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Lapatulllka [165]
3 years ago
6

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.
Chemistry
1 answer:
Temka [501]3 years ago
6 0
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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The pressure of a sample of helium in a 200. ml container is 2.0 atm. If the helium is compressed to a pressure of 40. atm witho
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Answer:

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Explanation:

Hello there!

In this case, according to the given information, it will be possible for us to solve this problem by using the Boyle's law as an inversely proportional relationship between pressure and volume:

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In such a way, we solve for the final volume, V2, and plug in the initial volume and pressure and final pressure to obtain:

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