Answer:
2Ag(s) + Zn²+(aq) + 2H2O(l) → 2Ag₂O(aq) + Zn(s) + 4OH-(aq)
Step-by-step explanation:
First, let's write the half-reactions for this redox reaction:
Oxidation Half-reaction: Ag(s) → Ag₂O(aq)
Reduction Half-reaction: Zn²+(aq) → Zn(s)
To balance the oxidation half-reaction, we first need to balance the number of oxygen atoms by adding H2O to the left side:
Ag(s) + H2O(l) → Ag₂O(aq)
Next, we need to balance the number of hydrogen atoms by adding OH- to the left side:
Ag(s) + H2O(l) + 2OH-(aq) → Ag₂O(aq) + 2OH-(aq)
To balance the reduction half-reaction, we first balance the zinc atoms by adding 2 electrons to the right side:
Zn²+(aq) + 2e- → Zn(s)
Now we have to balance the number of electrons between the two half-reactions. To do this, we multiply the oxidation half-reaction by 2 and the reduction half-reaction by 1 and add them together:
2Ag(s) + 2H2O(l) + 4OH-(aq) + Zn²+(aq) → 2Ag₂O(aq) + 2OH-(aq) + Zn(s)
Finally, we cancel out the OH- ions on both sides of the equation and simplify:
2Ag(s) + Zn²+(aq) + 2H2O(l) → 2Ag₂O(aq) + Zn(s) + 4OH-(aq)
Therefore, the balanced redox reaction in basic conditions is:
2Ag(s) + Zn²+(aq) + 2H2O(l) → 2Ag₂O(aq) + Zn(s) + 4OH-(aq)