Correct Answer is Option (C)

During the electrolysis of dil. sulphuric acid using Pt electrodes following reaction will take place.
At cathode :
$4H_{(aq)}^{+}+4e^{-}\to 2H_{2(g)}$
At anode :
$2H_2{O}_{(I)}\to O_{2(g)}+4H_{(aq)}^{+}+4e^{-}$
At the anode oxygen gas will be released.

Correct Answer is Option (D)

1 equivalent of any substance is deposited by 1F of charge.
We have 20g calcium.
No. of equivalent = $\frac{givenmass}{equivalentmass}$
Ca²⁺ + 2e^$-$ $\to$ Ca(s)
v.f. = 2
According to Faraday's 1st law
Charge passed in Faradey = g. equivalent of product
= $\frac{20}{40}\times 2$ = 1 F
So, 1 F of charge is required.

Correct Answer is Option (B)

At cathode : 2Na⁺ + 2e^– $\to$ 2Na

At anode : 2Cl^– $\to$ Cl₂ + 2e^–
----------------------------------------------

Net reaction: 2Na⁺ + 2Cl^– $\to$ 2Na + Cl₂

From Faraday’s first law of electrolysis,

w = Z$\times$I$\times$t

= $\frac{E}{96500}$$\times$I$\times$t

No. of moles of Cl₂ gas × Mol. wt. of Cl₂ gas

= $\frac{Eq.wt.ofC{l}_{2}gas\times I\times t}{96500}$

$\Rightarrow$ 0.10 $\times$ 71 = $\frac{35.5\times 3\times t}{96500}$

$\Rightarrow$ t = $\frac{0.10\times 71\times 96500}{35.5\times 3}$

= 6433.33 sec

= 107.22 min $\simeq$ 110 min

Correct Answer is Option (C)

Q = I × t

Q = 1 × 60 = 60 C

Now, 1.60 × 10^–19 C $\equiv$ 1 electron

$\therefore$ 60 C $\equiv$ $\frac{60}{1.6\times {10}^{-19}}$

= 3.75 $\times$ 10²⁰ electrons

Correct Answer is Option (C)

$\underset{0.1mole}{\overset{-6}{Mn{O}_4^{2-}}}$ $\to$ $\stackrel{-7}{Mn{O}_4^{-}}$ + e^-

Quantity of electricity required = 0.1F

= 0.1 × 96500 = 9650 C

Correct Answer is Option (D)

We know, from Faraday’s second law

$\frac{W_{Ag}}{E_{Ag}}=\frac{W_{O_2}}{E_{O_2}}$

$\Rightarrow$ $\frac{W_{Ag}}{108}=\frac{\frac{5600}{22400}\times 32}{8}$

$\Rightarrow$ $\frac{W_{Ag}}{108}=\frac{8}{8}$

$\Rightarrow$ $W_{Ag}=108$ g

Correct Answer is Option (B)

W = $\frac{IEt}{96500}$

= $\frac{10\times 109\times 60\times 59}{96500}$

= 20

Correct Answer is Option (A)

E = Z × 96500

$\Rightarrow$ $\frac{27}{3}$ = Z $\times$ 96500

$\Rightarrow$ Z = $\frac{9}{96500}$

Now applying the formula, W = Z × I × t

W = $\frac{9}{96500}$ $\times$ 4 $\times$ 10⁴ $\times$ 6 $\times$ 60 $\times$ 60

= 8.1 × 10⁴ g

Correct Answer is Option (D)

Faraday second law of electrolysis

$\frac{M_{Al}}{M_H}=\frac{E_{Al}}{E_H}$

$\Rightarrow$ $\frac{4.5}{M_H}=\frac{\frac{27}{3}}{1}$

$\Rightarrow$M_H = 0.5 g

We know, Volume of 2 g H₂ at STP = 22.4 L

$\therefore$ Volume of 0.5 g H₂ at STP

= $\frac{22.4\times 0.5}{2}$ = 5.6 L

Correct Answer is Option (B)

In electrolysis of NaCl when Pt electrode is taken then H₂ liberated at cathode while with Hg cathode it forms sodium amalgam because more voltage is required to reduce H⁺ at Hg than Pt.