Question

При электролиза раствора серной кислоты расходуется мощность 37 Вт. Определить сопротивление электролита, если за время 500 мин выделяется водород, масса которого 0,3 г."Электрохимический эквивалент водорода 10^-8

Answer 1

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Answer 2

Calculation of Resistance of the Electrolyte

To determine the resistance of the electrolyte in the process of electrolysis of sulfuric acid, we can use the formula:

Resistance (R) = Voltage (V) / Current (I)

In this case, the power (P) consumed during the electrolysis is given as 37 W, and the time (t) taken to produce hydrogen gas is 500 minutes. The mass of the hydrogen gas produced is 0.3 g, and the electrochemical equivalent of hydrogen is given as 10^-8.

To calculate the resistance, we need to find the current (I) flowing through the electrolyte. We can use the formula:

Current (I) = Charge (Q) / Time (t)

The charge (Q) can be calculated using the formula:

Charge (Q) = Mass (m) / Electrochemical Equivalent (E)

Substituting the given values, we have:

Charge (Q) = 0.3 g / 10^-8

Now, we can calculate the current (I) using the formula:

Current (I) = Charge (Q) / Time (t)

Substituting the values, we have:

Current (I) = (0.3 g / 10^-8) / (500 minutes)

Next, we can calculate the voltage (V) using the formula:

Power (P) = Voltage (V) * Current (I)

Substituting the values, we have:

37 W = Voltage (V) * Current (I)

Now, we can rearrange the equation to solve for the voltage (V):

Voltage (V) = 37 W / Current (I)

Finally, we can calculate the resistance (R) using the formula:

Resistance (R) = Voltage (V) / Current (I)

Substituting the values, we have:

Resistance (R) = (37 W / Current (I)) / Current (I)

Now, let's calculate the resistance.

Calculation:

Using the given values, we can calculate the resistance of the electrolyte.

Charge (Q) = 0.3 g / 10^-8 = 3 * 10^7 C

Current (I) = (3 * 10^7 C) / (500 minutes) = 6 * 10^4 C/min

Voltage (V) = 37 W / (6 * 10^4 C/min) = 0.617 V

Resistance (R) = (0.617 V) / (6 * 10^4 C/min) = 1.028 * 10^-5 Ω

Therefore, the resistance of the electrolyte in the process of electrolysis of sulfuric acid is approximately 1.028 * 10^-5 Ω.

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