Question

A Cu2+ solution is prepared by dissolving a 0.4749 g piece of copper wire in acid. The solution is then passed through a Walden reductor, reducing Cu2+ to Cu+ . The resulting Cu+ solution required 40.15 mL of each of the titrants to reach the endpoint. Calculate the concentration of each titrant.

Answer 1

Answer:

Concentration of $Cr_2O_7^{2-}$ = 0.03101 M

Concentration of $MnO_4^-$ = 0.03721 M

Explanation:

A)

The reduction for $Cr_2O_7^{2-}$ is;

$Cr_2O_7^{2-} + 14 H ^+ _{(aq)} + 6 e^- -----> 2 Cr^{3+} _{(aq)}+7H_2O _{(l)}$

$Cu^+_{(aq)} -----> Cu^{2+} _{(aq)} + 1 e^-$

6 moles of $Cu ^+$ = 1 mole of $Cr_2O_7^{2-}$

number of moles of Cu reacted = $\frac{mass \ of \ Cu \ wire }{ molecular weigh tof \ Cu wire }$

number of moles of Cu reacted = $\frac{0.4749}{63.55}$

number of moles of Cu reacted = 0.00747 mole

number of moles of $Cr_2O_7^{2-}$reacted = $\frac{0.00747}{6}$

number of moles of $Cr_2O_7^{2-}$reacted = 0.001245 mole

Concentration of $Cr_2O_7^{2-}$ = $\frac{number \ of moles }{Volume}$

Given that the volume = 40.15 mL = $40.15 *10^{-3}$; we have:

Concentration of $Cr_2O_7^{2-}$ = $\frac{0.001245}{40.15*10^{-3}}$

Concentration of $Cr_2O_7^{2-}$ = 0.03101 M

B)

The reduction for $MnO_4^-$ is;

$MnO_4^- + 8H^+ + 5 e^- -----> Mn^{2+} + 4H_2O$

$Cu^+_{(aq)} -----> Cu^{2+} _{(aq)} + 1 e^-$

5 moles of $Cu ^+$ = 1 mole of $Cr_2O_7^{2-}$

number of moles of Cu reacted = $\frac{mass \ of \ Cu \ wire }{ molecular weigh tof \ Cu wire }$

number of moles of Cu reacted = $\frac{0.4749}{63.55}$

number of moles of Cu reacted = 0.00747 mole

number of moles of $MnO_4^-$ reacted = $\frac{0.00747}{5}$

number of moles of $MnO_4^-$ reacted = 0.001494 mole

Concentration of $MnO_4^-$ = $\frac{number \ of moles }{Volume}$

Given that the volume = 40.15 mL = $40.15 *10^{-3}$; we have:

Concentration of $MnO_4^-$ = $\frac{0.001494 }{40.15*10^{-3}}$

Concentration of $MnO_4^-$ = 0.03721 M