# The physical intuition behind some simplifications in electrochemistry

Find the standard reduction potential of

$$\ce{AgIO3(s) + e- <=> Ag(s) + IO3-}.$$

$$K_\mathrm{sp} = \pu{3.1E-8},$$ and the standard reduction potential of $$\ce{Ag +e- <=> Ag(s)}$$ is $$E = \pu{0.799 V}.$$

So, $$E^\circ$$ will be standard reduction potential and $$E$$ will be the nonstandard.

\begin{align} E(\ce{AgIO3/Ag}) &= E^\circ(\ce{AgIO3/Ag}) - 0.0257\ln\frac{1}{[\ce{Ag}]}\\ &= E^\circ(\ce{AgIO3/Ag}) - 0.0257\ln\frac{[\ce{IO3}]}{\pu{3.1E-8}} \end{align}

$$\pu{0.799 V} = E^\circ(\ce{AgIO3/Ag}) - 0.0257\ln\frac{1}{\pu{3.1E-8}}$$

Now, my concern is that I understand the math and like most of the principles behind all this, but how come when we set $$[\ce{IO3}] = 1,$$ we can also say $$E (\ce{AgIO3/Ag})$$ is equal to $$E(\ce{Ag})?$$

tl;dr: how come setting the concentration of the anion to $$\pu{1 M}$$ allows us to assume that variable change for the electrode?

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