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copper can undergo a series of reactions known as the cycle of copper. in this cycle, the initial copper reactant is changed into different compounds before being recovered as copper metal in the last step. the series of reactions is represented by the following equations.
reactions in the cycle of copper
equation i
cu(s) + 4hno₃(aq) → cu(no₃)₂(aq) + 2no₂(g) + 2h₂o(l)
equation ii
cu(no₃)₂(aq) + 2naoh(aq) → cu(oh)₂(s) + 2nano₃(aq)
equation iii
cu(oh)₂(s) → cuo(s) + h₂o(l)
equation iv
cuo(s) + h₂so₄(aq) → cuso₄(aq) + h₂o(l)
equation v
cuso₄(aq) + zn(s) → znso₄(aq) + cu(s)
the oxidation - reduction reactions in the equations give above are
a. i and ii only
b. i, ii, iii, and iv
c. i and v only
d. i, iii, and v
clear my choice

Explanation:

Step1: Identify oxidation - reduction reactions

Oxidation - reduction (redox) reactions involve a change in oxidation state of elements. In Equation I: $Cu(s)+4HNO_3(aq)
ightarrow Cu(NO_3)_2(aq) + 2NO_2(g)+2H_2O(l)$, copper goes from 0 oxidation state to + 2 oxidation state, and nitrogen in $HNO_3$ is reduced from + 5 to + 4 in $NO_2$, so it is a redox reaction. In Equation II: $Cu(NO_3)_2(aq)+2NaOH(aq)
ightarrow Cu(OH)_2(s)+2NaNO_3(aq)$ is a double - displacement reaction with no change in oxidation states. In Equation III: $Cu(OH)_2(s)
ightarrow CuO(s)+H_2O(l)$ is a decomposition reaction with no change in oxidation states. In Equation IV: $CuO(s)+H_2SO_4(aq)
ightarrow CuSO_4(aq)+H_2O(l)$ is an acid - base reaction with no change in oxidation states. In Equation V: $CuSO_4(aq)+Zn(s)
ightarrow ZnSO_4(aq)+Cu(s)$, zinc is oxidized from 0 to + 2 and copper is reduced from + 2 to 0, so it is a redox reaction.

Answer:

a. I and V only