Chemistry B Mrs. Frost Oxidation-Reduction Homework Packet 12. Use the changes in oxidation numbers to identify which atoms are oxidized and which are reduced in each reaction. A. \[ 2 \mathrm{Na}_{(s)}+\mathrm{Cl}_{2(g)} \rightarrow 2 \mathrm{NaCl}_{(s)} \] Oxidized Reduced B. \[ 2 \mathrm{HNO}_{3(a q)}+6 \mathrm{HI}_{(a q)} \rightarrow 2 \mathrm{NO}_{(g)}+3 \mathrm{I}_{2(s)}+4 \mathrm{H}_{2} \mathrm{O}_{(\ell)} \] Oxidized Reduced C. \[ 3 \mathrm{H}_{2} \mathrm{~S}_{(g)}+2 \mathrm{HNO}_{3(a q)} \rightarrow 3 \mathrm{~S}_{(s)}+2 \mathrm{NO}_{(g)}+4 \mathrm{H}_{2} \mathrm{O}_{(\hbar)} \] Oxidized Reduced D. \[ 2 \mathrm{PbSO}_{4(s)}+2 \mathrm{H}_{2} \mathrm{O}_{(l)} \rightarrow \mathrm{Pb}_{(s)}+\mathrm{PbO}_{2(s)}+2 \mathrm{H}_{2} \mathrm{SO}_{4(a q)} \] Oxidized Reduced 13. For each reaction in Problem $\#$, identify the reducing agent and the oxidizing agent. A. \[ 2 \mathrm{Na}_{(s)}+\mathrm{Cl}_{2(g)} \rightarrow 2 \mathrm{NaCl}_{(s)} \] Oxidizing Agent Reducing Agent B. \[ 2 \mathrm{HNO}_{3(a q)}+6 \mathrm{HI}_{(a q)} \rightarrow 2 \mathrm{NO}_{(g)}+3 \mathrm{I}_{2(s)}+4 \mathrm{H}_{2} \mathrm{O}_{(l)} \] Oxidizing Agent Reducing Agent C. \[ 3 \mathrm{H}_{2} \mathrm{~S}_{(g)}+2 \mathrm{HNO}_{3(a q)} \rightarrow 3 \mathrm{~S}_{(s)}+2 \mathrm{NO}_{(g)}+4 \mathrm{H}_{2} \mathrm{O}_{(t)} \] Oxidizing Agent Reducing Agent D. \[ 2 \mathrm{PbSO}_{4(s)}+2 \mathrm{H}_{2} \mathrm{O}_{(\hbar)} \rightarrow \mathrm{Pb}_{(s)}+\mathrm{PbO}_{2(s)}+2 \mathrm{H}_{2} \mathrm{SO}_{4(a q)} \] Oxidizing Agent Reducing Agent

Step 1 ：First, we need to find the oxidation numbers of each element in the reactants and products.

Step 2 ：For reaction A: \(2 \mathrm{Na}_{(s)} + \mathrm{Cl}_{2(g)} \rightarrow 2 \mathrm{NaCl}_{(s)}\), the oxidation numbers are: Na: 0 to +1, Cl: 0 to -1. So, Na is oxidized and Cl is reduced.

Step 3 ：For reaction B: \(2 \mathrm{HNO}_{3(a q)}+6 \mathrm{HI}_{(a q)} \rightarrow 2 \mathrm{NO}_{(g)}+3 \mathrm{I}_{2(s)}+4 \mathrm{H}_{2} \mathrm{O}_{(\ell)}\), the oxidation numbers are: N: +5 to +2, I: -1 to 0, H: +1 (no change), O: -2 (no change). So, N is reduced and I is oxidized.

Step 4 ：For reaction C: \(3 \mathrm{H}_{2} \mathrm{~S}_{(g)}+2 \mathrm{HNO}_{3(a q)} \rightarrow 3 \mathrm{~S}_{(s)}+2 \mathrm{NO}_{(g)}+4 \mathrm{H}_{2} \mathrm{O}_{(\hbar)}\), the oxidation numbers are: S: -2 to 0, N: +5 to +2, H: +1 (no change), O: -2 (no change). So, S is oxidized and N is reduced.

Step 5 ：For reaction D: \(2 \mathrm{PbSO}_{4(s)}+2 \mathrm{H}_{2} \mathrm{O}_{(\hbar)} \rightarrow \mathrm{Pb}_{(s)}+\mathrm{PbO}_{2(s)}+2 \mathrm{H}_{2} \mathrm{SO}_{4(a q)}\), the oxidation numbers are: Pb: +2 to 0 and +4, S: +6 (no change), O: -2 (no change), H: +1 (no change). So, Pb is both oxidized and reduced.

Step 6 ：Next, we identify the reducing and oxidizing agents for each reaction.

Step 7 ：For reaction A: Oxidizing Agent: \(\mathrm{Cl}_{2}\), Reducing Agent: \(\mathrm{Na}\)

Step 8 ：For reaction B: Oxidizing Agent: \(\mathrm{HNO}_{3}\), Reducing Agent: \(\mathrm{HI}\)

Step 9 ：For reaction C: Oxidizing Agent: \(\mathrm{HNO}_{3}\), Reducing Agent: \(\mathrm{H}_{2}\mathrm{S}\)

Step 10 ：For reaction D: Oxidizing Agent: \(\mathrm{PbSO}_{4}\), Reducing Agent: \(\mathrm{PbSO}_{4}\)