For the redox reaction xP_4 + yHNO_3 to H_3PO | vedclass

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Question

(C) The unbalanced equation is $P_4 + HNO_3 	o H_3PO_4 + NO_2 + H_2O$. First,balance the oxidation and reduction half-reactions: Oxidation: $P_4^0

Vedclass · Accepted Answer

$x = 1, y = 20$

Vedclass · Answer

(C) The unbalanced equation is $P_4 + HNO_3 	o H_3PO_4 + NO_2 + H_2O$. First,balance the oxidation and reduction half-reactions: Oxidation: $P_4^0 	o 4P^{+5} + 20e^-$. Reduction: $N^{+5} + 1e^- 	o N^{+4}$. To balance electrons,multiply the reduction half-reaction by $20$: $20N^{+5} + 20e^- 	o 20N^{+4}$. Adding these gives: $P_4 + 20HNO_3 	o 4H_3PO_4 + 20NO_2 + 4H_2O$. Comparing this with $xP_4 + yHNO_3$,we get $x = 1$ and $y = 20$.

For the redox reaction $xP_4 + yHNO_3 \to H_3PO_4 + NO_2 + H_2O$,find the values of $x$ and $y$.

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