7. 2Sr + O2 2SrO
a) Determine what is oxidized and what is reduced
b) Identify the oxidizing agent and the reducing agent
8. 2Cs + Br2 2CsBr
a) Determine what is oxidized and what is reduced
b) Identify the oxidizing agent and the reducing agent
Can you write the states of each compound? I'll teach you how to do the first one and you can try the second one
7. Well, it looks like Sr is being oxidized because it's going from an oxidation state of 0 to +2 in SrO. And on the flip side, O2 is being reduced from an oxidation state of 0 to -2 in SrO. So we've got Sr being all oxidized and O2 getting reduced.
As for the oxidizing agent and reducing agent, the oxidizing agent is the one causing the oxidation, so in this case it would be O2. And the reducing agent is the one causing reduction, so that would be Sr. They're like the Batman and Robin of chemical reactions, working together to create SrO.
8. Ah, another chemical reaction for us to tackle! Here we have Cs being oxidized because it's going from an oxidation state of 0 to +1 in CsBr. And Br2 is getting reduced because it's going from an oxidation state of 0 to -1 in CsBr. So Cs is going through the oxidation dance and Br2 is doing the reduction shuffle.
Now, for the oxidizing agent and reducing agent. The oxidizing agent is the one that causes oxidation, so that would be Br2 in this case. And the reducing agent is the one causing reduction, which is Cs. They're like the peanut butter and jelly of chemical reactions, coming together to form CsBr.
To determine what is oxidized and what is reduced in a chemical reaction, you need to assign oxidation numbers to each element in the reactants and products.
Let's break down each reaction and find the oxidation numbers for the elements involved:
7. In the reaction 2Sr + O2 -> 2SrO, the oxidation number of Sr changes from 0 to +2, and the oxidation number of O changes from 0 to -2.
a) Based on the change in oxidation numbers, we can conclude that Sr is oxidized, and O is reduced.
b) The oxidizing agent is the species that causes oxidation, and the reducing agent is the species that causes reduction. In this case, O2 causes the oxidation of Sr, so O2 is the oxidizing agent. Sr is being reduced, so it is the reducing agent.
8. In the reaction 2Cs + Br2 -> 2CsBr, the oxidation number of Cs changes from 0 to +1, and the oxidation number of Br changes from 0 to -1.
a) Based on the change in oxidation numbers, we can conclude that Cs is oxidized, and Br is reduced.
b) The oxidizing agent is the species that causes oxidation, and the reducing agent is the species that causes reduction. In this case, Br2 causes the oxidation of Cs, so Br2 is the oxidizing agent. Cs is being reduced, so it is the reducing agent.
In summary:
7. a) Sr is oxidized, and O is reduced.
b) The oxidizing agent is O2, and the reducing agent is Sr.
8. a) Cs is oxidized, and Br is reduced.
b) The oxidizing agent is Br2, and the reducing agent is Cs.
Dgis shot
7. Sr zero/Sr+2 lost two electrons
O in O2 zero/ O -2 gained two electrons
8. Cs zero/Cs +1 lost an electron
Br in Br2 zero/Br-1 gained an electron