In the following electrochemical cell : Zn|Zn2+||H⊕|(H2)Pt Ecell=⊖cell. This will be when :
Cell reaction : Zn(s)+2H⊕(aq)→Zn2+(aq)+H2(g)
Ecell=E⊖cell−0.05912log(Qcell)
For Ecell=E⊖, Qcell=1
Ecell=E⊖cell,Qcell=1.0
a. Qcell=1×112=1
b. Qcell=1×0.01(0.1)2=1
c. Qcell=1×1(0.1)2=100
Hence, options A and B are correct.