Question

Consider the following balanced redox reaction: 2CrO2-(aq) + 2H2O(l) + 6ClO-(aq) LaTeX: \longrightarrow⟶ 2CrO42-(aq) + 3Cl2(g) + 4OH-(aq) 1. Which species is being oxidized? 2. Which species is being reduced? 3. Which species is the oxidizing agent? 4. Which species is the reducing agent? 5. How many electrons are being transferred? Hint: If you were to balance this equation how many electrons would be in each half-reaction? That is how many electrons are transferred.

Answer 1

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:

$2(Cr^{+3}O^{-2}_2)^-(aq) + 2H_2O(l) + 6(Cl^{+1}O^{-2})^-(aq)\longrightarrow 2(Cr^{+6}O^{-2}_4)^{2-}(aq) + 3Cl^0_2(g) + 4OH^-(aq)$

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:

$2(Cr^{+3}O^{-2}_2)^-(aq) \longrightarrow 2(Cr^{+6}O^{-2}_4)^{2-}(aq)+6e^-$

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:

$6(Cl^{+1}O^{-2})^-+6e^-\longrightarrow 3Cl^0_2(g)$

Therefore, there are 12 electrons that are being transferred, 6 for chromium and 6 for chlorine.

Best regards.