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### 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