1
MCQ (Single Correct Answer)

NEET 2017

In the electrochemical cell :
$$Zn\left| {ZnS{O_4}\left( {0.01\,M} \right)} \right|$$$$\left| {CuS{O_4}\left( {1.0M} \right)} \right|Cu,$$
the emf of this Daniell cell is E1. When the concentration of ZnSO4 is changed to 1.0 M and that of CuSO4 changed to 0.01 M, the emf changes to E2. From the followings, which one is the relationship between E1 and E2? (Given, RT/F = 0.059)
A
E1 < E2
B
E1 > E2
C
E2 = 01E1
D
E1 = E2

Explanation

Ecell = Eo - $${{0.059} \over n}\log {{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}$$

E1 = Eo - $${{0.059} \over 2}\log {{0.01} \over 1}$$

= Eo - $${{0.059} \over 2}\left( { - 2} \right)\log 10$$

E1 = Eo + 0.059

E2 = Eo - $${{0.059} \over 2}\log {1 \over {0.01}}$$

$$ \Rightarrow $$ E2 = Eo - $${{0.059} \over 2}\log 100$$

$$ \Rightarrow $$ E2 = Eo - 0.059

$$ \therefore $$ E1 > E2
2
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
3
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.
4
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

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