Discussion
In the following redox reaction P4(s) + 3OH-(aq)+3H2O(l)→PH3(g) + 3H2PO2-(aq), the oxidation state of phosphorus changes from
Answer: D
Phosphorus changes 0 to -3 and 0 to +1.
This is a disproportionation: elemental phosphorus, P4, is in the 0 state.
- In PH3: \(x + 3(+1) = 0\), so P is -3.
- In H2PO2-: \(2(+1) + x + 2(-2) = -1\), so P is +1.
So part of the phosphorus is reduced to -3 and part oxidised to +1. The option 0 to +3 would need phosphite, H2PO3-, and 0 to +5 would need phosphate; neither is formed here.
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