Understanding Standard Hydrogen Electrode
This is a concentration cell question where 2 electrodes of same stuff are dipped in their own ions and if there is non equal concentrations of anything a "potential" develops called electrode potential (to be understood in qualitative way)(kind of more concentration of reactant forces product formation)
as you can understand this would be a redox reaction where reduction would occur in one and oxidation in another, Now, by standard it means that all the components of the reaction (participants, i.e. reactants and products) are in standard state i.e. $1M, 1$ $bar$ etc, so, we know that concentration of $\ce{H+}$ in standard one is $1M$, we need to create biggest gap between them, as the concentration is really high and we cant compete with it, just reduce our own $\ce{H+}$ to increase their difference and force ours to undergo Oxidation to gas and standard undergoes reduction

Just for extra information,
$E=E_0-{RT\over nF}\ln Q$ and can be derived by $G=G_0+RT\ln Q,\;G=-nFE$ {$n$ is number of electrons being exchanged ex- in $\ce{2H+-&gt;H2},\;n=2$; $F$ is the charge on 1 mol electrons} and Standard Hydrogen is the standard bearer so, we assign its standard value as $0$, $E_0=0$.

EDIT : I can't comment due to reputation 
, basically, assume complete dissociation and and concentration of $\ce{H+}$ in $\ce{HCl}$ would be $0.1M$ , $\ce{H2SO4}$'s second ionization is not complete like others but can be assumed for a good approximation, so concentration becomes $0.2 M$ , and basic thing to remember is 

IF ONE SIDE  UNDERGOES OXIDATION THEN OTHER MUST UNDERGO REDUCTION 

and if we strengthen either reaction, potential increases, concentration is only in a solvent, how can a solid would have active $\ce{H+}$ ions (everything is activity not concentration or partial pressure flat out, they are approximated, solids are assumed to have activity as unity) and log is defined as the power to $10$ that would give us that number