Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V

An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:

Ni(s)VO2+(aq,0.083M)+2H+(aq,1.1M)+e−→→Ni2+(aq,2.5M)+2e−VO2+(aq,2.5M)+H2O(l)

Calculate the cell potential under these nonstandard concentrations.

Express the cell potential to two decimal places and include the appropriate units.

Answers

Answer 1
Answer:

Answer:

Cell potential under non standard concentration is 4.09 v

Explanation:

To solve this problem we need to use Nernst Equation because concentrations of the components of the chemical reactions are differents to 1 M (normal conditions:  1 M , 1 atm).

Nernst equation at 25ºC is:

E = E^(0) - [((0.0592)/(n) · log Q)]

where

E: Cell potential (non standard conditions)

E^(0) = Cell potential (standard conditions)

n: Number of electrons transfered in the redox reaction

Q: Reaction coefficient (we are going to get it from the balanced chemical reaction)

For example, consider the following general chemical reaction:

aA + bB --> cC + dD

where

a, b, c, d: coefficient of balanced chemical reaction

A,B,C,D: chemical compounds in the reaction.

Using the previous general reaction, expression of Q is:

Q = (C^(c) * D^(d) )/(A^(a)*B^(b)  )

Previous information is basic to solve this problem. Let´s see the Nernst equation, we need to know: E^(0), n and Q

Let´s  calcule potential in nomal conditions  (E^(0)):

1. We need to know half-reactions (oxidation and reduction), we take them from the chemical reaction given in this exercise:

Half-reactions:                               Eo (v):

(VO_(2))^(2+) + e- --> (VO_(2))^(+)                -0.23

Ni --> Ni^(2+)  + 2 e-                   +0.99    

Balancing each half-reaction, first we are going to balance mass and then we will balance charge in each half-reaction and then charge between half-reactions:

Half-reactions:                                                                Eo (v):

2 * [ (VO_(2))^(2+) + e- --> (VO_(2))^(+)]                                          -0.23

1 * [Ni --> Ni^(2+)  + 2 e-     ]                                         +0.99  

-------------------------------------------------------------------         -------------

2 (VO_(2))^(2+) + 2e- + Ni  --> 2(VO_(2))^(+) + Ni^(2+) + 2e-              0.76 v  

               

Then global balanced chemical reaction is:

2 (VO_(2))^(2+) + Ni  --> 2(VO_(2))^(+) + Ni^(2+)      

 and the potential in nomal conditions is:

E^(0) = 0.76 v

Also from the balanced reaction, we got number of electons transfered:

n = 2

2. Calculate Q:

Now using previous information, we can establish Q expression and we can calculate its value:

Q = ([(VO_(2)+]^(2)* [Ni^(2+)   )/([VO_(2+)]^(2) * Ni )]

From the exercise we know:

[VO_(2) ^(2+)] = 2.5 M

[VO_(2)+] = 0.083 M

[Ni^(2+)] = 2.5 M

[Ni] = 1 M, it is a pure solid, so its activity in Q is unit (1). It is also applied for pure liquids.

Q = ((2.5)^(2)* 2.5 )/((0.083)^(2) * 1) = 2,268.11

3. Use Nernst equation:

Finally, we replace all these results in the Nernst equation:

E = E^(0) - ((0.0592)/(n) - log Q)\n  \nE = 0.76 - ((0.0592)/(2)-log (2,268.11) \nE = 0.76 - (0.0296 - 3.36)\nE = 4.09 v

Cell potential under non standard concentration is 4.09 v

Answer 2
Answer:

Final answer:

To calculate the cell potential under nonstandard conditions, we need to apply the Nernst Equation. This involves finding the reaction quotient (Q) from the given concentrations and then subtracting a value derived from Q and the number of electrons transferred, from the cell potential under standard conditions.

Explanation:

For calculating the cell potential under nonstandard conditions for an electrochemical cell, we need to use the Nernst equation. In this case, the Nernst Equation is Ecell = E∘cell - (0.0592/n) * logQ, where Q, the reaction quotient, is the ratio of the concentrations of the products to the reactants raised to their stoichiometric coefficients.

Given the half-cell reduction potentials, we can calculate the cell potential under standard conditions (E°cell) by subtracting the potential of the anode from the potential of the cathode (E°cell = Ecathode - Eanode = 0.99V - (-0.23V), resulting in E°cell = 1.22V.

