how presence of two gasese in electrolysis of water can be tested
Answers
Creating an electric potential through water causes positive ions, including the inherent hydrogen ions H3O+, to move towards the negative electrode (cathode) and negative ions, including the inherent hydroxide ions OH-, to move towards the positive electrode (anode). With a sufficient potential difference, this may cause electrolysis with oxygen gas being produced at the anode and hydrogen gas produced at the cathode (see [1878] for current reviews). The electrolysis of water usually involves dilute, or moderately concentrated, salt solutions in order to reduce the power loss driving the current through the solution, but the presence of salt is not a requirement for electrolysis.
Thus,
Anode +ve 6H2O(l) --> O2(g) + 4H3O+(aq) + 4e-(to anode) b E° = +1.229 V, pH 0 d E°' = +0.815 V
Cathode -ve 4e-(from cathode) + 4H2O(l) --> 2H2(g) + 4OH-(aq) E° = -0.828 V, pH 14 E°' = -0.414 V
where (l), (g) and (aq) show the states of the material as being a liquid, a gas or an aqueous solution and with the electrical circuit passing the electrons back from the anode to the cathode. The reactions are heterogeneous, taking place at the boundary between the electrode and the electrolyte with the aqueous boundary layer subject to concentration and electrical potential gradients with the presence of the generated gaseous nanobubbles and microbubbles.