given the two half - reactions, what must be done in the next step before the reaction can be balanced…
given the two half - reactions, what must be done in the next step before the reaction can be balanced? au^{3+}→au i^{-}→i_{2} how many electrons are produced/used in this redox reaction when it is balanced? 2 3 6 done
Answer
Explanation:
Step1: Balance atoms in half - reactions
For $Au^{3+}\to Au$, it is already balanced in terms of atoms. For $I^-\to I_2$, we need to put a coefficient of 2 in front of $I^-$ to get $2I^-\to I_2$.
Step2: Balance charges in half - reactions
For $Au^{3+}\to Au$, the charge on the left is +3 and on the right is 0. So we add 3 electrons to the left side: $Au^{3+}+ 3e^-\to Au$. For $2I^-\to I_2$, the charge on the left is - 2 and on the right is 0. So we add 2 electrons to the right side: $2I^-\to I_2 + 2e^-$.
Step3: Make number of electrons equal
To make the number of electrons equal in both half - reactions, we multiply the first half - reaction by 2 and the second by 3. First half - reaction: $2Au^{3+}+6e^-\to 2Au$. Second half - reaction: $6I^-\to 3I_2+6e^-$.
Answer:
6