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for each reaction in the table below, write the chemical formulae of an…

Question

for each reaction in the table below, write the chemical formulae of any reactants that will be oxidized in the second column of the table. write the chemical formulae of any reactants that will be reduced in the third column. reaction reactants oxidized reactants reduced $\ce{pb(s) + 2cl_{2}(g) -> pbcl_{4}(s)}$ $\ce{o_{2}(g) + cr(s) -> cro_{2}(s)}$ $\ce{8mg(s) + s_{8}(s) -> 8mgs(s)}$

Explanation:

Step1: Determine oxidation states

  • For \(Pb(s)+2Cl_{2}(g)\to PbCl_{4}(s)\):
  • Oxidation state of \(Pb\) in \(Pb(s)\) is \(0\), in \(PbCl_{4}\) is \(+4\) (since \(Cl\) is \(- 1\) and \(x + 4\times(-1)=0\) gives \(x = + 4\)).
  • Oxidation state of \(Cl\) in \(Cl_{2}(g)\) is \(0\), in \(PbCl_{4}\) is \(-1\).
  • \(Pb\) is oxidized (loses electrons, oxidation state increases), \(Cl_{2}\) is reduced (gains electrons, oxidation state decreases).

Step2: For \(O_{2}(g)+Cr(s)\to CrO_{2}(s)\):

  • Oxidation state of \(Cr\) in \(Cr(s)\) is \(0\), in \(CrO_{2}\) is \(+4\) (since \(x+2\times(-2) = 0\) gives \(x=+4\)).
  • Oxidation state of \(O\) in \(O_{2}(g)\) is \(0\), in \(CrO_{2}\) is \(-2\).
  • \(Cr\) is oxidized, \(O_{2}\) is reduced.

Step3: For \(8Mg(s)+S_{8}(s)\to8MgS(s)\):

  • Oxidation state of \(Mg\) in \(Mg(s)\) is \(0\), in \(MgS\) is \(+2\) (since \(x+(-2)=0\) gives \(x = + 2\)).
  • Oxidation state of \(S\) in \(S_{8}(s)\) is \(0\), in \(MgS\) is \(-2\).
  • \(Mg\) is oxidized, \(S_{8}\) is reduced.

Answer:

reactionreactants oxidizedreactants reduced
\(O_{2}(g)+Cr(s)\to CrO_{2}(s)\)\(Cr\)\(O_{2}\)
\(8Mg(s)+S_{8}(s)\to8MgS(s)\)\(Mg\)\(S_{8}\)