Zn(OH)42- + As -> Zn + H2ASO- basic solution

Answer 1

To balance the given equation, we need to ensure that the number of atoms of each element is equal on both sides of the equation.

First, we write the oxidation half-reaction and reduction half-reaction:

Oxidation half-reaction: As → H2AsO-4 + 4H+ + 4e-

Reduction half-reaction: Zn(OH)4^2- + 4e- → Zn + 4OH-

Next, we balance the half-reactions:

Oxidation half-reaction: As + 4OH- → H2AsO-4 + 4H2O + 4e-

Reduction half-reaction: Zn(OH)4^2- + 4e- → Zn + 4OH-

Now, we balance the number of electrons transferred in each half-reaction by multiplying the oxidation half-reaction by 4:

4As + 16OH- → 4H2AsO-4 + 16H2O + 16e-

Zn(OH)4^2- + 4e- → Zn + 4OH-

Finally, we can add the two half-reactions together and cancel out any common species to obtain the balanced equation:

4As + Zn(OH)4^2- + 16OH- → Zn + 4H2AsO-4 + 16H2O

Therefore, the balanced equation for the reaction in basic solution is:

4As + Zn(OH)4^2- + 16OH- → Zn + 4H2AsO-4 + 16H2O.