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.
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