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.
1 answer:
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
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