The solution in a beaker turns blue ifA. Cu electrode is placed in `ZnSO_(4)` solutionB. Cu electrode is placed in `AgNO_(3)` solutionC. Cu electrode is placed in `Al_(2)(SO_(4))_(3)` solutionD. Cu electrode is placed in `FeSO_(4)` solution

The solution in a beaker turns blue ifA. Cu electrode is placed in `Zn

1.	The solution in a beaker turns blue ifA. Cu electrode is placed in `ZnSO_(4)` solutionB. Cu electrode is placed in `AgNO_(3)` solutionC. Cu electrode is placed in `Al_(2)(SO_(4))_(3)` solutionD. Cu electrode is placed in `FeSO_(4)` solution
Answer» Correct Answer - B Cu being more reactive than Ag due to lower electrode potential, displaces Ag from `AgNO_(3)` solution. It is dissolved in the solution in the form of `Cu^(2+)` ions.

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The solution in a beaker turns blue ifA. Cu electrode is placed in `ZnSO_(4)` solutionB. Cu electrode is placed in `AgNO_(3)` solutionC. Cu electrode is placed in `Al_(2)(SO_(4))_(3)` solutionD. Cu electrode is placed in `FeSO_(4)` solution

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