Question

fill in the missing number of electrons for each of the following half - reactions: \\(\ce{co^{3+}(aq)} + \square \ce{e^-} \
ightarrow \ce{co(s)}\\) \\(\ce{na(s)} \
ightarrow \ce{na^+(aq)} + \square \ce{e^-}\\)

Explanation:

Step1: Analyze the first half - reaction

For the half - reaction \(Co^{3 + }(aq)+?e^-\to Co(s)\), we need to balance the charge. The oxidation state of \(Co\) in \(Co^{3+}\) is \( + 3\) and in \(Co(s)\) is \(0\). To go from an oxidation state of \(+3\) to \(0\), the \(Co^{3+}\) ion must gain electrons. The change in oxidation state is \(0-( + 3)=- 3\). Since each electron has a charge of \(-1\), the number of electrons gained is \(3\). So the number of electrons in the first half - reaction is \(3\).

Step2: Analyze the second half - reaction

For the half - reaction \(Na(s)\to Na^{+}(aq)+?e^-\), the oxidation state of \(Na\) in \(Na(s)\) is \(0\) and in \(Na^{+}\) is \( + 1\). The change in oxidation state is \(+1 - 0=+1\). Since \(Na\) is losing electrons (oxidation), and each electron has a charge of \(-1\), the number of electrons lost is \(1\). So the number of electrons in the second half - reaction is \(1\).

Answer:

For \(Co^{3+}(aq)+ \square e^-\to Co(s)\), the missing number is \(3\); for \(Na(s)\to Na^{+}(aq)+\square e^-\), the missing number is \(1\).