Question

Balance the half reaction in basic solution: (Cr₂O₇²- → Cr³+)

a. (Cr₂O₇²- + 14H⁺ + 6e⁻ → 2Cr³+ + 7H₂O)
b. (Cr₂O₇²- + 6H₂O + 6e⁻ → 2Cr³+ + 14OH⁻)
c. (Cr₂O₇²- + 6e⁻ → 2Cr³- + 7OH⁻)
d. (Cr₂O₇²- + 14OH⁻ → 2Cr³+ + 6H₂O + 6e⁻)

Answer 1

The balanced half-reaction in basic solution for the conversion of (Cr₂O₇²- → Cr³+) is (Cr₂O₇²- + 14OH⁻ → 2Cr³+ + 6H₂O + 6e⁻).

Step-by-step explanation:

The balanced half-reaction in basic solution for the conversion of (Cr₂O₇²- → Cr³+) is option d: (Cr₂O₇²- + 14OH⁻ → 2Cr³+ + 6H₂O + 6e⁻).

To balance the half-reaction, the charges on both sides of the equation need to be equal. In this case, the left side of the equation has a charge of 2- from the dichromate ion, while the right side has a charge of 6+ from the two chromium ions. By adding 6 OH⁻ ions (hydroxide ions) to the left side, the charges are balanced and the reaction is correctly balanced in basic solution.

Therefore, option d is the correct balanced half-reaction in basic solution for the conversion of (Cr₂O₇²- → Cr³+).