question 2\nnot yet answered\nmarked out of 1.00\np flag question\ncopper can undergo a series of reactions…

question 2\nnot yet answered\nmarked out of 1.00\np flag question\ncopper 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.\nreactions in the cycle of copper\nequation i\ncu(s) + 4hno₃(aq) → cu(no₃)₂(aq) + 2no₂(g) + 2h₂o(l)\nequation ii\ncu(no₃)₂(aq) + 2naoh(aq) → cu(oh)₂(s) + 2nano₃(aq)\nequation iii\ncu(oh)₂(s) → cuo(s) + h₂o(l)\nequation iv\ncuo(s) + h₂so₄(aq) → cuso₄(aq) + h₂o(l)\nequation v\ncuso₄(aq) + zn(s) → znso₄(aq) + cu(s)\nthe oxidation - reduction reactions in the equations give above are\na. i and ii only\nb. i, ii, iii, and iv\nc. i and v only\nd. i, iii, and v\nclear my choice

question 2\nnot yet answered\nmarked out of 1.00\np flag question\ncopper 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.\nreactions in the cycle of copper\nequation i\ncu(s) + 4hno₃(aq) → cu(no₃)₂(aq) + 2no₂(g) + 2h₂o(l)\nequation ii\ncu(no₃)₂(aq) + 2naoh(aq) → cu(oh)₂(s) + 2nano₃(aq)\nequation iii\ncu(oh)₂(s) → cuo(s) + h₂o(l)\nequation iv\ncuo(s) + h₂so₄(aq) → cuso₄(aq) + h₂o(l)\nequation v\ncuso₄(aq) + zn(s) → znso₄(aq) + cu(s)\nthe oxidation - reduction reactions in the equations give above are\na. i and ii only\nb. i, ii, iii, and iv\nc. i and v only\nd. i, iii, and v\nclear my choice

Answer

Explanation:

Step1: Identify oxidation - reduction reactions

Oxidation - reduction (redox) reactions involve a change in oxidation state of elements. In Equation I: $Cu(s)+4HNO_3(aq)\rightarrow Cu(NO_3)_2(aq) + 2NO_2(g)+2H_2O(l)$, 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: $Cu(NO_3)_2(aq)+2NaOH(aq)\rightarrow Cu(OH)_2(s)+2NaNO_3(aq)$ is a double - displacement reaction with no change in oxidation states. In Equation III: $Cu(OH)_2(s)\rightarrow CuO(s)+H_2O(l)$ is a decomposition reaction with no change in oxidation states. In Equation IV: $CuO(s)+H_2SO_4(aq)\rightarrow CuSO_4(aq)+H_2O(l)$ is an acid - base reaction with no change in oxidation states. In Equation V: $CuSO_4(aq)+Zn(s)\rightarrow ZnSO_4(aq)+Cu(s)$, 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