Question

which side will the electrons be added to in the half - reaction? \\(\ce{h_{2}c_{2}o_{4} \
ightarrow 2co_{2} + 2h^{+}}\\) \\(\ce{a}\\). add electrons to the reactant side \\(\ce{b}\\). add electrons to the product side

Explanation:

1. First, determine the oxidation state of carbon in \(H_2C_2O_4\) and \(CO_2\). In \(H_2C_2O_4\), let the oxidation state of \(C\) be \(x\). The oxidation states of \(H\) is \( + 1\) and \(O\) is \(-2\). So, \(2\times( + 1)+2x + 4\times(-2)=0\), which gives \(2 + 2x-8 = 0\), and \(2x=6\), so \(x = + 3\). In \(CO_2\), let the oxidation state of \(C\) be \(y\). Then \(y+2\times(-2)=0\), so \(y = + 4\).
2. Since the oxidation state of \(C\) increases from \(+3\) (in \(H_2C_2O_4\)) to \(+4\) (in \(CO_2\)), \(H_2C_2O_4\) is being oxidized. Oxidation is the loss of electrons. In a half - reaction for oxidation, electrons are lost by the reactant, which means electrons should be added to the product side to balance the charge. For the reaction \(H_2C_2O_4

ightarrow2CO_2 + 2H^+\), the left - hand side (reactant side) has a net charge of \(0\) (since \(H_2C_2O_4\) is neutral). The right - hand side (product side) has a net charge of \(+2\) (from \(2H^+\)). To balance the charge, we need to add electrons to the product side (because oxidation involves loss of electrons, and the electrons lost by the reactant will appear as products in the half - reaction).

Answer:

B. Add electrons to the product side