Answer:
Concentration of [tex]Cr_2O_7^{2-}[/tex] = 0.03101 M
Concentration of [tex]MnO_4^-[/tex] = 0.03721 M
Explanation:
A)
The reduction for [tex]Cr_2O_7^{2-}[/tex] is;
[tex]Cr_2O_7^{2-} + 14 H ^+ _{(aq)} + 6 e^- -----> 2 Cr^{3+} _{(aq)}+7H_2O _{(l)}[/tex]
[tex]Cu^+_{(aq)} -----> Cu^{2+} _{(aq)} + 1 e^-[/tex]
6 moles of [tex]Cu ^+[/tex] = 1 mole of [tex]Cr_2O_7^{2-}[/tex]
number of moles of Cu reacted = [tex]\frac{mass \ of \ Cu \ wire }{ molecular weigh tof \ Cu wire }[/tex]
number of moles of Cu reacted = [tex]\frac{0.4749}{63.55}[/tex]
number of moles of Cu reacted = 0.00747 mole
number of moles of [tex]Cr_2O_7^{2-}[/tex]reacted = [tex]\frac{0.00747}{6}[/tex]
number of moles of [tex]Cr_2O_7^{2-}[/tex]reacted = 0.001245 mole
Concentration of [tex]Cr_2O_7^{2-}[/tex] = [tex]\frac{number \ of moles }{Volume}[/tex]
Given that the volume = 40.15 mL = [tex]40.15 *10^{-3}[/tex]; we have:
Concentration of [tex]Cr_2O_7^{2-}[/tex] = [tex]\frac{0.001245}{40.15*10^{-3}}[/tex]
Concentration of [tex]Cr_2O_7^{2-}[/tex] = 0.03101 M
B)
The reduction for [tex]MnO_4^-[/tex] is;
[tex]MnO_4^- + 8H^+ + 5 e^- -----> Mn^{2+} + 4H_2O[/tex]
[tex]Cu^+_{(aq)} -----> Cu^{2+} _{(aq)} + 1 e^-[/tex]
5 moles of [tex]Cu ^+[/tex] = 1 mole of [tex]Cr_2O_7^{2-}[/tex]
number of moles of Cu reacted = [tex]\frac{mass \ of \ Cu \ wire }{ molecular weigh tof \ Cu wire }[/tex]
number of moles of Cu reacted = [tex]\frac{0.4749}{63.55}[/tex]
number of moles of Cu reacted = 0.00747 mole
number of moles of [tex]MnO_4^-[/tex] reacted = [tex]\frac{0.00747}{5}[/tex]
number of moles of [tex]MnO_4^-[/tex] reacted = 0.001494 mole
Concentration of [tex]MnO_4^-[/tex] = [tex]\frac{number \ of moles }{Volume}[/tex]
Given that the volume = 40.15 mL = [tex]40.15 *10^{-3}[/tex]; we have:
Concentration of [tex]MnO_4^-[/tex] = [tex]\frac{0.001494 }{40.15*10^{-3}}[/tex]
Concentration of [tex]MnO_4^-[/tex] = 0.03721 M
