What is the product of electrolysis of nitric acid with an inert electrode?

Electrolysis of nitric acid is equal to electrolysis of water.

The cathode product is hydrogen.

The anode product is oxygen.

You can remember it this way, cathode: the cation moves to the cathode (the opposite sex wants to suck), so only hydrogen ions are discharged.

Anode: The anion moves to it, and the oxygenate is discharged after the hydroxide so that oxygen is generated.

Electrolysis of strong alkalis and oxygenated acids are electrolysis of water.

Supplemental Discharge Sequence:

1) The anion discharge sequence on an inert electrode (such as C, PT, AU, etc.) is:

S2 i br cl oh no3 so42 (all oxygenates are after OH).

The order of discharge of anions is related to reducibility and is electron loss.

2) The sequence of cation discharge on the inert electrode is:

The firing sequence of ag hg2 cu2 h fe2 zn2 cation is related to oxidation and is electron gain.

3) If the active electrode (such as Cu, Zn, Fe, Ag, etc.) is used as the anode, the anode is dissolved.

Sequence of discharge in aqueous solution:

s2- >cl- >oh- >f- ,no3- ,so4-

i.e. f-, no3-, so4- are not discharged in aqueous solution.

The cathode produces H2

The anode produces O2

Equivalent to electrolyzed water.

The discharge sequence of oh- in aqueous solution is before NO3-.

On the anode is an anion discharge.

S2 i br cl oh so42 no3 f The stronger the anion reduction, the first discharge.

In fact, both of them are electrolyzed water.

Anode: 2H2O - 4E- =O2 (gas) +4H + Cathode: 2H+ +2E- =H2 (gas without Zheng body) Total equation.

2H2O ==2H2 (Gas) +O2 (Gas) Question Supplement: Oxidation of nitrate under acidic conditions.

It must be stronger than hydrogen ions, and at the cathode, will it be nitrate that goes to get electrons instead of hydrogen ions that go to get electrons?

Answer: Remember, in the electrolysis reaction, the oxynate ions are discharged in the order after the water so as long as there is still water in the solution, it will not be the turn of the oxynate discharge, so it must be the water that gets the electrons.

In fact, both are electrolyzed water.

Anode: 2H2O - 4E- == O2 (Gas) +4H + Cathode: 2H+ +2E- == H2 (Gas) Total Equation:

2H2O == 2H2 (Gas) +O2 (Gas) Question Supplement: The oxidation of nitrate under acidic conditions is definitely stronger than that of hydrogen ions, will it be nitrate to get electrons instead of hydrogen ions at the cathode?

A: No, it won't. Remember, in the electrolysis reaction, the oxynate ions are discharged in the order after the water, so as long as there is still water in the solution, it will not be the turn of the oxynate discharge, so it must be the water that gets the electrons.

Electrolysis of nitric acid is the electrolysis of water As for the electrolysis of sodium sulfite, the hydrogen ion discharge of cathode water ionization The anode should be a sulfite discharge.

Male drain pole: 2H2O - 4E = 4H+ +O2

Negative search and leakage: ag+ +e = ag

Total reaction: 4AGnO3 + 2H2O = 4AG+ O2 + 4HNO3

Anode leak search: 2H2O - 4E = 4H+ +O2

Cathode: AG+ +E = AG

Total reaction: Leak key 4AGnO3 + 2H2O = 4AGg + O2 + 4HNO3

In fact, both are electrolyzed water.

Anode: 2H2O4E-

O2 (gas).

4H + Cathode: 2H + 2E-

H2 (gas).

Total equation: 2h2o

2h2 (gas).

O2 (gas).

Question: The oxidation of nitrate under acidic conditions is definitely stronger than that of hydrogen ions, will it be nitrate to get electrons instead of hydrogen ions at the cathode?

A: No, it won't. Remember, in an electrolysis reaction, the oxynate ions are discharged in order.

Water. After that, so as long as there is water in the solution, it will not be the turn of the oxynate discharge, so it must be the water to be electrified.

Analysis of the valence state of NH2OH shows that NO3 is reduced to NH2OH, so NO3 NH2OH is on the cathode

The electrode equation for the cathode is NO3 6E 7H NH2OH 2H2O Anode: 2H2O 4E 4HO2

Total 2HNO3 2H2O Electrolytic 2NH2OH 3O2 Hope it helps you!

Hope for talent!

a. Electrolysis of sodium sulfate, the essence is electrolysis of water, the anode generates oxygen, and the electrode reaction is: 4oh-4e-2h2

O+O2, the cathode generates hydrogen, and the electrode reaction is 2H+

2e-h2, the electron conservation calculation shows that the ratio of the amount of matter of the precipitated product on the cathode and the anode is 2:1, so a is wrong;

b. Electrolysis of copper chloride solution, chlorine gas is generated on the anode, the electrode reaction is: 2cl-2e-cl2, copper is generated on the cathode, the electrode reaction is: Cu2+2E-Cu, and the ratio of the amount of substances precipitated on the cathode and the anode is 1:

1, so b is correct;

c. Electrolysis of sulfuric acid solution essentially electrolyzed water, the concentration of the solution increases, and the pH of the solution decreases, so C is wrong;

d. The electrolytic dilute sodium hydroxide solution is essentially electrolyzed water, and the concentration of sodium hydroxide solution increases, the concentration of hydroxide ions increases, and the pH of the solution increases, so D is wrong;

Therefore, choose B