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: