or each reaction, write the chemical formulae of the oxidized reactants in the space provid rovid ed.

Question

Accepted Answer

# Explanation:
## Step1: Determine oxidation states
Oxidation is the loss of electrons (increase in oxidation state), reduction is the gain of electrons (decrease in oxidation state).
For $FeSO_4(aq)+Mg(s)\to Fe(s)+MgSO_4(aq)$:
- In $Mg(s)$, oxidation state of $Mg$ is $0$. In $MgSO_4$, oxidation state of $Mg$ is $+ 2$. So $Mg$ is oxidized.
- In $FeSO_4$, oxidation state of $Fe$ is $+2$. In $Fe(s)$, oxidation state of $Fe$ is $0$. So $Fe^{2 +}$ (from $FeSO_4$) is reduced.

For $Ca(s)+ZnCl_2(aq)\to Zn(s)+CaCl_2(aq)$:
- In $Ca(s)$, oxidation state of $Ca$ is $0$. In $CaCl_2$, oxidation state of $Ca$ is $+2$. So $Ca$ is oxidized.
- In $ZnCl_2$, oxidation state of $Zn$ is $+2$. In $Zn(s)$, oxidation state of $Zn$ is $0$. So $Zn^{2+}$ (from $ZnCl_2$) is reduced.

For $FeSO_4(aq)+Zn(s)\to Fe(s)+ZnSO_4(aq)$:
- In $Zn(s)$, oxidation state of $Zn$ is $0$. In $ZnSO_4$, oxidation state of $Zn$ is $+2$. So $Zn$ is oxidized.
- In $FeSO_4$, oxidation state of $Fe$ is $+2$. In $Fe(s)$, oxidation state of $Fe$ is $0$. So $Fe^{2+}$ (from $FeSO_4$) is reduced.

# Answer:
- For $FeSO_4(aq)+Mg(s)\to Fe(s)+MgSO_4(aq)$:
  - reactants oxidized: $Mg$
  - reactants reduced: $FeSO_4$
- For $Ca(s)+ZnCl_2(aq)\to Zn(s)+CaCl_2(aq)$:
  - reactants oxidized: $Ca$
  - reactants reduced: $ZnCl_2$
- For $FeSO_4(aq)+Zn(s)\to Fe(s)+ZnSO_4(aq)$:
  - reactants oxidized: $Zn$
  - reactants reduced: $FeSO_4$

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QUESTION IMAGE

Question

Explanation:

Step1: Determine oxidation states

Answer: