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 $\ce{Ag2CO3(aq)}$.




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


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


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


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