I’m pretty sure it’s abode and cathode
=<span><span><span><span><span><span><span><span><span>148.413159d</span><span>f2</span></span>i</span>l</span><span>o4</span></span>s</span>y</span>+<span><span>a<span>l2</span></span><span>o3</span></span></span>+<span>a<span>l is the answer and have a nice day :)</span></span></span>
Answer: Li is the reducing agentg and O is the oxidizing agent.
Explanation:
1) The oxidizing agent is the one that is reduced and the reducing agent is the one that is oxidized.
2) The given reaction is:
4Li(s) + O₂ (g) → 2 Li₂O(s)
3) Determine the oxidation states of each atom:
Li(s): oxidation state = 0 (since it is alone)
O₂ (g): oxidation state = 0 (since it is alone)
Li in Li₂O (s) +1
O in Li₂O -2
That because 2× (+1) - 2 = 0.
4) Determine the changes:
Li went from 0 to + 1, therefore it got oxidized and it is the reducing agent.
O went from 0 to - 2, therefore it got reduced and it is the oxidizing agent.
