1. ⇒ (MHT CET 2023 11th May Morning Shift )

Find the number of faradays of electricity required to produce $45 g$ of $Al$ from molten ${Al}_{2} O_{3}$ .

(At. mass of $Al = 27$ )

A. 1 F

B. 3 F

C. 5 F

D. 7 F

Correct Option is (C)

${Al}^{3 +} + 3 e^{-} ⟶ Al$

The equation shows that 3 moles of electrons are required to produce 1 mole of $Al$ (i.e., $27 g$ of $Al$ ).

$∴ 3 F$ of electricity is required to produce $27 g$ of $Al$ from molten ${Al}_{2} O_{3}$ .

$∴$ Faradays of electricity required to produce $45 g$ of $Al = \frac{3}{27} \times 45 = 5 F$

2. ⇒ (MHT CET 2023 11th May Morning Shift )

Which among the following is CORRECT formula for determination of cell constant?

A. $\frac{l}{a} = \frac{k}{R}$

B. $\frac{l}{a} = k \cdot R$

C. $\frac{l}{a} = \frac{R}{k}$

D. $\frac{l}{a} = \frac{1}{R}$

Correct Option is (B)

$\begin{array}{ll} k = G \frac{l}{a} = \frac{1}{R} \frac{l}{a} \\ ∴ & Cell constant = \frac{l}{a} = k \cdot R \end{array}$

3. ⇒ (MHT CET 2023 10th May Morning Shift )

What mass of $Mg$ is produced during electrolysis of molten ${MgCl}_{2}$ by passing $2 amp$ current for 482.5 second?

(Molar mass $Mg = 24 g {mol}^{- 1}$ )

A. 0.12 g

B. 0.24 g

C. 1.2 g

D. 0.4 g

Correct Option is (A)

$\begin{aligned} {Mg}_{(s)}^{2 +} + 2 e^{-} ⟶ {Mg}_{(s)} \\ Mole ratio = \frac{1 mol}{2 {mole}^{-}} \\ W = \frac{I (A) \times t (s)}{96500 (C / mol e^{-})} \times mole ratio \times molar mass \\ W = \frac{2 \times 482.5}{96500 (C / mol e^{-})} \times \frac{1 mol}{2 mol e^{-}} \times 24 g {mol}^{- 1} \\ W = 0.12 g \end{aligned}$

4. ⇒ (MHT CET 2023 9th May Morning Shift )

Calculate current in ampere required to deposit $4.8 g Cu$ from it's salt solution in 30 minutes. [Molar mass of $Cu = 63.5 g {mol}^{- 1}$ ]

A. 8.1 ampere

B. 6.4 ampere

C. 10.5 ampere

D. 12.3 ampere

Correct Option is (A)

$\begin{aligned} {Cu}_{(s)}^{2 +} + 2 e^{-} ⟶ {Cu}_{(s)} \\ Mole ratio = \frac{1 mol}{2 {mole}^{-}} \\ W = \frac{I (A) \times t (s)}{96500 (C / mol e^{-})} \times mole ratio \times molar mass \\ 4.8 g = \frac{I (A) \times 30 \times 60}{96500 (C / mol e^{-})} \times \frac{1 mol}{2 mol e} \times 63.5 g {mol}^{- 1} \\ I (A) = \frac{4.8 \times 96500 \times 2}{63.5 \times 30 \times 60} = 8.1 A \end{aligned}$

5. ⇒ (MHT CET 2021 21th September Evening Shift )

What is the number of moles of electrons passed when current of 5 ampere is passed through a solution of FeCl $_{3}$ for 20 minutes?

A. $6.25 \times 10^{- 2}$

B. $1.56 \times 10^{- 2}$

C. $3.12 \times 10^{- 2}$

D. $4.25 \times 10^{- 2}$

Correct Option is (A)

$\begin{aligned} I = 5 A, t & = 20 \min = 20 \times 60 = 1200 s \\ Q & = I \times t = 5 A \times 1200 s \\ = 6000 C \end{aligned}$

Moles of electrons actually passed

$\begin{aligned} = \frac{Q (C)}{965000 (C / mol e^{-})} \\ = \frac{6000}{96500} \\ = 6.22 \times 10^{- 2} mol e^{-} \end{aligned}$

⇒Electrochemistry