how much will potential of a hydrogen electrode change when its solution initially at ph=0 is neutralized to ph=7

Solution is as follows:

H⁺ (aq) + e^- $\to$ $\frac{1}{2}$ H₂(g)
Applying the Nernat Equation:

 
 E = E⁰ - $\frac{0.0591}{n}\log Q$$\frac{0.0591}{n}\log Q$
  = 0.0 - $$\begin{gathered} \frac{0.0591}{n}\log \frac{p_{H_2}}{\left[H^{+}\right]} \\ =- \frac{0.0591}{1}\log \frac{1}{{10}^{-7}} \\ = -0.413 V \end{gathered}$$

Thus, potential decreases to 0.413 V when pH goes from 0 to 7.