Two half cells are AI^(3+)(aq)/AI and Mg^(2+)(aq)/M The reduction potential of these half cells are -1.66 V and -2.36 V respectively calculate the cell potential write the cell reaction also

Two half cells are AI^(3+)(aq)/AI and Mg^(2+)(aq)/M The reduction pot

1.	Two half cells are AI^(3+)(aq)/AI and Mg^(2+)(aq)/M The reduction potential of these half cells are -1.66 V and -2.36 V respectively calculate the cell potential write the cell reaction also
Answer» Solution :Since `MG^(2+)(aq)//"Mgelectorde" =-2.36V` is at a lower potential than `AI^(3+)(aq)/(AI) "electrode" = 1.66V therefore Mg^(2+)(aq)//Mq` electrode ACTS as the anode and `AI^(3+)(aq)//AI` acts as the cathode in other words Mg loses electrons and `AI^(3+)` in accepts electrons THUS the cell reaction is `3Mg+2AI^(3+)rarr3Mg^(2+)+2AI"and"E_("cell")^(@)=E_(AI)^(@),AI-E_(Mg)^(2+),Mg^(@)=-1.66-(-2.36)=+0.70V`