Find the electrode potential of a solution

I'm stuck with the next exercise.


What is the electrode potential (against NHE) in the solution prepared by adding 90 mL of $\ce{ KI }$ 5.0 M to 10 mL of $H2O2$ 0.10 M, with a buffer at pH 2.0?


First I wrote the reduction half-reactions


$$ \ce {I2(s) +2e- ->2I-} $$
$$ \ce {H2O2 +2H+ +2e- ->2H2O} $$


then I wrote the complete reaction where $ \ce {H2O2} $ reduces and $ \ce {I-}$ oxides with the number of initial and final moles


$$ \ce {H2O2 +2H+ +2I- ->2H2O + I2(s)} $$
$ \ce {Initial: 0.001 n H2O2 , 0.045n 2I- } $


$ \ce {Final: x H2O2 , 0.043+x 2I- } $


and as the products are solid or liquid they have standard concentrations


Since it is not easy to know x, I proceeded to use Nernst's equation to find the potential of the electrode with the pair of iodes and knowing that the $E$ for this is $0.5345 V$


$$\ce {E = 0.5345 - \frac{0.05916}{2} log({\frac{0.043n}{0.1L}}})^2 = 0.5305V$$
but when I see the correct answer it appears as $0.419V$ and I don't know where is my mistake :c thanks for read. <3


