Given : $$E°_{Fe^{3+}/Fe} = -0.036$$ V, $$E°_{Fe^{2+}/Fe} = -0.439$$ V. The value of standard electrode potential for the change, $$Fe^{3+}_{(aq)} + e^- \to Fe^{2+}(aq)$$ will be :

The two given standard reduction couples share the common species $$Fe(s)$$. Hence we can convert each potential into its standard Gibbs energy change, combine the reactions, and finally reconvert the net $$\Delta G^{\circ}$$ back to an electrode potential.

Formula used: $$\Delta G^{\circ} = -\,nF E^{\circ}$$, where
  • $$n$$ = number of electrons in the half-reaction
  • $$F$$ = Faraday constant (same for every step, so it cancels in the ratio)

Step 1: Convert each couple to $$\Delta G^{\circ}$$
For $$Fe^{3+} + 3e^- \rightarrow Fe(s)$$, $$n = 3$$:
$$\Delta G^{\circ}_1 = -\,3F(-0.036\text{ V}) = +\,0.108F$$

For $$Fe^{2+} + 2e^- \rightarrow Fe(s)$$, $$n = 2$$:
$$\Delta G^{\circ}_2 = -\,2F(-0.439\text{ V}) = +\,0.878F$$

Step 2: Obtain $$\Delta G^{\circ}$$ for $$Fe^{3+} + e^- \rightarrow Fe^{2+}$$
Write reaction 1 forward and reaction 2 backward:

Reaction 1: $$Fe^{3+} + 3e^- \rightarrow Fe$$

Reverse of reaction 2: $$Fe \rightarrow Fe^{2+} + 2e^-$$  (its $$\Delta G^{\circ} = -\Delta G^{\circ}_2$$)

Add them; the solid iron cancels, leaving

$$Fe^{3+} + e^- \rightarrow Fe^{2+}$$

Therefore
$$\Delta G^{\circ}_3 = \Delta G^{\circ}_1 - \Delta G^{\circ}_2 = 0.108F - 0.878F = -\,0.770F$$

Step 3: Convert $$\Delta G^{\circ}_3$$ back to an electrode potential

For this half-reaction, $$n = 1$$, so

$$E^{\circ}_{Fe^{3+}/Fe^{2+}} = -\dfrac{\Delta G^{\circ}_3}{nF} = -\dfrac{-0.770F}{1F} = 0.770\text{ V}$$

Thus the standard electrode potential for $$Fe^{3+}_{(aq)} + e^- \rightarrow Fe^{2+}_{(aq)}$$ is

$$\boxed{0.770\ \text{V}}$$

Option C which is: $$0.770\text{ V}$$