Question

Oxidation of Cu3P to CuSO4, and H3PO4 involves the loss of
(A) 3 electrons (B) 5 electrons (C) 9 electrons (D) 11 electron ​

Answer 1

Oxidation of $Cu_3P$ to $CuSO_4$ and $H_3PO_4$ involves the change of 9 electrons. Thus option C is correct.

Explanation:

• The reaction involved in the given case is as follows:

$Cu_3P+Cr_2O^{2-}\rightarrow Cu^{2+}+H_3PO_4+Cr^{3+}$

• As per the equation, the following conversion has occurred:

1. $Cu_3P\rightarrow Cu^{2+}$
2. $Cu_3P\rightarrow H_3PO_4$

• In the case of equation 1, $Cu_3$ is getting oxidized to $Cu^{2+}$. Thus the difference of electrons, in this case, is 1.
• On the other hand, in equation 2, P is getting oxidized and changing the oxidation state from -3 to +5. Thus, in this case, the difference electrons, in this case, is 8.
• Thus when the difference electrons are added up, it gives 9 electrons. Hence it is inferred that oxidation of $Cu_3P$ to $CuSO_4$ and $H_3PO_4$ involves the change of 9 electrons.

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