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Z n | Z n S O 4 ( 0.01 M ) | | C u S O 4 ( 1.0 M ) | C u is a Daniel cell. The ‘emf ’ of this cell at 298K is E 1 . when the concentration of Z n S O 4 changes to 1.0 M and that of C u S O 4 is 0.01 M, the ‘emf ’ changed to E 2 . What is the relationship between E 1 and E 2

Options

  1. AE 2 = 0 ≠ E 1
  2. BE 1 > E 2
  3. CE 1 < E 2
  4. DE 1 = E 2

Correct answer

B. E 1 > E 2

Step-by-step solution

E 1 = E o - 0.0591 2 l o g 0.01 1 = E o + 0.0591 2 × 2 E 2 = E o &#x2212; 0.0591 2 log 1 0.01 = E o &#x2212; 0.0591 2 &#x00D7; 2 ∴ E 1 > E 2 .

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