Redox Reactions and Electrochemistry — CBSE Class 12 · 2019 Board Question with Solution

(a) A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series, containing electrolytes FeSO

_{4}

and ZnSO

_{4}

, until 2.8 g of Fe deposited at the cathode of cell X. How long did the current flow? Calculate the mass of Zn deposited at the cathode of cell Y. (Molar mass: Fe = 56 g mol

^{- 1}

, Zn = 65.3 g mol

^{- 1}

, 1 F = 96500 C mol

^{- 1}

)
(b) In the plot of molar conductivity (Λ_m) vs square root of concentration (c

^{1/2}

), the following curves are obtained for two electrolytes A and B (see figure). Answer the following:
(i) Predict the nature of electrolytes A and B.
(ii) What happens on extrapolation of Λ_m to concentration approaching zero for electrolytes A and B?

Structure Molar conductivity vs square root of concentration for electrolytes A and B — Molar conductivity vs square root of concentration for electrolytes A and B

Solution

(a) Moles of Fe:

n = \frac{2.8}{56} = 0.05 mol

^{2 +}

+ 2e

^{-} \to

Fe, so charge:

Q = 0.05 \times 2 \times 96500 = 9650 C

Time:

t = \frac{9650}{2}

t = 4825 s \approx 80.4 min

Series cells pass the same charge; Zn

^{2 +}

+ 2e

^{-} \to

Zn also needs 2 electrons, so moles of Zn = 0.05:

m_{Z n} = 0.05 \times 65.3

m_{Z n} = 3.27 g

(b)(i) A, varying linearly with a gentle slope, is a strong electrolyte; B, shooting up steeply as

c \to 0

, is a weak electrolyte.
(ii) For A, extrapolating the straight line to

c^{1/2} = 0

gives the limiting molar conductivity

Λ_{m}^{\circ}

. For B the curve rises too steeply for extrapolation — its

Λ_{m}^{\circ}

must be obtained from Kohlrausch's law of independent migration of ions.

CBSE Class 12 Board · 2019 · Subjective

CBSE Class 12 · 2019 · Redox Reactions and ElectrochemistryLong Answer

Solution

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