# Electrode potential for saturated silver carbonate solution

An electrochemical cell is set up to measure the electrode potential $$E(\ce{Ag^+}/\ce{Ag})$$ using the saturated $$\ce{Ag2CO3(aq)}$$ solution $$(K_\mathrm{sp}(\ce{Ag2CO3(aq)},\pu{25 °C}) = \pu{6.3E-12})$$ with a standard hydrogen electrode.

Equilibrium concentration of $$\ce{Ag+}$$ in the saturated solution is $$\pu{2.33E-4 M}.$$

Electrode potential can be calculated using Nernst equation:

$$E(\ce{Ag^+}/\ce{Ag}) = \pu{0.8 V} + \frac{\pu{0.059 V}}{1}\log\frac{[\ce{Ag+}]}{[\ce{Ag(s)}]}.$$

But we can't use the equilibrium concentration of $$\ce{Ag+}$$ we got above for the concentration of oxidised species to calculate E. Instead, we use the value half of that, or just the concentration of $$\ce{Ag2CO3}$$. Why is that?

Reference: Original question, page 14, question 7(a)(iv): https://papers.gceguide.com/A%20Levels/Chemistry%20(9701)/2020/9701_s20_qp_41.pdf

Mark scheme: page 13, question 7(a)(iv) https://bestexamhelp.com/exam/cambridge-international-a-level/chemistry-9701/2020/9701-s20-ms-41.php

Worked solution here, at 2:08:20, https://youtube.com/watch?v=rDYnDi4k5pU But I don't understand the reason mentioned: Because concentration of silver ions is sparingly soluble, so we are using the concentration of $$\ce{Ag2CO3(aq)}$$.

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• But I don't understand the reason mentioned. Quote the reason you do not understand. Apr 29, 2022 at 8:11

## 1 Answer

Where is the problem?

Let's call $$s$$ the solubility of $$\ce{Ag2CO3}$$. As a consequence $$[\ce{Ag+}] = 2s,$$ and $$[\ce{CO3^2-}] = s.$$ So the solubility product is

$$K_\mathrm{s} = [\ce{Ag+}]^2\,[\ce{CO3^2-}] = 4 s^3 = \pu{6.3E-12},$$

which gives

$$s^3 \approx \pu{1.6E-12 M^3} \quad\implies\quad s \approx \pu{1.2E-4 M}$$

$$[\ce{Ag+}] = 2s\approx\pu{2.4E-4 M}$$

• The problem is with the calculation of E, they are using half the value of your answer, which I am not sure why. Apr 29, 2022 at 8:01
• @radastro If A asks B why C wrote anything wrong then B cannot provide the answer. Only C can answer that, but only in cases it was the intention and not an error. Even the most reputable sources can make accidental errors. Apr 29, 2022 at 8:06
• @Maurice I'm afraid you are using MathJax, namely \pu{…} and \ce{…} macros incorrectly. Please look up their intended use — we've had this discussion several times before and you seem to get a good grasp on it, but nothing seems to change which is… disturbing. Also, I personally would refrain from using a non-standard symbol for a variable (solubility $s$) that can also be mixed up with another frequently used quantity in this context (mole fraction $x$). Apr 29, 2022 at 8:12
• @Andselisk You are right. I have some trouble with using Mathjax. My main problem is how to introduce italics characters in a text written without italics. By the way, how do you call a character which is "not italics" ? And also how to print the symbol percent (%) using MathJax ? Apr 29, 2022 at 11:30
• @Maurice The text is upright without any markup. MathJax switches on math mode to enter variables that should be italicized. Math functions, labels, names should be upright and every type should be handled using appropriate macro. Seriously, read up the documentation and check out examples of decent code posted elsewhere, say, on TeX.SE. There is a quick official guide on mhchem for MathJax, and for the most pat MathJax follows the same principles that LaTeX markup that's been around for decades. Apr 29, 2022 at 12:01