The solution in a beaker turns blue if

Question

The solution in a beaker turns blue if (A) Cu electrode is placed in ZnSO4 solution (B) Cu electrode is placed in AgNO3 solution (C) Cu electrode is placed in Al2(SO4)3 solution (D) Cu electrode is placed in FeSO4 solution.

Tardigrade Admin · Accepted Answer

Cu being more reactive than Ag due to lower electrode potential, displaces Ag from AgNO3 solution. It is dissolved in the solution in the form of Cu2+ ions.

Q. 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.

$C u$ being more reactive than Ag due to lower electrode potential, displaces $A g$ from $A g N O_{3}$ solution. It is dissolved in the solution in the form of $C u^{2 +}$ ions.

Q. The solution in a beaker turns blue if

Cu electrode is placed in ZnSO4​ solution

Cu electrode is placed in AgNO3​ solution

Cu electrode is placed in Al2​(SO4​)3​ solution

Cu electrode is placed in FeSO4​ solution.

Cu being more reactive than Ag due to lower electrode potential, displaces Ag from AgNO3​ solution. It is dissolved in the solution in the form of Cu2+ ions.

$C u$ 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.

$C u$ being more reactive than Ag due to lower electrode potential, displaces $A g$ from $A g N O_{3}$ solution. It is dissolved in the solution in the form of $C u^{2 +}$ ions.