Why blue colour of `CuSO_(4)` solution gets discharged when zinc rod is dipped in it ? Given, `E_(Cu^(+2)//Cu)^(@)=0.34 V and E_(Zn^(+2)//Zn)^(@)=-0.76V`

Explanation

Since `E_(Zn,Zn^(2+))^(@)=-E_(Zn^(2+),Zn)^(@)=-0.76V` is lower than `E_(Cu^(2+),Cu)^(@)` therefore Zn will lose electrons and copper will accept them in other words the following reaction will occur `Zn+Cu^(2+)rarrZn^(2+)+Cu` since blue `Cu^(2+)` ions are consumed and colourless `Zn^(2+)` ions are produced during the above rections therefore colour of `CuSO_(4)` solution gets discharged when zinc rod is dipped in it