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cestrela7 [59]
1 year ago
5

Write the cell notation for the voltaic cell that incorporates each of the following redox reactions:(b) Cu²⁺(aq) + SO₂(g) + 2H₂

O}(l) →Cu(s) + SO₄²⁻(aq) + 4H⁺(a q)
Chemistry
1 answer:
yawa3891 [41]1 year ago
8 0

The cell notation for the given voltaic cell is

SO₂(g) | SO₄²⁻(aq) || Cu²⁺(aq) | Cu (s)

<h3>What is a voltaic cell?</h3>

A voltaic cell often called a galvanic cell, is a type of electrochemical cell.

It is a combination of two metal rods called electrodes.

A half cell is formed when both electrodes are submerged in a solution containing a particular mix of ions.

The electrolyte is the liquid in which both electrodes are submerged and acts as a salt bridge connecting the half cells.

The cell undergoes a chemical reaction known as a redox reaction.

One of the electrodes serves as an anode and undergoes oxidation, while the other electrode serves as a cathode and undergoes reduction.

To write the cell notation, write the half-reaction of the cells.

At the anode, oxidation occurs

SO_2(g) \longrightarrow SO_4^{2-}(aq) + 4H^+ (aq) + 2e^-

At cathode

Cu^{2+}(aq) + 2e^- \longrightarrow Cu(s)

In the cell notation, the anode is written on the left side, and the cathode is written on the right side. A vertical line differentiates the phases and the double vertical line represents the salt bridge.

The cell notation for the given voltaic cell is

SO₂(g) | SO₄²⁻(aq) || Cu²⁺(aq) | Cu (s)

Learn more about cell notation:

brainly.com/question/17218591

#SPJ4

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Use the successive ionization energies for this unknown element to identify the family it belongs to.
NNADVOKAT [17]

Answer: Belongs to the group 2A

Explanation:

As you can see, the first two ionization energies are close and low, meaning that this element ionizates easily.

Not only loses easily the first electron, but the second too

To remove the third electron you requiered a huge amount of energy

Now, elements easily ionizable are the ones from group IA, group 2A and transition metals.

The last ones have mixed characteristics in matter of how many electrons you can remove from them, so they are not a family.

Now the question: group I or group II ?

The elements of group I have low  ionization energies  for the first electron but high energies for the second ones.

Being all that said, the unknown element belongs to the Group 2A

4 0
3 years ago
What do cyanobacteria use to produce energy and by what process do they produce energy
andrezito [222]
As they contain, green pigement they can perform "Photosynthesis" and obtains energy from that process

Hope this helps!
7 0
3 years ago
Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re
borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

                        A            +            B           ⇄        C

where;

numbers of moles = 0.386 mol C  (g)

Volume =  7.29 L

Molar concentration of C = \frac{0.386}{7.29}

= 0.053 M

                        A            +            B           ⇄        C

Initial               0                           0                      0.530    

Change          +x                          +x                       - x

Equilibrium      x                           x                      (0.0530 - x)

K = \frac{[C]}{[A][B]}

where

K is given as ; 78.2 atm-1.

So, we have:

78.2=\frac{[0.0530-x]}{[x][x]}

78.2= \frac{(0.0530-x)}{(x^2)}

78.2x^2= 0.0530-x

78.2x^2+x-0.0530=0  

Using quadratic formula;

\frac{-b+/-\sqrt{b^2-4ac} }{2a}

where; a = 78.2 ; b = 1 ; c= - 0.0530

= \frac{-b+\sqrt{b^2-4ac} }{2a}   or \frac{-b-\sqrt{b^2-4ac} }{2a}

= \frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}  or \frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}

= 0.0204  or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

     = 0.0530 - 0.0204

     = 0.0326

Total number of moles at equilibrium = 0.0204 +  0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

V = \frac{nRT}{P}

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K  (fixed constant temperature )

V (volume) = ???

V=\frac{(0.0734*0.0821*273.15)}{(1.00)}

V = 1.64604

V ≅ 1.65 L

3 0
3 years ago
The half-reaction occurring at the cathode in the balanced reaction shown below is __________?2Ag(s) + 1/2 O2(g) + 2H+(aq) → 2Ag
Shtirlitz [24]

Answer:

Correct choice are C and D (they are both, the same).

Explanation:

Cathode is the positive(+) electrode where a reduction occurs.

Reduction is the chemical reaction where the oxidation state is reduced.

2Ag(s) + 1/2 O2(g) + 2H+(aq) → 2Ag+(aq) + H2O (l)

A. 2H2O (l) → O2 (g) + 4H+ (aq) + 4e-

B. 2Ag (s) → 2Ag+ (aq) + 2e-

C. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)

D. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)

C or D, are ok. They are the same equation.

Oxygen from ground state reduce the oxidation state from 0 to -2

3 0
3 years ago
what is the limiting reactant when 1.50g of lithium and 1.50 g of nitrogen combine to form lithium nitride, a component of advan
PSYCHO15rus [73]
<h3>Answer:</h3>

Limiting reactant is Lithium

<h3>Explanation:</h3>

<u>We are given;</u>

  1. Mass of Lithium as 1.50 g
  2. Mass of nitrogen is 1.50 g

We are required to determine the rate limiting reagent.

  • First, we write the balanced equation for the reaction

6Li(s) + N₂(g) → 2Li₃N

From the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.

  • Second, we determine moles of Lithium and nitrogen given.

Moles = Mass ÷ Molar mass

Moles of Lithium

Molar mass of Li = 6.941 g/mol

Moles of Li = 1.50 g ÷ 6.941 g/mol

                   = 0.216 moles

Moles of nitrogen gas

Molar mass of Nitrogen gas is 28.0 g/mol

Moles of nitrogen gas = 1.50 g ÷ 28.0 g/mol

                                     = 0.054 moles

  • According to the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
  • Therefore, 0.216 moles of lithium will require 0.036 moles (0.216 moles ÷6) of nitrogen gas.
  • On the other hand, 0.054 moles of nitrogen, would require 0.324 moles of Lithium.

Thus, Lithium is the limiting reagent while nitrogen is in excess.

7 0
3 years ago
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