[Part 3]- C6- electrolysis💧 Flashcards
What are aqueous solutions ?
Hint: _____ molecules ionise, forming what ions?
Hint: hydroxide
- aqueous solutions are dissolved in water- meaning they form when substances are dissolved in water
- and water molecules ionise/dissociate [split], forming hydrogen ion and hydroxide ions.
What can you tell me about the formula of the copper sulfate solution, during electrolysis ? [of an aqueous solution]
- copper sulfate solution, has the formula CuSO4 (aq)- containing the copper ion: CU^2+ and the sulfate ion: SO4^2-
- because it’s dissolved in water, the hydrogen ion H^+ and the hydroxide ion OH^- needs to be considered.
During the electrolysis of copper sulfate solution [an aqueous solution], what’s attracted to the cathode ?
Hint: rule- reactivity of hydrogen
- the copper sulfate solution, contains two positive ions that would be attracted to the cathode.
- ; the rule, is that hydrogen is (only) produced at the cathode, if the metal is more reactive than hydrogen.
- and because copper is less reactive than hydrogen, copper will be produced at the cathode.
During the electrolysis of copper sulfate solution [an aqueous solution], what’s attracted to the anode ?
Hint: why
- and at the anode, oxygen gas is made- [because oxygen is usually produced at the anode, when aqueous solutions are electrolysed]
Why are the electrodes inert ?
- the electrodes are inert [unreactive] because, it’s important that the electrodes don’t react, with the chemicals that are being made in electrolysis.
- otherwise, the electrodes will take part in the reaction
What is the half-equation at the cathode, during the electrolysis of copper sulfate solution
Hint: 2e
- Cu^2+ 2e^- ————-> Cu
- this is a reduction reaction, as the copper ions are gaining electrons, to form copper atoms.
What is the half-equation at the anode, during the electrolysis of copper sulfate solution ?
Hint: 4OH^- —> O2
4OH^- ———> O2 + 2H2O + 4e^-
OR
4OH^- ➖ 4e^- ———-> O2 + 2H2O
- four hydroxide ions react to form oxygen gas and water, using a total of four electrons in the process- therefore this is an oxidation reaction.
What’s the rule to use, when predicting what will be produced at the cathode ?
- ; the rule, is that hydrogen is (only) produced at the cathode, if the metal is more reactive than hydrogen.
What can you tell me about the formula of sodium chloride solution during electrolysis ? [of an aqueous solution]
Hint:
- sodium chloride solution, has the formula: NaCl^- containing the sodium ion: Na and the chloride ion Cl^-
- because it’s dissolved in water, the hydrogen ion H^+ and the hydroxide ion OH^- needs to be considered.
During the electrolysis of sodium chloride solution, what’s attracted to the cathode
Hint: why?
- the sodium chloride solution, contains two positive ions that would be attracted to the cathode. [the sodium ion Na^+ and the hydroxide ion H^+]
- ; the rule, is that hydrogen is (only) produced at the cathode, if the metal is more reactive than hydrogen.
- and because sodium is more reactive than hydrogen, hydrogen gas will be produced at the cathode.
During the electrolysis of sodium chloride solution, what’s attracted to the anode?
- and at the anode, there are two negative ions that could be attracted to it- [the chloride ion: Cl^- and the hydroxide ion OH^-]
- ; the rule for the anode is: that if the aqueous solution contains halide ions, then halogen will be produced at the anode. [otherwise it would typically be oxygen gas]
- therefore at the anode, chlorine gas is made- since the sodium chloride solutions, contains a halide ion [chloride]
What is the half-equation at the cathode, during the electrolysis of sodium chloride solution ?
- H^+ ➕ e^- ———> H
- this is a reduction reaction, as the hydrogen ion is gaining one electron, to form a hydrogen atom.
During the electrolysis of sodium chloride solution, what can else can you tell me about the cathode ?
- Since hydrogen atoms immediately pair up to form a hydrogen molecule [H2]- as hydrogen is a diatomic molecule, the half equation is doubled and now looks like this:
- 2H^+ ➕ 2e^- ———> H2
What is the half-equation at the anode, during the electrolysis of sodium chloride solution ?
- Cl^- ———-> Cl + e^-
- ; as chloride atoms always pair to form a double molecule, the half-equation must be doubled:
- 2Cl^- ———-> Cl2 + 2e^-
OR
- 2Cl^- ➖ 2e^- ———-> Cl2
- this is an oxidation reaction, as chlorine loses one electron, to form a chlorine atom.
What’s the rule to use, when predicting what will be produced at the anode ?
Hint: halide
- the rule for the anode is: that if the aqueous solution contains halide ions, then halogen will be produced at the anode.
- [otherwise it would typically be oxygen gas, as the hydroxide ions would be attracted to the anode]
- these halides ions are: fluoride (F −), chloride ( Cl −), bromide ( Br −), iodide ( I −) and astatide ( At −)