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Question
what is the oxidation number of phosphorous in na3po4? +1 -5 -1 +5 question 26 which of the following is an oxidation half - reaction? o2 + 4h+ + 4e− ----> 2h2o sn2+ ----> sn4+ + 2e− cl2 + 2e− ----> 2cl− fe3+ + e− ----> fe2+
Step1: Recall oxidation - number rules for Na₃PO₄
Sodium (Na) has an oxidation number of +1 and oxygen (O) has an oxidation number of -2 in most compounds. Let the oxidation number of phosphorus (P) be \(x\).
Step2: Set up the equation based on the net - charge of the compound
The net - charge of Na₃PO₄ is 0. So, \(3\times(+ 1)+x + 4\times(-2)=0\).
Step3: Solve the equation for \(x\)
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Step4: Recall the definition of oxidation half - reaction
Oxidation is the loss of electrons. In an oxidation half - reaction, the oxidation number of the species increases.
For \(O_2 + 4H^++4e^-\longrightarrow2H_2O\), oxygen's oxidation number decreases from 0 in \(O_2\) to -2 in \(H_2O\), so it is a reduction half - reaction.
For \(Sn^{2+}\longrightarrow Sn^{4+}+2e^-\), the oxidation number of tin increases from +2 to +4, so it is an oxidation half - reaction.
For \(Cl_2 + 2e^-\longrightarrow2Cl^-\), the oxidation number of chlorine decreases from 0 in \(Cl_2\) to -1 in \(Cl^-\), so it is a reduction half - reaction.
For \(Fe^{3+}+e^-\longrightarrow Fe^{2+}\), the oxidation number of iron decreases from +3 to +2, so it is a reduction half - reaction.
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For the first question: +5
For the second question: \(Sn^{2+}\longrightarrow Sn^{4+}+2e^-\)