However, SO42¯ ions will not be released since their discharge potential is more as compared to OH¯ ions. Predict the products of electrolysis in each of the following: (i) An aqueous solution of AgNO 3 with silver electrodes. Question 18.

It’s very helpful mam. Note that 2H2O(l) + 2e- ----> H2(g) + 2OH-(aq), E = -.83 V O2(g) + 4H+(aq) + 4e- ---> 2H2O(l), E = +1.23 V I think the answer is either Al and H2, or H2 and Br2. Electrolysis reactions involving H+ ions are fairly common in acidic solutions.

Electrolysis is very important commercially as a stage in the separation of elements from naturally occurring sources, such as ores, using an electrolytic cell. The answer is Cu2+ and SO42-.

Water is only weakly ionized but in the presence of an acid, its degree of ionization increases. The key process of electrolysis is the interchange of atoms and ions by the removal or addition of electrons to the external circuit. 2Cu2+ (aq) + 2 H2O (l) ———> Cu(s) + 4 H+ + O2 (g), 2Cu2+ (aq) + 2SO42¯ + 2 H2O (l) ——-> 2Cu(s) + 4 H+ + O2 (g) + 2SO42¯. When electric current is passed through the solution, Cl2 gas is evolved at the anode and hydrogen is evolved at the cathode. This extra voltage required is called, Thus, the aqueous solution of sodium chloride contains Na, In concentrated solution of NaCl, oxidation of chloride ions is preferred than water at anode and therefore Cl, Thus, during the electrolysis of aqueous sodium chloride, H, The electrolysis may be carried out by taking solid lead bromide in a silica crucible. The slowness of electrode reaction creates electrical resistance at the electrode surface. The slowness of electrode reaction creates electrical resistance at the electrode surface. (iii) A dilute solution of H 2 SO 4 with platinum electrodes. A direct current (DC) supply: provides the energy necessary to create or discharge the ions in the electrolyte. (iv) An aqueous solution of CuCl 2 with platinum electrodes. She has started this educational website with the mindset of spreading Free Education to everyone. Check out a sample textbook solution. As a result, sodium metal is obtained at the cathode. For example, it is possible to oxidize ferrous ions to ferric ions at the anode: $Fe^{2+} (aq) \rightarrow Fe^{3+} (aq) + e^-$. It is also possible to reduce ferricyanide ions to ferrocyanide ions at the cathode: $Fe(CN)^{3-}_6 + e^- \rightarrow Fe(CN)^{4-}_6$. An electrolyte: a substance containing free ions, which are the carriers of electric current in the electrolyte. When a D.C. voltage source is applied, no current is observed. The reaction with higher value of Eø is preferred and therefore, the reaction at the cathode during is : But H+ (aq) ions are produced by the dissociation of water as: Therefore, the net reaction at the cathode may be written as: At the anode, the following reactions are possible : Cl¯ (aq) ——–> ½ Cl2(g) + e¯   Eø = + 1.36 V, 2 H2O (l) ——–> O2 (g) + 4 H+ (aq) + 4 e¯ Eø = +1.23 V. In concentrated solution of NaCl, oxidation of chloride ions is preferred than water at anode and therefore Cl2 gas is liberated. However, when the crucible is heated so that lead bromide metals the current is found to pass. The reaction at this electrode is: $Cu (s) \rightarrow Cu^{2+} (aq) + 2e^-$.

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