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+//Chapter 6: Electrochemistry
+//Problem: 14
+clc;
+
+//Declaration of Variables
+T = 25 // C
+Eo = - 0.8277 // V
+
+// Solution
+mprintf("The reversible reaction,\n")
+mprintf(" 2H2O <--> H3O+ + OH-\n")
+mprintf(" May be divided into two parts.\n")
+mprintf(" 2H2O + e- --> 1/2 H2 + OH- (cathode) Eo = -0.8277 V\n")
+mprintf(" H2O + 1/2 H2 --> H3O+ + e- (anode) Eo = 0")