Sure! To balance the redox reaction in basic solution, you can follow these steps:
1. Separate the reaction into two half-reactions: the oxidation half-reaction and the reduction half-reaction.
Oxidation half-reaction: Zn -> Zn(OH)4^(-2)
Reduction half-reaction: H2 -> 2OH^(-)
2. Balance the atoms other than oxygen and hydrogen in each half-reaction.
Oxidation half-reaction: Zn -> Zn(OH)4^(-2)
Reduction half-reaction: H2 -> 2OH^(-) + 2e^-
3. Balance the oxygen atoms by adding H2O molecules to the appropriate side of the half-reaction.
Oxidation half-reaction: Zn + 4H2O -> Zn(OH)4^(-2)
Reduction half-reaction: H2 -> 2OH^(-) + 2e^-
4. Balance the hydrogen atoms by adding H^+ ions to the appropriate side of the half-reaction.
Oxidation half-reaction: Zn + 4H2O -> Zn(OH)4^(-2) + 8H^+
Reduction half-reaction: H2 + 2OH^- -> 2H2O + 2e^-
5. Balance the charges by adding electrons to the appropriate side of each half-reaction.
Oxidation half-reaction: Zn + 4H2O + 8e^- -> Zn(OH)4^(-2) + 8H^+
Reduction half-reaction: H2 + 2OH^- -> 2H2O + 2e^-
6. Multiply the half-reactions by appropriate coefficients to equalize the number of electrons in both reactions.
To balance the electrons, we need to multiply the reduction half-reaction by 4 and the oxidation half-reaction by 2.
Oxidation half-reaction: 2Zn + 8H2O + 16e^- -> 2Zn(OH)4^(-2) + 16H^+
Reduction half-reaction: 4H2 + 8OH^- -> 8H2O + 8e^-
7. Combine the half-reactions and cancel out the species that appear on both sides of the equation.
2Zn + 8H2O + 16e^- + 4H2 + 8OH^- -> 2Zn(OH)4^(-2) + 16H^+ + 8H2O + 8e^-
Simplifying the equation:
2Zn + 16OH^- + 4H2 -> 2Zn(OH)4^(-2) + 16H^+
Therefore, the balanced redox reaction in basic solution is:
2Zn + 16OH^- + 4H2 -> 2Zn(OH)4^(-2) + 16H^+
Can you give me the solution of zn=zn(oh)4-2+H2 in basic solution of redox balance reaction
1 answer