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?

  • 3
    $\begingroup$ I took a liberty to format your question for better readability and proper notations; if you are interested further in technical details, feel free to visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. $\endgroup$ – andselisk Nov 16 '19 at 21:19

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