1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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
You might be interested in
What part of a chemical is represented by the letters
bonufazy [111]

Answer:

It's compound? It's chemical compound would be represented my letters or numbers

3 0
2 years ago
Read 2 more answers
Cuántos moles hay en 17.5 gramod de ZnCl2​
Aliun [14]

Explanation:

Zn=65

Cl2= 35+35=70

65+70=135g

1 mole ZnCl2 = 135g

x mole = 17.5g

17.5g × 1 mole/ 135g= 0.129 moles en 17.5g de ZnCl2

4 0
3 years ago
What problem can fossil fuel create for life on earth?
Darya [45]
Pollution and rising temperatures
4 0
3 years ago
An atom of argon has a radius rar = 88 pm and an average speed in the gas phase at 25°C of 172 m/s.
Rudik [331]

Answer:

1.2* 10³ rNe.

Explanation:

Given speed of neon=350 m/s

Un-certainity in speed= (0.01/100) *350 =0.035 m/s

As per heisenberg uncertainity principle

Δx*mΔv ≥\frac{h}{4\pi }

4π

h

..................(1)

mass of neon atom =\frac{20*10^{-3} }{6.22*10^{-23} } =3.35*10^{-26} kg

6.22∗10

−23

20∗10

−3

=3.35∗10

−26

kg

substituating the values in eq. (1)

Δx =4.49*10^{-8}10

−8

m

In terms of rNe i.e 38 pm= 38*10^{-12}10

−12

Δx=\frac{4.49*10^{-8} }{38*10^{-12} }

38∗10

−12

4.49∗10

−8

=0.118*10^{4}10

4

* (rNe)

=1.18*10³ rN

= 1.2* 10³ rNe.

Explanation:

This is the answer

7 0
2 years ago
If 20 atoms of aluminum react with 45 molecules of chlorine gas, which reactant is limiting and how many more atoms/molecules wo
saul85 [17]
Al:ch2 an molar ratio.

2:3 and an x. X is 30.  20/2 = 10 so an answer is 

45/3 = 15. Both ratios are used 

45 - 30 = 15 CL2 left.

Your Welcome :)

4 0
3 years ago
Read 2 more answers
Other questions:
  • What coefficient of h+ balances the atoms in this half-reaction h+ + mno2 → mn2+ + h2o?
    13·2 answers
  • Why do elements in the same family generally have a similar properties ?
    7·1 answer
  • A double colvalent bond involves two atoms sharing two electrons. True or false?
    6·1 answer
  • Why can people on one side of a room smell the scent of an air freshener sprayed on the opposite side?
    11·1 answer
  • Read the list of common household substances.
    13·2 answers
  • An atom of aluminum (AI) has an atomic number of 13 and a mass number of
    13·1 answer
  • How many mL of a 0.375 M solution can be made from 35 g of calcium phosphate?
    14·1 answer
  • Help please I don’t know
    9·1 answer
  • Sodium (Na), lithium (Li), and potassium (K) are grouped together in one column on the periodic table. Which property do they sh
    13·1 answer
  • Calculate the molarity of 0.060 moles NaHCO3 in 1500. mL of solution
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!