1

### 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

### 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

### 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

### 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