Chapter U1, Problem 17STP

Interpretation Introduction

Interpretation:

The correct statement for the copper ion in copper cycle needs to be determined.

Concept introduction:

In copper cycle, it is observed that mass of copper is conversed in a chemical reaction.

Answer to Problem 17STP

(B) only copper (II) is present in the ionic compounds.

Explanation of Solution

Reason for correct option:

STEP-1 :

In copper cycle, copper metal react with nitric acid HNO₃ to give copper (II) nitrate $Cu{\left(N{O}_3\right)}_2$, and NO₂ and H₂O asbi-product.

The reaction is represented as follows:

$Cu\left(s\right)+2HN{O}_3\left(aq\right)+3O\to Cu{\left(N{O}_3\right)}_2\left(aq\right)+N{O}_2\left(g\right)+H_{2}O\left(l\right)$

STEP-2 :

Copper (II) nitrate $Cu{\left(N{O}_3\right)}_2$ react with sodium hydroxide NaOH to give copper (II) hydroxide $Cu{\left(OH\right)}_2$.

The reaction is represented as follows:

$Cu{\left(N{O}_3\right)}_2\left(aq\right)+NaOH\left(aq\right)\to Cu{\left(OH\right)}_2\left(s\right)+NaN{O}_3\left(aq\right)$

STEP-3:

When copper (II) hydroxide $Cu{\left(OH\right)}_2$ is heated,it gives copper (II) oxide CuO and then react in presence of acid H₂SO₄ to give copper (II) sulfate CuSO₄. In last step of this cycle, copper (II) sulfate CuSO₄isreduced in presence of magnesium Mg to give copper metal and magnesium sulfate MgSO₄as a bi-product.

$Cu{\left(OH\right)}_2\left(s\right)\to CuO\left(s\right)+H_{2}O\left(l\right)$

$CuO\left(s\right)+H_{2}S{O}_4\left(aq\right)\to CuS{O}_4\left(aq\right)+H_{2}O\left(l\right)$

$CuS{O}_4\left(aq\right)+Mg\left(s\right)\to Cu\left(s\right)+MgS{O}_4\left(aq\right)$

Thus, Cu is only present in +2 oxidation state.

Conclusion

Reasons for incorrect options:

(A) From the above cycle it seems that only copper (II) participate in the copper cycle.

(C) Only copper (II) present in Ionic compound.

(D) Both elemental copper and copper (II)present in the copper cycle.

Thus, the option (A), (B) and (C) are incorrect.