Question

which half - cell will the sodium ions travel into for the spontaneous cell shown? standard reduction potentials half - reaction e° (v) mg²⁺ + 2e⁻ → mg - 2.37 fe²⁺ + 2e⁻ → fe - 0.44 options: fe half - cell, mg half - cell

Explanation:

Step1: Determine anode and cathode

In a spontaneous electrochemical cell, the half - reaction with the more negative standard reduction potential will be the oxidation (anode) reaction, and the one with the less negative (more positive) potential will be the reduction (cathode) reaction. The standard reduction potential of \(Mg^{2 +}+2e^-
ightarrow Mg\) is \(E^{\circ}=- 2.37\ V\) and for \(Fe^{2+}+2e^-
ightarrow Fe\) is \(E^{\circ}=-0.44\ V\). Since \(-2.37<- 0.44\), \(Mg\) will be oxidized (anode: \(Mg
ightarrow Mg^{2+}+2e^-\)) and \(Fe^{2+}\) will be reduced (cathode: \(Fe^{2+}+2e^-
ightarrow Fe\)).

Step2: Determine ion flow

In an electrochemical cell, cations (like \(Na^+\) from the salt bridge, and also the cations produced at the anode) move towards the cathode. The cathode is the \(Fe/Fe^{2+}\) half - cell (since \(Fe^{2+}\) is being reduced here). So sodium ions (cations) will travel into the \(Fe\) half - cell.

Answer:

Fe half - cell