o Net reactions may be summarized as : NaCl ( aq ) -- > Na + ( aq ) + Cl – ( aq ) Cathode : H2O ( l ) + e – -- > ½ H2 ( g ) + OH – ( aq ) Anode : Cl – ( aq ) -- > ½ Cl2 ( g ) + e – Net reaction : NaCl ( aq ) + H2O ( l ) -- > Na + ( aq ) + OH – ( aq ) + ½H2 ( g ) + ½Cl2 ( g )
Problem : Given the standard electrode potentials , K + / K = - 2.93V , Ag + / Ag = 0.80V , Hg 2 + / Hg = 0.79V Mg 2 + / Mg = - 2.37 V , Cr 3 + / Cr = - 0.74V Arrange these metals in their increasing order of reducing power .
Solution :
Lower the reduction potential leads to higher reducing power . The given standard electrode potentials increases in the following order :
K + / K < Mg 2 + / Mg < Cr 3 + / Cr < Hg 2 + / Hg < Ag + / Ag . Hence , reducing power of the given metals increases in the following order :