⁺+B^-,E°_cell=+0.20V

Cell reaction

Cathode: H₂O(l) + $\frac{1}{2}$O₂(g) + 2e^– $\underset{}{\overset{}{\to }}$ 2OH^–(aq.)

Anode: H₂(g) $\underset{}{\overset{}{\to }}$ 2H⁺ (aq.) + 2e^–

H₂O(l) + $\frac{1}{2}$O₂(g) $\underset{}{\overset{}{\to }}$ 2H⁺ (aq.) + 2OH^–(aq.)

Also we have

H₂(g)(l) + $\frac{1}{2}$O₂(g) $\underset{}{\overset{}{\to }}$ H₂O(l)ΔG°_f = –237.2 kj/mol

H₂O(l) $\underset{}{\overset{}{\to }}$ H⁺ (aq.) + OH^–(aq.)ΔG° = 80 kj/mol

Hence for cell reaction

ΔG° = –77.20 kj/mole

So, E = – $\frac{77200}{2\hspace{0.17em}\times \hspace{0.17em}96500}$ = 0.40 V

E_cell = – $\frac{0.0591}{1}\mathrm{\hspace{0.17em}}\log \mathrm{\hspace{0.17em}}\frac{[{\mathrm{H}}^{\mathrm{+}}{\mathrm{]}}_{\mathrm{a}}}{[{\mathrm{H}}^{\mathrm{+}}{\mathrm{]}}_{\mathrm{c}}}$

For E_cell to be highest [H⁺]_a should be lower and [H⁺]_c should be higher and that why anode compartment should be more basic and cathodic compartment should be acidic.

anode : Ag(s) $\underset{}{\overset{}{\to }}$ Ag+ (aq) + 1e^–

Cathode : Ag+ (aq) + 1e^– $\underset{}{\overset{}{\to }}$ Ag

Net : ${\mathrm{Ag}}_{\left(\mathrm{AgBr}\right)}^{\mathrm{+}}\mathrm{\hspace{0.17em}}\stackrel{1{\mathrm{e}}^{-}}{\to }\hspace{0.17em}{\mathrm{Ag}}_{({\mathrm{Ag}}_{\mathrm{2}}{\mathrm{Co}}_3\mathrm{)}}^{\mathrm{+}}$

0 = 0 + $\frac{0.059}{1}\mathrm{\hspace{0.17em}}\log \mathrm{\hspace{0.17em}}\frac{\left(\frac{{\mathrm{K}}_{\mathrm{SP}}\mathrm{AgBr}}{\left[{\mathrm{Br}}^{-}\right]}\right)}{\sqrt{\frac{{\mathrm{K}}_{\mathrm{SP}}{\mathrm{Ag}}_2{\mathrm{Co}}_3}{\left[{\mathrm{CO}}_3^{2-}\right]}}}$ ⇒ $\frac{{\mathrm{K}}_{\mathrm{SP}}\mathrm{AgBr}}{\left[{\mathrm{Br}}^{-}\right]}$ = $\sqrt{\frac{{\mathrm{K}}_{\mathrm{SP}}{\mathrm{Ag}}_2{\mathrm{Co}}_3}{\left[{\mathrm{CO}}_3^{2-}\right]}}$

⇒ $\frac{4\mathrm{\hspace{0.17em}}\times \mathrm{\hspace{0.17em}}{10}^{-13}}{\sqrt{8\mathrm{\hspace{0.17em}}\times \mathrm{\hspace{0.17em}}{10}^{-12}}}\mathrm{\hspace{0.17em}}=\mathrm{\hspace{0.17em}}\frac{\left[{\mathrm{Br}}^{-}\right]}{\sqrt{\left[{\mathrm{CO}}_3^{2-}\right]}}$ ⇒ $\frac{\left[{\mathrm{Br}}^{-}\right]}{\sqrt{\left[{\mathrm{CO}}_3^{2-}\right]}}$ = $\sqrt{2}\mathrm{\hspace{0.17em}}\times \mathrm{\hspace{0.17em}}{10}^{-7}$

The linear dependence of voltage on the log of the square root of hydrogen pressure and the cube root of M^(x+) concentration suggests that the balanced reaction involves the transfer of 3 electrons and the formation of 1.5 H2 molecules, which implies x = 3.

