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Using ΞGΒ°=βnFβEΒ° (GΒ° additive), for:
- $$\Delta G^\circ_{M^{n+}/M}=-nF\,E^\circ_{M^{n+}/M},$$
- $$\Delta G^\circ_{M^{m+}/M}=-mF\,E^\circ_{M^{m+}/M}.$$
Reverse the second reaction (oxidation) to add +mFβEΒ°; summing gives:
$$-nF\,E^\circ_{M^{n+}/M}+mF\,E^\circ_{M^{m+}/M}=-(n-m)F\,E^\circ_{M^{n+}/M^{m+}}$$Thus:
$$E^\circ_{M^{n+}/M^{m+}}=\frac{n\,E^\circ_{M^{n+}/M}-m\,E^\circ_{M^{m+}/M}}{\,n-m\,}$$
For the half-reaction: $$M^{n+}+(n-m)e^-\rightarrow M^{m+}.$$
