$$FeO_{4}^{2-}\xrightarrow{+2.0_{V}} Fe^{3+}\xrightarrow{0.8_{V}} Fe^{2+}\xrightarrow{-0.5_{V}}Fe^{0}$$ In the above diagram, the standard electrode potentials are given in volts (over the arrow). The value of $$E_{FeO_{4}^{2-}/Fe^{2+}}^{0}$$ is

We are given the following standard electrode potentials:

$$FeO_4^{2-} \xrightarrow{+2.0\text{ V}} Fe^{3+} \xrightarrow{+0.8\text{ V}} Fe^{2+} \xrightarrow{-0.5\text{ V}} Fe^0$$

We need to find $$E^0_{FeO_4^{2-}/Fe^{2+}}$$.

We use the relationship between standard Gibbs free energy and electrode potential:

$$ \Delta G^0 = -nFE^0 $$

For $$FeO_4^{2-} \to Fe^{3+}$$: Fe goes from +6 to +3, so $$n_1 = 3$$ and $$ \Delta G_1^0 = -3F(2.0) = -6.0F $$.

For $$Fe^{3+} \to Fe^{2+}$$: Fe goes from +3 to +2, so $$n_2 = 1$$ and $$ \Delta G_2^0 = -1F(0.8) = -0.8F $$.

Adding these gives for $$FeO_4^{2-} \to Fe^{2+}$$ (where Fe changes from +6 to +2, $$n = 4$$): $$ \Delta G^0 = \Delta G_1^0 + \Delta G_2^0 = -6.0F - 0.8F = -6.8F $$.

Thus,

$$ E^0_{FeO_4^{2-}/Fe^{2+}} = \frac{-\Delta G^0}{nF} = \frac{6.8F}{4F} = 1.7 \text{ V} $$. The correct answer is Option 2: 1.7 V.