During the electrolysis of an aqueous solution of sodium chloride, gas evolved at the anode, the volume is 5.6 L. Determine the volume of evolved gas at the cathode and the mass of the formed alkali in the solution.
First, let’s write the equation for the dissociation of sodium chloride in water:
NaCL = Na (+) + Cl (-)
Sodium is a very active metal and its ions will not be reduced at the cathode since the decomposition of water begins to form hydrogen. At the anode, chlorine ions will be oxidized to form free chlorine.
Let us write down the equations of electrode reactions:
К (-): 2H2O + 2e = H2 + 2OH (-)
A (+): 2Cl (-) – 2e = Cl2
Overall electrode reaction: 2NaCl + 2H2O = 2NaOH + H2 + Cl2
Through the proportion, we find the mass of alkali and the volume of hydrogen:
22.4-2 * 40
x = 5.6 * 2 * 40 / 22.4 = 20 grams
22.4 – 22.4
y = 5.6 * 22.4 / 22.4 = 5.6 l
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