Reaction is H_c⁺ $\stackrel{1\mathrm{e}-}{\to }\hspace{0.17em}{{\mathrm{H}}_{\mathrm{A}}}^{\mathrm{+}}$

∴ E = $\frac{0.059}{\mathrm{1}}\mathrm{\hspace{0.17em}}\log \mathrm{\hspace{0.17em}}\frac{2.1\mathrm{\hspace{0.17em}}\times \mathrm{\hspace{0.17em}}{10}^{-4}}{1.8\mathrm{\hspace{0.17em}}\times \mathrm{\hspace{0.17em}}{10}^{-5}}$ = 0.0629 V 

${\mathrm{E}}_{{\mathrm{Br}}^{\mathrm{-}}/\mathrm{AgBr}/\mathrm{Ag}}^{\mathrm{0}}$ = ${\mathrm{E}}_{{\mathrm{Ag}}^{\mathrm{+}}/\mathrm{Ag}/\mathrm{Ag}}^{\mathrm{0}}\mathrm{\hspace{0.17em}}+\mathrm{\hspace{0.17em}}\frac{0.059}{1}$ log K_SPAgBr = ${\mathrm{E}}_{{\mathrm{Ag}}^{\mathrm{+}}/\mathrm{Ag}}^{\mathrm{0}}$ – 0.7257

and ${\mathrm{E}}_{{\mathrm{Cr}}^{\mathrm{-}}/\mathrm{AgCl}/\mathrm{Ag}}^{\mathrm{0}}$ = ${\mathrm{E}}_{{\mathrm{Ag}}^{\mathrm{+}}/\mathrm{Ag}}^{\mathrm{0}}\mathrm{\hspace{0.17em}}+\mathrm{\hspace{0.17em}}\frac{0.059}{1}$ log K_SPAgCl = ${\mathrm{E}}_{{\mathrm{Ag}}^{\mathrm{+}}/\mathrm{Ag}}^{\mathrm{0}}$ – 0.59

Now cell reaction is

Ag + Br^– $\stackrel{}{\to }$ AgBr + 1e^–

AgCl + 1e^– $\stackrel{}{\to }$ AgBr + Cl^–

Br^–AgCl $\stackrel{1e^{-}}{\to }$ Cl^– + AgBr 

0 = (0.7257–0.59) + $\frac{0.059}{1}\mathrm{\hspace{0.17em}}\log \mathrm{\hspace{0.17em}}\frac{[{\mathrm{Br}}^{-}\mathrm{]}}{[{\mathrm{Cl}}^{-}\mathrm{]}}$ 

⇒ $\frac{[{\mathrm{Br}}^{-}\mathrm{]}}{[{\mathrm{Cl}}^{-}\mathrm{]}}$ = 0.005

${\Lambda }_m={\Lambda }_m^{\infty }-b \sqrt{c}$

260 = ${\Lambda }_m^{\infty }$ – 0.5 b…

250 = ${\Lambda }_m^{\infty }$ – b … (2)

On solving & (2), we get

${\Lambda }_m^{\infty }$ = 270 Ω^– cm² mol^–1

AgCl(s) + e $\underset{}{\overset{}{\to }}$ Ag(s) + Cl^–E° = 22 V

1/₂Cl₂ + e $\underset{}{\overset{}{\to }}$ Cl^– E° = 1.36 V

We get

Ag(s) + $\frac{1}{2}$Cl₂(g) $\underset{}{\overset{}{\to }}$ AgCl(s)E°_cell = 1.14 V

∴ ΔG = –nEF° = – (96500) (1.4) = –110 KJ/mol

if reduction potential of metal ion is greater then O₂/H₂O couple, the ion is stable in water. So Co³⁺ is stable in water