Question

The preparation of LiOH by the electrolysis of a 35% solution of LiCl using a platinum anode led to a current efficiency of 80%. What weight of LiOH was formed by the passage of 2.5 A for 4825 s?

Answer 1

To find the weight of LiOH formed by the passage of 2.5 A for 4825 s, calculate the charge generated and then the amount of LiOH produced.

Step-by-step explanation:

To find the weight of LiOH formed by the passage of 2.5 A for 4825 s, we need to calculate the charge generated and then the amount of LiOH produced.

The charge generated can be found using the formula Q = It, where Q is the charge, I is the current, and t is the time. Plugging in the values, Q = (2.5 A)(4825 s) = 12062.5 C.

The molar mass of LiOH is 23.95 g/mol. To find the weight of LiOH, we can use the formula Weight of LiOH = (Q / F) * (Molar mass of LiOH / 2), where F is the Faraday constant.

Since the current efficiency is 80%, we can multiply the weight of LiOH by 0.8 to get the final weight of LiOH formed.