The standard cell potential (e°) of a voltaic cell constructed using the cell reaction below is 0.76 v: zn (s) + 2h+ (aq) → zn2+ (aq) + h2 (g) with ph2 = 1.0 atm and [zn2+] = 1.0 m, the cell potential is 0.56 v. the concentration of h+ in the cathode compartment is m.

Cell reaction overall is Zn(s)+2H+(aq)→Zn²+(aq)+H2(g).
The half-reaction of oxidation is Zn(s)→Zn²+(aq)+2E- and E°zn2+/zn=0.76v
Half reaction reduction  2H+(aq)+2∈-→H2(g) and E°H+/H2+=0.00v
Cell potential is E°cell=E°cathode-E°anode
=E°H+/H2e-E°zn2+/zn
=0.00v-(-0.76v)
=0.76v
Nernst equation
Ecell = -0.059W/N log [zn²+]PH2/[Zn][H+]²
Ecell = 0.66v
[zn²+]=1.0M
1=[Zn]
PH2=1atm
[H+]=?
n = number of moles of electrons transfered in the cell=2mol
Ecell=E°cell -0.059W/n log [Zn²+]/[Zn][H+]²
0.66v = 0.76 - 0.059W/2 log 1.0×1.0/1.0×[H+]²
0.059W/2 log 1/[H+]² = 0.76v-0.66v = 0.10v
log 1/[H+]² =0.10v×2/0.059W =3.4
-2log [H+] = 3.4
log [H+] = 1.7
[H+] =10-n
=0.020
=2.0×10-²M
The concenrtation in cathodic compartments is 2.0×10-²M

Hey how did you manage to type such a big lab report

Answerhave membrane-bound organelles;