Here we are considering a concentration cell as both the electrodes are made of same element.
For a concentration cell,
Ecell=0.059nlogC1C2
At anode; H2→2H++2e− [H+]=10−6 M
At cathode; 2H++2e−→H2
So,
Ecell=0.0591log[CH+10−6]
⇒0.118=0.0591log[CH+10−6]
⇒log[CH+10−6]=0.1180.059=2
Therefore, [CH+]=10−4 M
So the vale of x will be 10