Chemistry Redox Revision

Half equations: Mg (s) → Mg²⁺ (aq) +2e^- and 2H⁺ (aq) + 2e^- → H_{2 (aq)}

Full equation = Mg (s) + 2H⁺ (aq) → Mg²⁺ (aq) + H₂ (aq)

Colour changes: Magnesium metal is a grey solid. It is oxidized to a colourless ion. H_⁺ from HCl is a colourless ion, when reduced it changes to a colourless gas H₂

Half equations: Cr₂O_7^2- (aq) +14H⁺ (aq) + 6e^- → 2Cr³⁺ (aq) + 7H₂O (l) and 2Br^- (aq) →Br₂ (aq) + 2e^-

Full equation: Cr₂O_7^2- (aq) + 14H⁺ (aq) +6Br^- (aq) →3Br₂ (aq) + 2Cr³⁺ (aq) + 7H₂0 (l)

Colour changes: Cr₂O_7^2- is orange, Br^- and K⁺ are colourless, so the initial solution is orange. The products are Cr³⁺ which is green and Br₂ which is brown. So the resulting colour is a mix of brown and green.

Half equations: Fe (s)→ Fe²⁺ (aq) + 2e^- and Cu²⁺ (aq) + 2e^- → Cu (s)

Full equations: Fe (s) + Cu²⁺ (aq) → Fe²⁺ (aq) + Cu (s)

Colour changes: Fe is a shiny grey metal, Cu²⁺ is blue in a solution, SO₄^2- is a colourless ion. The pinky brown substance is due to Cu metal forming and the blue colour fades because the Cu²⁺ is turning into the pinky brown substance. The pale green colour is due to the formation of Fe²⁺ ions- which are pale green.

Half equations: HSO₃^-/H⁺ (aq) + H₂O (l) → SO₄^2- (aq) +3H⁺ (aq) + 2e^- and I₂ (aq) + 2e^- → 2I^- (aq)

Full equations: HSO₃^-/H⁺ (aq) + H₂O (l) + I₂ (aq)→ SO₄^2- (aq)+ 3H⁺ (aq) + 2I^- (aq)

Colour changes: K⁺, HSO₄^- and SO₄^2- are all colourless, so the colour change comes from I₂ ( brown colour) being reduced to 2I^_- (colourless).

Half equations: Br₂ (aq) + 2e^- → 2Br^- (aq) and 2I^- (aq)→ I₂ (aq) + 2e^-

Full equations: Br₂ (aq) +2I^- (aq) → 2Br^- (aq)+ I₂ (aq)

Colour changes: Br₂ (Orange/brown) is reduced to Br^- (colourless). The reason the overall colour change goes to brown is due to the I^- (colourless) being oxidized to I₂ (brown).

Half equations: Cr₂O_7^2- (aq) +14H⁺ (aq) + 6e^- → 2Cr³⁺ (aq) + 7H₂O (l) and SO₂ (g) + 2H₂O (l) → SO₄^2- (aq) + 4H⁺ (aq) +2e^-

Full equations: Cr₂O_7^2- (aq) + 2H⁺ (aq) + 3SO₂ (g) → 2Cr³⁺ (aq) + H₂O (l) + 3SO₄^2- (aq)

Colour changes: K⁺ in K₂Cr₂0₇ is a colourless ion, and dichromate (Cr₂O₇^2-) is an orange solution. SO₂ is a colourless gas, so the intial solution is orange.
The solution turns green because the Cr₂O₇^2- is reduced to Cr³⁺ which is a green ion- so the solution turns from orange to green.

Half equations: 2Fe³⁺ (aq) + 2e^- → 2Fe²⁺ (aq) and SO₂ (aq) + 2H₂O (l) + 2Fe³⁺ (aq) → SO_4^2- (aq) + 4H⁺ (aq) + 2e^-

Full equations: SO₂ (aq) + 2H₂O (l) + 2Fe³⁺ (aq) → SO_4^2- (aq) + 4H⁺ (aq) + 2Fe²⁺

Colour changes: Both NO₃^- and SO₂ are colourless, but Fe³⁺ is orange. When Fe(NO₃)₃ is reduced to Fe²⁺ the solution turns a pale green as Fe²⁺ is a green ion and SO_4^2- is a colourless ion, so is H⁺.

Half equations: Fe²⁺ (aq) → Fe³⁺ (aq) + e^- and MnO_4^2- (aq) + 8H⁺(aq) + 5e^- → Mn²⁺ (aq) + 4H₂O (I)

Full equations: MnO_4^2- (aq) + 8H⁺(aq) + 5Fe²⁺ (aq) → Mn²⁺ (aq) + 4H₂O (I) + Fe³⁺ (aq)

Colour change: The K⁺ ion in KMnO₄ is colourless and the permanganate solution is purple. The iron sulfate is a colourless solution. When MnO₄^- is reduced to Mn²⁺ the solution changes to a colourless solution. The Fe²⁺ is oxidized to Fe³⁺ ions which are colourless.