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^(2+)|Zn)^(0)` = `- E_(Zn|Zn_(2+))^(0)` = -0.76 V is lower than `E_(Cu^(2+)| Cu)^(0)` therefore, Zn will lose electrons and copper will accept them. In other words, the following reaction will occur, `Zn + Cu^(2+) rarr Zn^(2+) + Cu`. Since, blue `Cu^(2+)` ions are consumed and colourless `Zn^(2+)` ions are produced during the above reaction, therefore, colour of `CuSO_(4)` solution gets discharged when zinc rod is dipped in it.