[Good score of 50 points] Senior one chemistry

The topic is wrong. The molecular formula of chromium hydroxide is Cr(OH)3, and chromium has no valence of +2.

Analysis: According to "H2O2→ O2", H2O2 is the reactant, H2O2 is the reducing agent (the valence of oxygen element increases), and the corresponding oxidation product is O2; Then the reaction at least lacks oxidant (valence reduction is needed), which can be obtained by calculating the valence of five substances in the system: the oxidant is H2CrO4, and the corresponding reduction product is Cr (OH) 3; Because H2O2 only produces O2, it can be seen from the chemical valence that H2O is neither an oxidation product nor a reduction product (the square formula shows that H2O is not a reactant, but a product), but a simple product.

So the equation of this reaction is: 3H2O2+2H2Cro4 = 2cr (OH) 3↓+2H2O+3O2 ↑.

The answer to this question can be drawn from this:

(1) The reducing agent in this reaction is (H2O2) (as shown in the above analysis);

(2) In this reaction, the process of reduction reaction is [H2CrO4→Cr(OH)3] (the oxidant is reduced and a reduction reaction occurs).

(3) Write the chemical equation of the reaction and mark the direction and quantity of electron transfer;

The reaction equation is (3H2O2+2H2Cro4 = 2cr (OH) 3↓+2H2O+3O2 =); (As shown in the above analysis)

The electron transfer direction is (from H2O2 to H2CrO4) (reducing agent loses electrons to generate oxidation products, and oxidant gains electrons to generate reduction products, so the electron transfer direction is from reducing agent to oxidant);

The number of electrons transferred is 6e- (as can be seen from the equilibrium equation, the total number of electrons lost by reducing agent is 6e-, and the total number of electrons gained by oxidizing agent is 6e-, so the number of electrons transferred is 6e-).

(4) If the reaction transfers 0.3mol of electrons, the volume of gas generated under standard conditions is (3.36L).

According to the reaction equation, for every 6mol of electrons transferred, 3mol of oxygen is generated, so the following proportional formula can be obtained:

Electron oxygen

6 moles and 3 moles

0.3 mole of nitrogen (oxygen)

6mol/3mol = 0.3mol/n(O2)； The solution is n (O2) = 0.1.5 mol;

Under standard conditions, v (O2) = n (O2) × VM = 0.1.5mol× 22.4l/mol = 3.36l;

Therefore, if the reaction transfers 0.3mol of electrons, the gas volume generated under standard conditions is 3.36L L.