you are given the reaction cu + hno₃ → cu(no₃)₂ + no + h₂o. half - reactions: 3cu → 3cu²⁺ + 6e⁻ 2n⁵⁺ + 6e⁻ →…

you are given the reaction cu + hno₃ → cu(no₃)₂ + no + h₂o. half - reactions: 3cu → 3cu²⁺ + 6e⁻ 2n⁵⁺ + 6e⁻ → 2n²⁺. complete the final balanced equation based on the half - reactions. □cu+□hno₃→□cu(no₃)₂+□no+□h₂o

you are given the reaction cu + hno₃ → cu(no₃)₂ + no + h₂o. half - reactions: 3cu → 3cu²⁺ + 6e⁻ 2n⁵⁺ + 6e⁻ → 2n²⁺. complete the final balanced equation based on the half - reactions. □cu+□hno₃→□cu(no₃)₂+□no+□h₂o

Answer

Explanation:

Step1: Combine half - reactions

The oxidation half - reaction is $3Cu\rightarrow3Cu^{2 +}+6e^{-}$ and the reduction half - reaction is $2N^{5 +}+6e^{-}\rightarrow2N^{2 +}$. Combining them gives $3Cu + 2N^{5+}\rightarrow3Cu^{2+}+2N^{2+}$.

Step2: Rewrite in terms of compounds

In the original reaction, $N^{5+}$ is from $HNO_3$ and $N^{2+}$ is in $NO$. Also, for every $Cu^{2+}$ in $Cu(NO_3)_2$, we need 2 nitrate ions from $HNO_3$. The $3Cu$ will form $3Cu(NO_3)_2$, which requires 6 additional nitrate ions from $HNO_3$. So the total number of $HNO_3$ molecules is 8 (2 for the reduction part and 6 for the nitrate part with $Cu$). The balanced equation is $3Cu+8HNO_3\rightarrow3Cu(NO_3)_2 + 2NO+4H_2O$. We can check the balance of atoms:

  • Cu atoms: 3 on the left and 3 on the right.
  • N atoms: 8 on the left ($8$ in $8HNO_3$) and $3\times2 + 2=8$ on the right ($6$ in $3Cu(NO_3)_2$ and $2$ in $2NO$).
  • O atoms: $8\times3 = 24$ on the left ($8HNO_3$), and $3\times(6)+2 + 4=24$ on the right ($3Cu(NO_3)_2$ has $3\times6$ O - atoms, $2NO$ has 2 O - atoms and $4H_2O$ has 4 O - atoms).
  • H atoms: $8\times1 = 8$ on the left and $4\times2=8$ on the right.

Answer:

$3Cu + 8HNO_3\rightarrow3Cu(NO_3)_2+2NO + 4H_2O$