Balance the following skeleton redox reactions using the half reaction method.
Ag2SeO3(aq) + H3PO3(aq) = Ag(s) + SeO3^2-(aq) + H3PO4 (aq)
To balance the redox reaction using the half-reaction method, we need to separate the reaction into two half-reactions: the oxidation half-reaction and the reduction half-reaction.
First, let's identify the oxidation states of the elements involved:
Ag2SeO3(aq) → Ag(s) + SeO3^2-(aq)
Ag: +1 (unchanged)
Se: +6 (unchanged)
O: -2 (unchanged)
H3PO3(aq) → H3PO4(aq)
P: +3 → +5
H: +1 (unchanged)
O: -2 (unchanged)
To balance the oxidation half-reaction:
Ag2SeO3(aq) → Ag(s) + SeO3^2-(aq)
Since the oxidation state of Ag remains unchanged, it is not oxidized. Therefore, the Ag2SeO3(aq) does not contribute to the oxidation half-reaction.
SeO3^2-(aq) → SeO3^2-(aq)
This is an example of a half-reaction where the species is not oxidized or reduced. We can ignore it for now.
To balance the reduction half-reaction:
H3PO3(aq) → H3PO4(aq)
We can see that phosphorus (P) is oxidized from +3 to +5. To balance the oxidation half-reaction, we add 2 electrons (e-) on the left side:
H3PO3(aq) + 2e- → H3PO4(aq)
Now, we need to balance the overall charge in both half-reactions. Multiply the reduction half-reaction by 2 to balance the electrons:
2H3PO3(aq) + 4e- → 2H3PO4(aq)
Now, let's combine the two half-reactions:
2H3PO3(aq) + 4e- → 2H3PO4(aq)
Ag2SeO3(aq) → Ag(s) + SeO3^2-(aq)
Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to balance the electrons:
6H3PO3(aq) + 12e- → 6H3PO4(aq)
3Ag2SeO3(aq) → 6Ag(s) + 3SeO3^2-(aq)
Finally, add the two half-reactions together and cancel out any common species on both sides of the equation:
6H3PO3(aq) + 3Ag2SeO3(aq) → 6Ag(s) + 6H3PO4(aq) + 3SeO3^2-(aq)
This is the balanced redox reaction using the half-reaction method.