Answer:
1. Chromium
2. Chlorine.
3. Chlorine.
4. Chromium.
5. 12 electrons.
Explanation:
Hello,
In this case, the given reaction with the appropriate oxidation states turns out:
[tex]2(Cr^{+3}O^{-2}_2)^-(aq) + 2H_2O(l) + 6(Cl^{+1}O^{-2})^-(aq)\longrightarrow 2(Cr^{+6}O^{-2}_4)^{2-}(aq) + 3Cl^0_2(g) + 4OH^-(aq)[/tex]
In such a way, the oxidation half-reaction is written for chromium as the reducing agent so it is oxidized from +3 to +6, nonetheless, since there are two chromiums undergoing such change, 6 electrons are being transferred as shown below:
[tex]2(Cr^{+3}O^{-2}_2)^-(aq) \longrightarrow 2(Cr^{+6}O^{-2}_4)^{2-}(aq)+6e^-[/tex]
On the other hand, chlorine's reduction half-reaction as the oxidizing agent result from the transfer of 6 electrons as well from +1 to 0, nonetheless, there are 6 chlorines undergoing such change:
[tex]6(Cl^{+1}O^{-2})^-+6e^-\longrightarrow 3Cl^0_2(g)[/tex]
Therefore, there are 12 electrons that are being transferred, 6 for chromium and 6 for chlorine.
Best regards.
