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2 years ago

If the concentration of fluoride anion and aluminum cation was increased to 5 m, by how much would the measured ecell change?.

1 answer:

6 0

If the concentration of fluoride anion and aluminum cation was increased to 5M, then there would be decrease in Ecell by 0.055 volts

The standard reduction potential of Fluoride anion and aluminum cation are -

2Al -----> 2Al3+ + 6e- $E^o$ cell = +1.66

6e- + 3F2 -----> 6F- $E^o$ cell = +2.87

The complete reaction is -

2Al + 3F2 ------> 2Al3+ + 6F- $E^o$ = +4.53

Using Nernst Equation :-

E = $E^o$ – 0.0592/n*log[Al3+]^2[F-]^6

n = 6 (n = number of transferred electrons)

E = +4.53 - 0.0592/6*log(5)^2(5)^6

E = +4.53 - 0.00987*log(25)(1.56 x 10^4)

E = +4.53 – 0.00987*log(3.9 x 10^5)

E = + 4.53 -0.00987(5.59)

E = + 4.53 - 0.055

E = +4.47

The change in Ecell = 5 - 4.47 = 0.055V

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Pls pls I need help to solve this questions

cupoosta [38]

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6 0

3 years ago

What does contact forces and non-contact forces mean?

Dmitrij [34]

A contact force is a force in which and object comes in contact with a another object. A non-contact force, is a force is an object applied by another body. (A good example of a non-contact force is gravity)

8 0

3 years ago

The vapor pressure of water at 25.0°c is 23.8 torr. determine the mass of glucose (molar mass = 180 g/mol) needed to add to 500.

svp [43]

Q: A
according to this formula, we can get the mole fraction of water (n):
P(solu) = n Pv(water)
when we have Pv(solu) = 22.8 and Pv(water) = 23.8 so by substitution:
22.8 = n * 23.8
n= 0.958
- we need to get the moles of glucose:
moles of water = 500 g(mass weight) / 18 (molar weight)= 27.7 mol
n = moles of water / ( moles of water + moles of glucose)
0.958 = 27.7 / ( 27.7+ moles of glucose)
0.958 moles of glucose + 26.5 = 27.7
0.968 moles of glucose = 1.2
moles of glucose = 1.253 mol
∴ the mass of glucose = no.of glucose moles x molar mass
= 1.253 x 180 = 225.5 g
Q: B
here we also need to get n (mole fraction of water )by using this formula:
Pv(solu) = n Pv(water)
when we have Pv(solu)=132 & Pv(water)=150 so, by substition:
132= n * 150
n = 0.88
so, mole fraction of solution = 1 - 0.88 = 0.12
and we can get after that the moles of water = (mass weight / molar mass)
- no.moles of water = 85 g / 18 g/mol = 4.7 moles
- total moles in solution = moles of water / moles fraction of water
= 4.7 / 0.88 = 5.34 moles
∴ moles of the solution = total moles in solu - moles of water
= 5.34 - 4.7 = 0.64 moles solute
∴ the molar mass of the solute = mass weight of solute / no.of moles of solute
= 53.8 / 0.64 = 84 g/mole

Q: C

moles of urea (NH2)2 CO = mass weight / molar mass
= 4.49 g / 60 g /mol
= 0.07 mol
moles of methanol = mass weight / molar mass
= 39.9 g / 32 g/mol = 1.25 mol
moles fraction of methanol = moles of methanol / (moles of methanol + moles of urea )
moles fraction of methanol = 1.25 / ( 1.25+0.07) = 0.95
by substitution in Pv formula we will be able to get the vapour pressure of the solu :
Pv(solu) = n P°v
Pv(solu) = 0.95 * 89 mm Hg
∴Pv(solu) = 84.55 mmHg

7 0

3 years ago

If i have 8 L of 0.25 M solution, how many miles of MgCl2 are present

qaws [65]

Answer:

2 moles

Explanation:

The following were obtained from the question:

Molarity = 0.25 M

Volume = 8L

Mole =?

Molarity is simply defined as the mole of solute per unit litre of solution. It is represented mathematically as:

Molarity = mole of solute/Volume of solution.

With the above equation, we can easily find the number of mole of MgCl2 present in 8 L of 0.25 M MgCl2 solution as follow:

Molarity = mole of solute/Volume of solution.

0.25 = mole of MgCl2 /8

Cross multiply to express in linear form

Mole of MgCl2 = 0.25 x 8

Mole of MgCl2 = 2 moles

Therefore, 2 moles of MgCl2 are present in 8 L of 0.25 M MgCl2 solution

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3 years ago

Cr2o2−7(aq)+i−(aq)→cr3+(aq)+io−3(aq) (acidic solution) express your answer as a chemical equation. identify all of the phases in

Rashid [163]

<span>Answer: Nothing is balanced in your final equation: not H, not O, not Cr, not I and your charges aren't either. Start with your 2 half reactions: I- --> IO3- Cr2O72- --> 2 Cr3+ Balance O by adding H2O: I- + 3 H2O --> IO3- Cr2O72- --> 2 Cr3+ + 7H2O Balance H by adding H+: I- + 3 H2O --> IO3- + 6 H+ Cr2O72- + 14 H+ --> 2 Cr3+ + 7H2O Balance charge by adding e-: I- + 3 H2O --> IO3- + 6 H+ + 6 e- Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7H2O Since the numbers of electrons in your two half reactions are the same, just add them and simplify to give: Cr2O72- + I- + 8 H+ --> IO3- + 2 Cr3+ + 4 H2O</span>

4 0

3 years ago