Final answer:
A balanced half-reaction for the reduction of aqueous vanadium(V) cations to solid vanadium is V5+ (aq) + 5e- → V(s), showing the gain of five electrons by the vanadium cation.
Step-by-step explanation:
To write a balanced half-reaction for the reduction of aqueous vanadium(V) cations to solid vanadium, you can follow these steps:
- Write the skeleton equation for the reduction process. For vanadium, this would be V5+ (aq) → V(s).
- Balance all elements except oxygen and hydrogen, which are already balanced in this case as there are no oxygen or hydrogen atoms involved in the reduction of vanadium.
- Balance charge by adding electrons. Since vanadium is being reduced, it gains electrons. Vanadium's oxidation state decreases from +5 to 0, meaning it gains 5 electrons. This gives us the equation V5+ (aq) + 5e- → V(s).
Both the charges and atoms are balanced, so the reduction half-reaction of aqueous vanadium(V) to solid vanadium is V5+ (aq) + 5e- → V(s).