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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")
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