for each reaction, write the chemical formulae of the oxidized reactants in the space provided. write the chemical formu provided.
$\mathrm{pb}(s)+\mathrm{hg}(\mathrm{no}_{3})_{2}(a q) \
ightarrow \mathrm{hg}(l)+\mathrm{pb}(\mathrm{no}_{3})_{2}(a q)$
$\begin{array}{|l|l|}hline & \text { reactants } \\ & \text { oxidized: } \\ hline & \text { reactants } \\ & \text { reduced: } \\ hline 2 \mathrm{fe}(s)+3 \mathrm{cucl}_{2}(a q) \
ightarrow 2 \mathrm{fecl}_{3}(a q)+3 \mathrm{cu}(s) & \text { reactants } \\ & \text { oxidized: } \\ hline & \text { reactants } \\ & \text { reduced: } \\ hline \mathrm{zncl}_{2}(a q)+2 \mathrm{na}(s) \
ightarrow \mathrm{zn}(s)+2 \mathrm{nacl}(a q) & \text { reactants } \\ & \text { oxidized: } \\ hline & \text { reactants } \\ & \text { reduced: } \\ hline \end{array}$

Question

Accepted Answer

# Explanation:
## Step1: Determine oxidation states
Oxidation state of $Pb$ in $Pb(s)$ is $0$, in $Pb(NO_3)_2(aq)$ is $+2$. Oxidation state of $Hg$ in $Hg(NO_3)_2(aq)$ is $+2$, in $Hg(l)$ is $0$. Oxidation state of $Fe$ in $Fe(s)$ is $0$, in $FeCl_3(aq)$ is $+3$. Oxidation state of $Cu$ in $CuCl_2(aq)$ is $+2$, in $Cu(s)$ is $0$. Oxidation state of $Na$ in $Na(s)$ is $0$, in $NaCl(aq)$ is $+1$. Oxidation state of $Zn$ in $ZnCl_2(aq)$ is $+2$, in $Zn(s)$ is $0$.
## Step2: Identify oxidized and reduced reactants
Oxidation is increase in oxidation state. Reduction is decrease in oxidation state.
- For $Pb(s)+Hg(NO_3)_2(aq)\to Hg(l)+Pb(NO_3)_2(aq)$: $Pb$ is oxidized ($0\to + 2$), $Hg$ is reduced ($+2\to0$).
- For $2Fe(s)+3CuCl_2(aq)\to2FeCl_3(aq)+3Cu(s)$: $Fe$ is oxidized ($0\to+3$), $Cu$ is reduced ($+2\to0$).
- For $ZnCl_2(aq)+2Na(s)\to Zn(s)+2NaCl(aq)$: $Na$ is oxidized ($0\to+1$), $Zn$ is reduced ($+2\to0$).

Reaction	Reactants oxidized	Reactants reduced
\(2Fe(s)+3CuCl_2(aq)\to2FeCl_3(aq)+3Cu(s)\)	\(Fe\)	\(CuCl_2\)
\(ZnCl_2(aq)+2Na(s)\to Zn(s)+2NaCl(aq)\)	\(Na\)	\(ZnCl_2\)

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QUESTION IMAGE

Question

Explanation:

Step1: Determine oxidation states

Step2: Identify oxidized and reduced reactants

Answer: