Question

"Электролизу подвергли водный раствор нитрата серебра.При этом на катоде выделилось 43.2 г металла.Вычислите объём газа(н.у.),выделившегося на аноде,и массу образовавшейся кислоты

Answer 1

AgNO3→Ag(+)+NO3(-)

K(-) --- A(+)

Ag(+)+e=Ag----------------NO3(-)

H2O---------------------2H2O-4e=O2+4H(+)


1) из уравнения следует что n(Ag) = n(O2)

m(Ag)/Mr(Ag)=V(O2)/Vr(O2)

V(O2)=43.2*5.6/108=2,24 л

m(Ag)/M(Ag)= m(HNO3)/M(HNO3)

m(HNO3)=43,2*63/108=25,2 г
Ответ: 2,24 л и 25,2 г 

Answer 2

Electrolysis of Silver Nitrate Solution

When a water solution of silver nitrate (AgNO3) is subjected to electrolysis, several reactions occur at the anode and cathode. At the cathode (negative electrode), silver ions (Ag+) are reduced and metallic silver (Ag) is deposited. At the anode (positive electrode), nitrate ions (NO3-) are oxidized, resulting in the formation of oxygen gas (O2) and water (H2O).

Given that 43.2 grams of silver metal is deposited at the cathode, we can calculate the volume of gas (at standard temperature and pressure) produced at the anode and the mass of acid formed.

Calculation of Gas Volume at the Anode

To calculate the volume of gas produced at the anode, we need to use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

Since the pressure and temperature are at standard conditions (1 atm and 273 K), we can simplify the equation to V = n * 22.4 L (at standard temperature and pressure).

To determine the number of moles of gas produced, we need to use the stoichiometry of the reaction. From the balanced equation:

2 AgNO3(aq) -> 2 Ag(s) + O2(g) + 2 NO2(g)

We can see that for every 2 moles of silver nitrate reacted, 1 mole of oxygen gas is produced. Therefore, the number of moles of oxygen gas produced is half the number of moles of silver metal deposited.

To find the number of moles of silver metal, we need to convert the mass of silver metal to moles using its molar mass. The molar mass of silver (Ag) is 107.87 g/mol.

Let's calculate the number of moles of silver metal deposited:

Step 1: Convert the mass of silver metal to moles: 43.2 g Ag * (1 mol Ag / 107.87 g Ag) = 0.400 mol Ag

Step 2: Calculate the number of moles of oxygen gas produced: 0.400 mol Ag * (1/2) = 0.200 mol O2

Step 3: Calculate the volume of gas at standard temperature and pressure: V = n * 22.4 L V = 0.200 mol O2 * 22.4 L/mol = 4.48 L

Therefore, the volume of gas (at standard temperature and pressure) produced at the anode is 4.48 liters.

Calculation of Mass of Acid Formed

To calculate the mass of acid formed, we need to determine the number of moles of nitrate ions (NO3-) that were oxidized at the anode. From the balanced equation, we can see that for every 2 moles of silver nitrate reacted, 2 moles of nitrate ions are oxidized.

Therefore, the number of moles of nitrate ions oxidized is equal to the number of moles of silver metal deposited.

Using the previously calculated number of moles of silver metal (0.400 mol Ag), we can calculate the mass of acid formed by multiplying the number of moles of nitrate ions by the molar mass of nitric acid (HNO3), which is 63.01 g/mol.

Let's calculate the mass of acid formed:

Mass of acid = 0.400 mol Ag * (2 mol NO3- / 2 mol AgNO3) * (63.01 g HNO3 / 1 mol HNO3) = 50.41 g

Therefore, the mass of acid formed is 50.41 grams.

In summary, when a water solution of silver nitrate is subjected to electrolysis, 43.2 grams of silver metal is deposited at the cathode, 4.48 liters of gas (at standard temperature and pressure) is produced at the anode, and 50.41 grams of acid is formed.