1. Tardigrade
  2. Question
  3. Chemistry
  4. In the electrochemical cell Zn|.ZnSO4(.0.01M.)||CuSO4(.1.0M.)|.Cu, the emf of this Daniel cell is E1 . When the concentration of ZnSO4 is changed to 1.0M and that of CuSO4 changed to 0.01M, the emf changes to E2 . From the followings, which one is the relationship between E1 and E2 ? (Given, (R T/F)=0.059 )

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Q. In the electrochemical cell

$Zn\left|\right.ZnSO_{4}\left(\right.0.01M\left.\right)\left|\right|CuSO_{4}\left(\right.1.0M\left.\right)\left|\right.Cu,$ the emf of this Daniel cell is $E_{1}$ . When the concentration of $ZnSO_{4}$ is changed to 1.0M and that of $CuSO_{4}$ changed to 0.01M, the emf changes to $E_{2}$ . From the followings, which one is the relationship between $E_{1}$ and $E_{2}$ ?

(Given, $\frac{R T}{F}=0.059$ )

NTA AbhyasNTA Abhyas 2020Electrochemistry

Solution:

$E = E ^{0}-\frac{0.059}{2} \log _{10} \frac{[ Zn ]^{2+}}{[ Cu ]^{2+}}$

$E _{1}= E ^{\circ}-\frac{0.059}{2} \log _{10} \frac{(0.01)}{(1)^{2}}$

$E _{2}= E ^{\circ}-\frac{0.059}{2} \log _{10} \frac{(1)}{(0.01)^{2}}$

Hence $E _{1}> E _{2}$