Question

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for the following electron - transfer reaction:
$\ce{ni(s) + f_{2}(g) -> ni^{2 + }(aq) + 2f^{-}(aq)}$
the oxidation half - reaction is:
the reduction half - reaction is:

Explanation:

Step1: Identify oxidation and reduction

Oxidation is loss of electrons. In \(Ni(s)\to Ni^{2 +}(aq)\), nickel loses 2 electrons. Reduction is gain of electrons. In \(F_{2}(g)\to 2F^{-}(aq)\), each fluorine atom gains 1 electron (total 2 electrons for \(F_{2}\)).

Step2: Write oxidation half - reaction

For \(Ni(s)\to Ni^{2 +}(aq)\), the oxidation half - reaction is \(Ni(s)\to Ni^{2+}(aq)+2e^{-}\)

Step3: Write reduction half - reaction

For \(F_{2}(g)\to 2F^{-}(aq)\), the reduction half - reaction is \(F_{2}(g)+2e^{-}\to 2F^{-}(aq)\)

Answer:

The oxidation half - reaction is \(Ni(s)\to Ni^{2+}(aq)+2e^{-}\)
The reduction half - reaction is \(F_{2}(g)+2e^{-}\to 2F^{-}(aq)\)