Lessons 33 · 1 lessons · S. Masharipov, A. Mutalibov, E. Murodov, H. Islomova. General chemistry, Grade 11, 1st edition. G‘afur G‘ulom Publishing and Printing House, Tashkent, 2018
33
Electrolysis problems and their solutions
Textbook: pp. 149–155
GoalSolve electrolysis calculation problems: masses released from the current, changes in solution concentration, and cases where the current is not enough for the salt.
New words
mass fraction of solution · eritma massa ulushiequivalent amount · ekvivalent miqdorelectrolytic cell · elektrolizyorquantity of current · tok miqdori
Explanation
In an electrolysis problem first determine the electrode reactions by the rules and write the overall equation. Then find the quantity of electricity in equivalents: n(eq) = I·t/96500. The equivalent amount of substance released at each electrode equals this number, and the mass is m = n(eq)·E. In a solution of an oxygen-containing salt of an active metal only water decomposes: the salt mass does not change, the solution mass falls by the mass of water decomposed, so the mass fraction rises. If the current is more than needed to decompose all the salt, the excess quantity of electricity decomposes water (for example in AgNO₃, once Ag is used up, H₂ is released at the cathode); if the current is insufficient, all of it goes to the salt. After electrolysis the solution mass is smaller than at the start by the mass of the substances released at the cathode and anode. Gas volumes: 1 equivalent of H₂ or Cl₂ is 11.2 L and of O₂ 5.6 L (under normal conditions).
Worked examples
A 540 g solution of 10 % Na₂SO₄ was electrolysed until it was 20 %. The salt is 54 g (unchanged); the final solution is 54·100/20 = 270 g; the water decomposed is 540 − 270 = 270 g; n(eq) = 270/9 = 30.
A current of 5 A passes for 19 300 s through CuSO₄ solution: n(eq) = 1. 32 g of Cu forms at the cathode, 8 g of O₂ (5.6 L) at the anode, and 49 g of H₂SO₄ forms in the solution; the solution mass falls by 32 + 8 = 40 g.
Class activity
Group work: each group chooses one salt (Na₂SO₄, CuSO₄, AgNO₃), writes a three-step plan (reaction equation – n(eq) – masses) and explains it to the class in three minutes.
Practice
1
A 540 g solution of 10 % Na₂SO₄ is electrolysed until it is 20 %. How many equivalents of electricity have passed?
30
2
3·96 500 C (3 mol of electrons) passes through a solution containing 2 mol of AgNO₃. What is the total mass in grams of the Ag and H₂ released at the cathode?
217
3
How many grams of H₂SO₄ (E = 49) form in the solution when 10 A flows for 9650 s through CuSO₄ solution?
49
4
If the salt mass does not change during the electrolysis of Na₂SO₄ solution, why does the mass fraction rise?