= Solution
The fixed nonproton <thermodynamic activity> ratio is $(0.1)^4(0.3)/(3\times10^{-6})=10$. Hence
$$
Q=10\,10^{11\,{\rm pH}},\qquad
E=1.945-\frac{RT\ln10}{4F}(1+11\,{\rm pH}).
$$
Using the supplied constants at $298\,\mathrm K$,
$$
\boxed{E\simeq1.9302-0.1626\,{\rm pH}\quad\mathrm V}.
$$
At <pH> 7, \b[$E\simeq0.792\,\mathrm V$]. This is the formal fixed-thermodynamic activity calculation requested; actual cerium speciation and thermodynamic activities would need to remain compatible with the chosen <pH> to realize it experimentally.
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