A galvanic cell is constructed using two Hydrogen electrodes. Emf is highest when the PH of the solutions at anode and cathode is _____ and _____ respectively.

pH= - log[H⁺]

$E_{cell}=E_{cell}^o-\frac{0.0591}{2}\log \frac{\left[reduction\right]}{\left[oxidation\right]}$

For option -1 

$$\begin{gathered} pH=4 when anode\left[H^{+}\right] = {10}^{-4} \\ pH\hspace{0.17em}= 1 when Cathode \left[H^{+}\right]={10}^{-1} \\ {E}_{cell}= -\frac{0.0591}{2}\log \frac{{10}^{-4}}{{10}^{-1}}= 0.0885V \end{gathered}$$

Similarly for 

option -2  E_cell = -0.059V 

option -3 E_cell = 0.03 V

option- 4 E_cell = 0.059 V