Question

Write a balanced half-reaction for the reduction of dichromate ion cr2o−27 to chromium ion cr 3 in basic aqueous solution. be sure to add physical state symbols where appropriate.

Answer 1

Half reaction: Cr₂O₇²⁻(aq) + 7H₂O(l) + 6e⁻ → 2Cr³⁺(aq) + 14OH⁻(aq).
Chromium change oxidation number from +6 in dichromate ion Cr₂O₇²⁻ to +3 in chromium cation.
Sum of the charges on the left side of the half reaction is -2 and on the left side is -8 (2·(+3) +14·(-1)), so six electrons must be added on the left side of half reaction.

Answer 2

Explanation:

When electrons are added in a reaction then it means reduction is taking place. Whereas removal of electrons from a chemical reaction is known as oxidation.  

Oxidation state of chromium in $Cr_{2}O^{2-}_{7}$ is +6. So, its reduction half-reaction will be as follows.

         $Cr_{2}O_{7}^{2-} + 3e^{-1} \rightarrow Cr^{3+}$

Since it is given that reaction is taking place in basic solution. So, we add $H_{2}O$ on reactant side and $OH^{-}$ on the product side.

        $Cr_{2}O_{7}^{2-} + 3e^{-1} + H_{2}O \rightarrow Cr^{3+} + OH^{-}$

Now balancing Cr atom and charges on both sides, we get the following.

       $Cr_{2}O_{7}^{2-} + 6e^{-1} + 7H_{2}O \rightarrow 2Cr^{3+} + 14OH^{-}$