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Electrochemistry question

2017 · Q90
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Electrochemistry question

2017 · Q90

NEETChemistryElectrochemistryMCQ+4 / −1
In the electrochemical cell :
Zn∣ZnSO4(0.01 M)∣Zn\left| {ZnS{O_4}\left( {0.01\,M} \right)} \right|Zn∣ZnSO4​(0.01M)∣∣CuSO4(1.0M)∣Cu,\left| {CuS{O_4}\left( {1.0M} \right)} \right|Cu,∣CuSO4​(1.0M)∣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)
  1. A
    E1 < E2
  2. B
    E1 > E2
  3. C
    E2 = 01E1
  4. D
    E1 = E2
View written solutionFree

Correct answer: B

Ecell = Eo - 0.059nlog⁡[Zn2+][Cu2+]{{0.059} \over n}\log {{\left[ {Z{n^{2 + }}} \right]} \over {\left[ {C{u^{2 + }}} \right]}}n0.059​log[Cu2+][Zn2+]​

E1 = Eo - 0.0592log⁡0.011{{0.059} \over 2}\log {{0.01} \over 1}20.059​log10.01​

= Eo - 0.0592(−2)log⁡10{{0.059} \over 2}\left( { - 2} \right)\log 1020.059​(−2)log10

E1 = Eo + 0.059

E2 = Eo - 0.0592log⁡10.01{{0.059} \over 2}\log {1 \over {0.01}}20.059​log0.011​

⇒\Rightarrow⇒ E2 = Eo - 0.0592log⁡100{{0.059} \over 2}\log 10020.059​log100

⇒\Rightarrow⇒ E2 = Eo - 0.059

∴\therefore∴ E1 > E2

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