How many electrons are involved in the following redox reaction?

${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+{\mathrm{Fe}}^{2+}+{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\to {\mathrm{Cr}}^{3+}+{\mathrm{Fe}}^{3+}+{\mathrm{CO}}_2$ (unbalanced)

The reduction of ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\text{ to }2{\mathrm{Cr}}^{3+}$ requires 6 electrons.

Explanation:

${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}\to 2{\mathrm{Cr}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$

Step 1: Oxidation and Reduction Half-Reactions

• Reduction Half-Reaction (Chromium):

  Cr₂O₇²⁻ → Cr³⁺

  - In dichromate (Cr₂O₇²⁻), the oxidation state of chromium is +6.

  - Each chromium atom is reduced to +3.

  - Total change in oxidation state for two chromium atoms:

  2 × (6 − 3) = 6 electrons

• Oxidation Half-Reaction (Iron and Oxalate):

  Iron:

  Fe²⁺ → Fe³⁺

  - Each iron atom loses 1 electron.

  Oxalate (C₂O₄²⁻):

  C₂O₄²⁻ → 2CO₂

  - Carbon in oxalate has an oxidation state of +3, and in CO₂, it is +4.

  - Total change for two carbons is 2 electrons.

Step 2: Balance the Half-Reactions

For chromium:

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

For iron and oxalate:

6Fe²⁺ → 6Fe³⁺ + 6e⁻

3C₂O₄²⁻ → 6CO₂ + 6e⁻

Step 3: Combine the Balanced Half-Reactions

Combine reduction and oxidation reactions, ensuring electrons cancel out:

Cr₂O₇²⁻ + 6Fe²⁺ + 3C₂O₄²⁻ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 6CO₂ + 7H₂O

Step 4: Count the Total Electrons

- Total electrons transferred in the reaction:

• 6 electrons from the reduction of chromium.
• 6 electrons from the oxidation of iron and oxalate.

Total electrons involved: 6