1. Given standard reduction potentials

The sequence is:

$\mathrm{FeO_4^{2-} \xrightarrow[ ]{+2.2\,V} Fe^{3+} \xrightarrow[ ]{+0.70\,V} Fe^{2+} \xrightarrow[ ]{-0.45\,V} Fe^0}$

We need to find:

$E^\circ_{\mathrm{FeO_4^{2-}/Fe^{2+}}}$

2. Write the relevant half-reactions

For ferrate to ferric:

$\mathrm{FeO_4^{2-} + 8H^+ + 3e^- \rightarrow Fe^{3+} + 4H_2O} \qquad E^\circ = 2.20\,V$

For ferric to ferrous:

$\mathrm{Fe^{3+} + e^- \rightarrow Fe^{2+}} \qquad E^\circ = 0.70\,V$

We must combine these to get:

$\mathrm{FeO_4^{2-} + 8H^+ + 4e^- \rightarrow Fe^{2+} + 4H_2O}$

3. Use $\Delta G^\circ = -nFE^\circ$

When combining half-reactions, Gibbs energies add:

$\Delta G^\circ_{\text{net}} = \Delta G^\circ_1 + \Delta G^\circ_2$

So,

$-nFE^\circ_{\text{net}} = -(3F)(2.20) -(1F)(0.70)$

For the net reaction, $n=4$. Therefore,

$-4F E^\circ_{\text{net}} = -6.6F -0.7F$

$4E^\circ_{\text{net}} = 7.3$

$E^\circ_{\text{net}} = \frac{7.3}{4} = 1.825\,V$

4. Match with the asked form

Given:

$E^\circ_{\mathrm{FeO_4^{2-}/Fe^{2+}}} = x \times 10^{-3}\,V$

Since

$1.825\,V = 1825 \times 10^{-3}\,V$

we get

$x = 1825$

5. Comparison with stored answer

Stored correct answer = 1825

This matches our derived value.