1
MCQ (Single Correct Answer)

NEET 2016 Phase 1

The pressure of H2 required to make the potential of H2-electrode zero in pure water at 298 K is
A
10$$-$$10 atm
B
10$$-$$4 atm
C
10$$-$$14 atm
D
10$$-$$12 atm

Explanation

2H+ (aq) + 2e $$ \to $$ H2(g)

Ecell = Eocell - $${{0.0591} \over 2}\log {{{p_{{H_2}}}} \over {{{\left[ {{H^ + }} \right]}^2}}}$$

$$ \Rightarrow $$ 0 = 0 - $${{0.0591} \over 2}\log {{{p_{{H_2}}}} \over {{{\left[ {{{10}^{ - 7}}} \right]}^2}}}$$

$$ \Rightarrow $$ $${{{p_{{H_2}}}} \over {{{\left[ {{{10}^{ - 7}}} \right]}^2}}} = 1$$

$$ \Rightarrow $$ PH2 = 10–14 atm
2
MCQ (Single Correct Answer)

NEET 2016 Phase 2

If the Eocell for a given reaction has a negative value, which of the following gives the correct relationships for the values of $$\Delta $$Go and Keq ?
A
$$\Delta $$Go > 0;   Keq < 1
B
$$\Delta $$Go > 0;   Keq > 1
C
$$\Delta $$Go < 0;   Keq > 1
D
$$\Delta $$Go < 0;   Keq < 1

Explanation

We know that

$$\Delta $$Go = –nFEocell

Eocell = -ve then $$\Delta $$Go = +ve or $$\Delta $$Go > 0

$$\Delta $$Go = -nRTlog Keq

For $$\Delta $$Go = +ve, Keq = -ve or Keq < 1.
3
MCQ (Single Correct Answer)

NEET 2016 Phase 2

During the electrolysis of molten sodium chloride, the time required to produce 0.10 mol of chlorine gas using a current of 3 amperes is
A
55 minutes
B
110 minutes
C
220 minutes
D
330 minutes

Explanation

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

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

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

= $${{Eq.\,wt.\,of\,C{l_2}\,gas \times I \times t} \over {96500}}$$

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

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

= 6433.33 sec

= 107.22 min $$ \simeq $$ 110 min
4
MCQ (Single Correct Answer)

NEET 2016 Phase 2

The number of electrons delivered at the cathode during electrolysis by a current of 1 ampere in 60 seconds is (charge on electron = 1.60 $$ \times $$ 10$$-$$19C)
A
6 $$ \times $$ 1023
B
6 $$ \times $$ 1020
C
3.75 $$ \times $$ 1020
D
7.48 $$ \times $$ 1023

Explanation

Q = I × t

Q = 1 × 60 = 60 C

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

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

= 3.75 $$ \times $$ 1020 electrons

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