Next, Q = [Ni2+]/([VO2+]×[H+]²), substituting the given concentrations, Q = (2.5)/(0.083×1.1²).

After calculating Q, we substitute all known values into the Nernst Equation and solve for Ecell. Hence, the cell potential under these nonstandard conditions can be calculated.

Learn more about Nernst Equation here:

brainly.com/question/31593791

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The correct electron configuration for magnesium is: 1s 22s 22p 63s 3 True False

Answers

Answer:

False

Explanation:

Magnesium is the element of second group and third period. The electronic configuration of magnesium is - 2, 8, 2 or 1s^22s^22p^63s^2

There are 2 valence electrons of magnesium.

Only the valence electrons are shown by dots in the Lewis structure.  

As, stated above, there are only two valence electrons of magnesium, so in the Lewis structure, two dots are made around the magnesium symbol.

Given that the electronic configuration is:- 1s^22s^22p^63s^3.

Orbital s cannot accommodate 3 electrons and also in magnesium it has 3s^2. Hence, the statement is false.

Which one of the following combinations cannot function as a buffer solution?A) HCN and KCN
B) NH3 and (NH4)2SO4
C) HNO3 and NaNO3
D) HF and NAF
E) HNO2 and NaNO2

Answers

Answer:

it is B

; )

Which one of the following combinations cannot function as a buffer solution?

A) HCN and KCN

B) NH3 and (NH4)2SO4

C) HNO3 and NaNO3

D) HF and NAF

E) HNO2 and NaNO2

The balanced equation below shows the products that are formed when pentane (C5H12) is combusted.C5H12 + 802 → 10CO2 + 6H20
What is the mole ratio of oxygen to pentane?

Answers

Answer:

8 : 1

Explanation:

The balanced equation for the reaction is given below:

C5H12 + 8O2 → 5CO2 + 6H2O

From the balanced equation above,

1 mole of C5H12 reacted with 8 moles of O2.

Thus the mole ratio of O2 to C5H12 is:

8 : 1

Answer:

8:1 !!!

Explanation:

I Just take the test

Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial chemist studying this reaction fills a 500. mL flask with 3.3 atm of sulfur dioxide gas and 0.79 atm of oxygen gas at 31.0 °C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of sulfur trioxide gas to be 0.47 atm Calculate the pressure equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to 2 significant digits.

Answers

Answer:

0.051

Explanation:

Let's consider the following reaction.

2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)

We can compute the pressures using an ICE chart.

         2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)

I             3.3         0.79           0

C           -2x            -x           +2x

E         3.3-2x      0.79-x         2x

The partial pressure of sulfur trioxide gas is 0.47 atm. Then,

2x = 0.47

x = 0.24

The pressures at equilibrium are:

pSO₂ = 3.3-2x = 3.3-2(0.24) = 2.82 atm

pO₂ = 0.79-x = 0.79-0.24 = 0.55 atm

pSO₃ = 0.47 atm

The pressure equilibrium constant (Kp) is:

Kp = pSO₃² / pSO₂² × pO₂

Kp = 0.47² / 2.82² × 0.55

Kp = 0.051

Which one of the following compounds will NOT be soluble in water? Which one of the following compounds will NOT be soluble in water? LiOH BaSO4 NaNO3 MgCl2 K2S

Answers

Answer:

BaSO_(4) will be not soluble in water

Explanation:

LiOH is a strong base. Hence it gets completely dissociated in aqueous solution.

NaNO_(3) is a strong electrolyte. Hence it gets completely dissociated in aqueous solution.

MgCl_(2) is a strong electrolyte. Hence it gets completely dissociated in aqueous solution.

K_(2)S is a strong electrolyte. Hence it gets completely dissociated in aqueous solution.

BaSO_(4) is a sparingly soluble salt. Hence it is not dissociated and hence dissolved in water. This is due to the fact that both Ba^(2+) and SO_(4)^(2-) ions are similar in size. Hence crystal structure of BaSO_(4) is quite stable. Hence BaSO_(4) is reluctant to undergo any dissociation in aqueous solution.

Which of the following should be measured with a meter stick, not a tape measure?A.
the length of ribbon needed to tie around a vase
B.
the size of a student's waist
C.
the distance from the ground to the top of a ramp
D.
the circumference of an orange

Answers

Answer:

C

the distance from the ground to the top of a ramp