To complete and balance the given equation, which involves the dichromate ion ($$Cr_2O_7^{2-}$$) reacting with ethylene glycol (C₂H₄O) to form acetic acid (C₂H₄O₂) and chromium (III) ion (Cr³⁺), we need to follow a systematic approach. First, let's write down the unbalanced equation:

$Cr_2O_7^{2-} + C_2H_4O \to C_2H_4O_2 + Cr^{3+}$

Next, we balance the equation in acidic solution using the half-reaction method, which involves balancing the reduction and oxidation half-reactions separately and then combining them.

Oxidation (Ethylene glycol to acetic acid):

$C_2H_4O \to C_2H_4O_2$

Reduction (Dichromate to Chromium (III)):

$Cr_2O_7^{2-} \to Cr^{3+}$

Oxidation doesn’t need balancing in this respect. For the reduction:

$Cr_2O_7^{2-} \to 2Cr^{3+}$

For the reduction half:

$Cr_2O_7^{2-} \to 2Cr^{3+} + 7H_2O$

For the reduction half:

$Cr_2O_7^{2-} + 14H^+ \to 2Cr^{3+} + 7H_2O$

The oxidation half does not involve oxygen or hydrogen atoms changing, so we skip balancing them directly here.

Oxidation half (2 electrons are needed to balance the change from +2 to +3 oxidation state for 2 Carbon atoms):

$C_2H_4O \to C_2H_4O_2 + 2e^-$

Reduction half (to balance it, we need to add 6 electrons):

$Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O$

To balance electrons, multiply the oxidation half-reaction by 3 and the reduction half-reaction by 1 (since 6 electrons are transferred in the reduction and 2 in the oxidation; the least common multiple is 6).

$3(C_2H_4O \to C_2H_4O_2 + 2e^-)$

$Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O$

Combine and simplify:

$Cr_2O_7^{2-} + 14H^+ + 3C_2H_4O \to 3C_2H_4O_2 + 2Cr^{3+} + 7H_2O$

$Cr_2O_7^{2-} + 14H^+ + 3C_2H_4O \to 3C_2H_4O_2 + 2Cr^{3+} + 7H_2O$

This represents the balanced chemical equation for the reaction between $$Cr_2O_7^{2-}$$ and ethylene glycol in an acidic medium, producing acetic acid and chromium (III) ions alongside water.