In the electrochemical cell Z n | Z n S O 4 ( 0.01 M ) C u S O 4 ( 1.0 M ) | C u , the emf of this Daniel cell is E 1 . When the concentration of Z n S O 4 is changed to 1.0M and that of C u S O 4 cha

Question

In the electrochemical cell Z n | Z n S O 4 ( 0.01 M ) C u S O 4 ( 1.0 M ) | C u , the emf of this Daniel cell is E 1 . When the concentration of Z n S O 4 is changed to 1.0M and that of C u S O 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, R T F = 0.059 )

TestHub · Accepted Answer

E =E o − 0.059 2 log 10 Zn 2 + Cu 2 + E 1 = E ° − 0.059 2 log 10 0.01 ( 1 ) 2 E 2 = E ° − 0.059 2 log 10 1 ( 0.01 ) 2 Hence E 1 > E 2

Chemistry - Electrochemistry Question with Solution | TestHub

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