(a) Conductivity of 2 × 10^{-3} {M} ...

CBSE Class 12 Chemistry 2023 Outside Delhi Set 1 Solved Paper

Question : 35 of 35

Marks: +1, -0

(a) Conductivity of

2 \times 10^{-3} \mathrm{M}

methanoic acid is 8

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

. Calculate its molar conductivity and degree of dissociation if

\Lambda^{0}_{m}

for methanoic acid is

404 \mathrm{S} \mathrm{cm}^2 \mathrm{mol}^{-1}

(b) Calculate the

\Delta_{G}^{0}

and

\log K_c

for the given reaction at

298 \mathrm{K}

\mathrm{Ni}_{(s)} + 2 \mathrm{Ag}_{(aq)}^{+} \rightleftharpoons \mathrm{Ni}_{(laq)}^{2+} + 2 \mathrm{Ag}_{(s)}

\text{Given : }E^0 \mathrm{Ni}^{2+} / \mathrm{Ni} = -0.25 \mathrm{V}

E^0_{\mathrm{Ag}^{+}/\mathrm{Ag}} = +0.80 \mathrm{V}

1 \mathrm{F} = 96500 \mathrm{C} \mathrm{mol}^{-1}

Solution: 👈: Video Solution

(a) Molar conductivity

\Lambda_{m}=\frac{\kappa \times 1000}{C}

=\frac{8 \times 10^{-5} \mathrm{S} \mathrm{cm}^{-1} \times 1000}{2 \times 10^{-3} \mathrm{mol} \mathrm{L}^{-1}}

=\frac{8 \times 10^{-2}}{2 \times 10^{-3}}=40 \mathrm{S} \mathrm{cm}^2 \mathrm{mol}^{-1}

Degree of dissociation

\frac{\Lambda_{m}}{\Lambda^{\circ}_{m}}=\frac{40}{404}=0.099

(b)

\mathrm{Ni} \rightarrow \mathrm{Ni}^{2+} (E_0 = -0.25 \mathrm{V})

( Oxidation half)

2 \mathrm{Ag}^{+} \rightarrow 2 \mathrm{Ag} (E_0 = 0.80 \mathrm{V})

(Reduction half)

E^{\circ} = E_c - E_a

= 0.80 - (-0.25) = 1.05 \mathrm{V}

\Delta G = n F E^{\circ}

= 2 \times 96500 \times 10.5

= 202.650 \mathrm{J} \mathrm{mol}^{-1}

= 202.650 \mathrm{kJ} \mathrm{mol}^{-1}

E^{\circ}_{\text{cell}} = \frac{0.0591}{n} \log K_c

\log K_c = \frac{1.05 \times 2}{0.0591} = 35.53

By taking Antilog

Antilog 35.35

= 10^{53} \times 3.38

\because \text{So } K_c = 3.38 \times 10^{53}

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