The solution in a beaker turns blue if-Cu electrode is placed in ZnSO-4 solutionCu electrode is placed in AgNO-3 solutionCu electrode is placed in Al-2-SO-4-3 solutionCu electrode is placed in FeSO-4 solution

Question

The solution in a beaker turns blue if-Cu electrode is placed in ZnSO-4 solutionCu electrode is placed in  AgNO-3 solutionCu electrode is placed in Al-2-SO-4-3 solutionCu electrode is placed in FeSO-4 solution

Toppr Admin · Accepted Answer

Order of reduction potential of given metals is:

$A l < Z n < F e < C u < A g$

Only $A g$ will get reduced easily than $C u$ . Therefore, copper metal in the presence of $A g$ will get oxidise as:

$C u + 2 A g^{+} \to C u^{2 +} + 2 A g$
where,
$A g N O_{3} \to A g^{+} + N O_{3}^{-}$

Thus obtained the solution of copper ions is blue in colour.

Therefore the solution in a beaker turns blue if $C$ u electrode is placed in $A g N O_{3}$ solution.

The solution in a beaker turns blue if:Cu electrode is placed in ZnSO4 solutionCu electrode is placed in AgNO3 solutionCu electrode is placed in Al2(SO4)3 solutionCu electrode is placed in FeSO4 solution

The solution in a beaker turns blue if:
$C u$ electrode is placed in $Z n S O_{4}$ solution
$C u$ electrode is placed in $A g N O_{3}$ solution
$C u$ electrode is placed in $A l_{2} (S O_{4})_{3}$ solution
$C u$ electrode is placed in $F e S O_{4}$ solution