If the metal is more reactive than H, hydrogen gas forms.
If the metal is less reactive than H, metal forms.
At the anode:
If the solution is concentrated and halide (contains halogen ion), halogen gas forms.
If the solution is dilute or not halide, oxygen gas forms.
Exercise #2
Determine the product observed at both cathode and anode for the electrolysis of:
Concentrated solution of sulfuric acid, H2SO4.
Dilute solution of potassium chloride, KCl.
Learning Objectives
Describe how ions move, compete, and react during electrolysis.
Explain electrolysis as a redox reaction.
Write half-equations for electrode reactions.
Molten Lead(II) Bromide - Ion Movement and Reactions
Ions are free to move in molten lead(II) bromide.
Opposite charges attract; positive lead ions (Pb^{2+}) move to the cathode (-), and negative bromide ions (Br^-) move to the anode (+).
At the cathode (-): Lead ions receive two electrons each and become lead atoms. The half-equation is: Pb^{2+}(l) + 2e^- \rightarrow Pb(l)
At the anode (+): Bromide ions each give up an electron and become atoms, which pair up to form molecules. The half-equation is: 2Br^-(l) \rightarrow Br_2(g) + 2e^-
Ions gain electrons: reduction.
Ions lose electrons: oxidation.
Overall, electrolysis is a redox reaction; reduction occurs at the cathode, and oxidation occurs at the anode.
Remember OILRIG: Oxidation Is Loss, Reduction Is Gain.
Electrolysis of Concentrated vs. Dilute Solutions
Concentrated Solution of Sodium Chloride
Ions from water are also present: (H^+) and (OH^-).
The solution contains (Na^+) and (Cl^-) ions from the salt and (H^+) and (OH^-) ions from water.
The positive ions go to the cathode, and the negative ions go to the anode.
At the cathode, the (H^+) ions accept electrons since hydrogen is less reactive than sodium: 2H^+(aq) + 2e^- \rightarrow H_2(g)
The hydrogen gas bubbles off.
The (Na^+) and (OH^-) ions remain, giving a solution of sodium hydroxide.
At the anode, the (Cl^-) ions give up electrons more readily than the (OH^-) ions do: 2Cl^-(aq) \rightarrow Cl_2(aq) + 2e^-
The chlorine gas bubbles off.
Remember RAC: Reduction At Cathode.
Dilute Solution of Sodium Chloride
The same ions are present as before, but the proportion of (Na^+) and (Cl^-) ions is lower.
At the cathode, hydrogen 'wins' as before and bubbles off: 4H^+(aq) + 4e^- \rightarrow 2H_2(g)
At the anode, (OH^-) ions give up electrons since not many (Cl^-) ions are present: 4OH^-(aq) \rightarrow O2(g) + 2H2O(l) + 4e^-
Half-Equations for Electrode Reactions
A half-equation shows the electron transfer at an electrode.
Steps in writing half-equations:
Name the ions present and the products.
Write each half-equation correctly:
Give the ion its correct charge.
Positive ions go to the cathode, and negative ions to the anode.
Write the correct symbol for the element that forms (e.g., Cl for chlorine, not CI).
The number of electrons in the equation should be the same as the total charge on the ion(s) in it.
Add the state symbols.
Example
Electrolysis of molten magnesium chloride:
Magnesium ions (Mg^{2+}) and chloride ions (Cl^-) are present; Magnesium and chlorine form.
Reduction at the cathode: Mg^{2+} + 2e^- \rightarrow Mg or Mg^{2+}(l) + 2e^- \rightarrow Mg(s)
Oxidation at the anode: 2Cl^- \rightarrow Cl2 + 2e^- or 2Cl^-(l) \rightarrow Cl2(g) + 2e^-