(a) Calculate E^_{\;{cell}\;} for the ...

CBSE Class 12 Chemistry 2016 Outside Delhi Set 1 Solved Paper

Question : 24 of 26

Marks: +1, -0

(a) Calculate

E^\circ_{\;\text{cell}\;}

for the following reaction at 298K :

\;2 \mathrm{Al}(\mathrm{s}) + 3 \mathrm{Cu}^{2+}(0.01 \mathrm{M}) \rightarrow

2 \mathrm{Al}^{3+}(0.01 \mathrm{M}) + 3 \mathrm{Cu}(\mathrm{s})

\;\;\text{ Given }\;: E^\circ_{\;\text{cell}\;} = 1.98 \mathrm{V}

(b) Using the

E^\circ

values of

A

and

B

, predict which is better for coating the surface of iron [

E^\circ_{(\mathrm{Fe}^{2+}/\mathrm{Fe})}

= -0.44 \mathrm{V}]

to prevent corrosion and why ?

Given :

E^\circ_{(\mathrm{A}^{2+}/\mathrm{A})} = -2.37 \mathrm{V}

\ : \ E^\circ_{(\mathrm{B}^{2+}/\mathrm{B})} = -0.14 \mathrm{V}

(a) The conductivity of

0.001 \mathrm{mol} \mathrm{L}^{-1}

solution of

\mathrm{CH}_3\mathrm{COOH}

3.905 \times 10^{-5} \mathrm{S} \mathrm{cm}^{-1}

. Calculate its molar conductivity and degree of dissociation

(\alpha)

Given

\lambda^0 (\mathrm{H}^{+}) = 349.6 \mathrm{S} \mathrm{cm} \mathrm{mol}^{-1}

and

\lambda^0

(\mathrm{CH}_3\mathrm{COO}^{-}) = 40.9 \mathrm{S} \mathrm{cm}^2 \mathrm{mol}^{-1}

(b) Define electrochemical cell. What happens if external potential applied becomes greater than

E^\circ_{\;\text{cell}\;}

of electrochemical cell ?

Solution:

(a)

E_{\text{cell}} = E^0_{\text{cell}} - \frac{0.0591}{n} \log \frac{[\mathrm{Al}^{3+}]^2}{[\mathrm{Cu}^{2+}]^3}

E^0_{\text{cell}} = E_{\text{cell}} + \frac{0.0591}{n} \log \frac{[\mathrm{Al}^{3+}]^2}{[\mathrm{Cu}^{2+}]^3}

E^0_{\text{cell}} = 1.98 \mathrm{V} + \frac{0.0591}{6} \log \frac{(0.01)^2}{(0.01)^3} \;\; 1

E^0_{\text{cell}} = 1.98 \mathrm{V} + \frac{0.0591}{6} \log 10^2

E^0_{\text{cell}} = 1.98 \mathrm{V} + \frac{0.0591}{6}

\times 2 \times \log 10 \left[ \because \log 10 = 1 \right]

E^0_{\text{cell}} = 1.98 \mathrm{V} + \frac{0.0591 \mathrm{V}}{6} \times 2

E^0_{\text{cell}} = 1.98 \mathrm{V} + 0.0197 \mathrm{V}

E^0_{\text{cell}} = 1.9997 \mathrm{V}

(b)

A

, because its

E^0

value is more negative

(a)

\Lambda m = \kappa \times 1000 / \mathrm{C}

= \frac{3.905 \times 10^{-5} \times 1000}{0.001}

= 39.05 \mathrm{cm}^2 / \mathrm{mol}

\alpha = \frac{\Lambda_m}{\Lambda^0_m}

= \frac{39.05}{390.5}

= 0.1

\mathrm{CH}_3\mathrm{COOH} \rightarrow \mathrm{CH}_3\mathrm{COO}^{-} + \mathrm{H}^{+}

\Lambda^0_{\mathrm{CH}_3\mathrm{COOH}} = \lambda^0_{\mathrm{CH}_3\mathrm{COO}^{-}} + \lambda^0_{\mathrm{H}^{+}}

= 40.9 + 349.6

\Lambda^0_{\mathrm{CH}_3\mathrm{COOH}} = 390.5 \mathrm{Sm}^2 / \mathrm{mol}

(b) Device used for the production of electricity from energy released during spontaneous chemical reaction and the use of electrical energy to bring about a chemical change.

The reaction gets reversed / It starts acting as an electrolytic cell & vice - versa.

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