The pressure of ${\mathrm{H}}_2$ required to make the potential of ${\mathrm{H}}_2$- electrode zero in pure water at $298 \mathrm{K}$ is

In the electrochemical cell :$\mathrm{Zn}\left|{\mathrm{ZnSO}}_4(0.01\mathrm{M})\Vert {\mathrm{CuSO}}_4(1.0\mathrm{M})\right|\mathrm{Cu},$ the $\mathrm{emf}$ of this Daniel cell is ${\mathrm{E}}_1$. When the concentration of ${\mathrm{ZnSO}}_4$ is changed to $1.0 \mathrm{M}$ and that of ${\mathrm{CuSO}}_4$ changed to $0.01\mathrm{M}$, the $\mathrm{emf}$ changes to ${\mathrm{E}}_2$. From the followings, which one is the relationship between ${\mathrm{E}}_1$ and ${\mathrm{E}}_2?\left(\text{ Given, }\frac{\mathrm{RT}}{\mathrm{F}}=0.059\right)$.

Consider the following ${\mathrm{E}}_{\mathrm{o}}$ values:

${\mathrm{E}}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}^0=+0.77\mathrm{V};{\mathrm{E}}_{{\mathrm{Sn}}^{2+}/\mathrm{Sn}}^0=-0.14\mathrm{V}$

Under standard conditions, the cell potential for the reaction given below is:

${\mathrm{Sn}}_{\left(9\right)}+2{\mathrm{Fe}}_{\left(\mathrm{aq}\right)}^{3+}\to 2{\mathrm{Fe}}_{\left(\mathrm{aq}\right)}^{2+}+{\mathrm{Sn}}_{\left(\mathrm{aq}\right)}^{2+}$