Explanation:
To balance the given redox reaction in basic solution:
1. Write the unbalanced equation:
\[ \text{CrO}_2^{-1} + \text{S}_2\text{O}_8^{-2} = \text{CrO}_4^{-2} + \text{SO}_4^{-2} \]
2. Identify the oxidation states of each element and determine which elements are oxidized and reduced.
3. Balance the atoms that are neither Cr nor S first. In this case, balance the O atoms by adding H₂O:
\[ \text{CrO}_2^{-1} + \text{S}_2\text{O}_8^{-2} = \text{CrO}_4^{-2} + \text{SO}_4^{-2} + \text{H}_2\text{O} \]
4. Balance the H atoms by adding H⁺ ions:
\[ \text{CrO}_2^{-1} + \text{S}_2\text{O}_8^{-2} + 6\text{H}^+ = \text{CrO}_4^{-2} + \text{SO}_4^{-2} + \text{H}_2\text{O} \]
5. Balance the charge by adding electrons (e⁻) to the more positive side:
\[ \text{CrO}_2^{-1} + \text{S}_2\text{O}_8^{-2} + 6\text{H}^+ = \text{CrO}_4^{-2} + \text{SO}_4^{-2} + \text{H}_2\text{O} + 6\text{e}^- \]
6. Balance Cr and S atoms by adding appropriate coefficients:
\[ 3\text{CrO}_2^{-1} + \text{S}_2\text{O}_8^{-2} + 6\text{H}^+ = 3\text{CrO}_4^{-2} + 2\text{SO}_4^{-2} + \text{H}_2\text{O} + 6\text{e}^- \]
Now, the equation is balanced in terms of atoms and charge.
