A student mixes aqueous solutions of iron(III) ions, iron(II) ions, iodide ions, and iodine under standard conditions. The standard electrode potentials for the two relevant half-cells are:
Fe3+(aq)+e−⇌Fe2+(aq)Eθ=+0.77 V \text{Fe}^{3+}(aq) + e^- \rightleftharpoons \text{Fe}^{2+}(aq) \quad E^\theta = +0.77\text{ V} Fe3+(aq)+e−⇌Fe2+(aq)Eθ=+0.77 V I2(aq)+2e−⇌2I−(aq)Eθ=+0.54 V \text{I}_2(aq) + 2e^- \rightleftharpoons 2\text{I}^-(aq) \quad E^\theta = +0.54\text{ V} I2(aq)+2e−⇌2I−(aq)Eθ=+0.54 VWhich statement correctly predicts and explains the thermodynamically feasible reaction that will occur?
Fe2+(aq)\text{Fe}^{2+}(aq)Fe2+(aq) reduces I2(aq)\text{I}_2(aq)I2(aq) because the standard electrode potential of the Fe3+/Fe2+\text{Fe}^{3+}/\text{Fe}^{2+}Fe3+/Fe2+ half-cell is more positive than that of the I2/I−\text{I}_2/\text{I}^-I2/I− half-cell.
I−(aq)\text{I}^-(aq)I−(aq) reduces Fe3+(aq)\text{Fe}^{3+}(aq)Fe3+(aq) because the standard electrode potential of the Fe3+/Fe2+\text{Fe}^{3+}/\text{Fe}^{2+}Fe3+/Fe2+ half-cell is more positive than that of the I2/I−\text{I}_2/\text{I}^-I2/I− half-cell.
Fe3+(aq)\text{Fe}^{3+}(aq)Fe3+(aq) reduces I−(aq)\text{I}^-(aq)I−(aq) because the standard electrode potential of the I2/I−\text{I}_2/\text{I}^-I2/I− half-cell is less positive than that of the Fe3+/Fe2+\text{Fe}^{3+}/\text{Fe}^{2+}Fe3+/Fe2+ half-cell.
I2(aq)\text{I}_2(aq)I2(aq) oxidises Fe2+(aq)\text{Fe}^{2+}(aq)Fe2+(aq) because the standard electrode potential of the I2/I−\text{I}_2/\text{I}^-I2/I− half-cell is less positive than that of the Fe3+/Fe2+\text{Fe}^{3+}/\text{Fe}^{2+}Fe3+/Fe2+ half-cell.