write the cathode & anode reaction for electrolysis of water
Answers
Answer:
Cathode: [latex]{H}_{2}O + 2 {e}^{-} \rightarrow {H}_{2}(g) + 2 {OH}^{-}[/latex]
Anode: [latex]2 {H}_{2}O \rightarrow {O}_{2}(g) + 4 {H}^{+} + 4 {e}^{-}[/latex]
Explanation:
Multiplying the cathode reaction by 2, in order to match the number of electrons transferred, results in this net equation, after OH– and H+ions combine to form water..Hydrogen will appear at the cathode, the negatively charged electrode, where electrons enter the water, and oxygen will appear at the anode, the positively charged electrode. The number of moles of hydrogen generated is twice the number of moles of oxygen, and both are proportional to the total electrical charge conducted by the solution. The number of electrons pushed through the water is twice the number of generated hydrogen molecules, and four times the number of generated oxygen molecules.
Answer:
during electrolysis h + I will migrate to the cathode except electron and is neutralized to form hydrogen atom to hydrogen atom then form molecular hydrogen
OH - Ion on the other hand migrate to word anode release electron to form which radical that finally form oxygen at the anode
At anode
OH^- - e = OH
4OH = 2h2O + O2
