# At what current strength does the electrolysis of copper sulfate occur if 8 grams are released

**At what current strength does the electrolysis of copper sulfate occur if 8 grams are released at the cathode in 20 minutes?**

Let’s translate the values from given to the SI system:

t = 20 min = 1200 s.

m = 8 g = 0.008 kg.

According to the first Faraday’s law, the mass of a substance that was released on the electrode during the time Δt at a current strength I is proportional to them.

m = k * I * Δt, where k is the electrochemical equivalent of a substance (the ratio of the mass of an ion to its charge)

According to the reference book, we determine the electrochemical equivalent of copper k = 3.28 * 10 ^ -7 kg / C, namely, it will be released at the cathode.

Let us express the current strength from this expression:

I = m / (k * Δt)

Substitute the numerical values:

I = m / (k * Δt) = 0.008 / (3.28 * 10 ^ -7 * 1200) = 20.32 A.

Answer: electrolysis occurs at a current strength of 22.32 A.