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alisha [4.7K]
4 years ago
6

A galvanic cell consists of a Cu(s)|Cu2+(aq) half-cell and a Cd(s)|Cd2+(aq) half-cell connected by a salt bridge. Oxidation occu

rs in the cadmium half-cell. The cell can be represented in standard notation as
Chemistry
1 answer:
Kryger [21]4 years ago
6 0

Answer:

Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)

Explanation:

A galvanic cell is composed of two electrodes immersed in a suitable electrolyte and connected via a salt bridge. One of the electrodes serves as a cathode where reduction or gain of electrons takes place. The other half cell functions as an anode where oxidation or loss of electrons occurs.

The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a salt bridge. Then the cathodic ion with its molar concentration is written and then the cathode.

As it is given that cadmium acts as anode, it must be on the left hand side and copper must be on right hand side.

Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)

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Identify the spectator ions in this reaction. Check all that apply.
balandron [24]
CN, LI are the only two
4 0
3 years ago
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When adjusted for any changes in ΔHΔH and ΔSΔS with temperature, the standard free energy change ΔG∘TΔGT∘Delta G_{T}^{\circ} at
STALIN [3.7K]

The equilibrium constant is 0.0022.

Explanation:

The values given in the problem is

ΔG° = 1.22 ×10⁵ J/mol

T = 2400 K.

R = 8.314 J mol⁻¹ K⁻¹

The Gibbs free energy should be minimum for a spontaneous reaction and equilibrium state of any reaction is spontaneous reaction. So on simplification, the thermodynamic properties of the equilibrium constant can be obtained as related to Gibbs free energy change at constant temperature.

The relation between Gibbs free energy change with equilibrium constant is ΔG° = -RT ln K

So, here K is the equilibrium constant. Now, substitute all the given values in the corresponding parameters of the above equation.

We get,

1.22 * 10^{5} = - 8.314* 2400 * ln K

\\ 1.22 * 10^{5} = -19953.6 * ln K

ln K = \frac{-1.22*10^{5} }{19953.6} =-6.114\\\\k =e^{-6.114}=0.0022

So, the equilibrium constant is 0.0022.

4 0
3 years ago
Three 1.0-l flasks, maintained at 308 k, are connected to each other with stopcocks. initially the stopcocks are closed. one of
Licemer1 [7]

Answer:

0.6103 atm.

Explanation:

  • We need to calculate the vapor pressure of each component after the stopcocks are opened.
  • Volume after the stopcocks are opened = 3.0 L.

<u><em>1) For N₂:</em></u>

P₁V₁ = P₂V₂

P₁ = 1.5 atm & V₁ = 1.0 L & V₂ = 3.0 L.

P₂ of N₂ = P₁V₁ / V₂ = (1.5 atm) (1.0 L) / (3.0 L) = 0.5 atm.

<u><em>2) For H₂O:</em></u>

Pressure of water at 308 K is 42.0 mmHg.

we need to convert from mmHg to atm: <em>(1.0 atm = 760.0 mmHg)</em>.

P of H₂O = (1.0 atm x 42.0 mmHg) / (760.0 mmHg) = 0.0553 atm.

We must check if more 2.2 g of water is evaporated,

n = PV/RT = (0.0553 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.00656 mole.

m = n x cmolar mass = (0.00656 mole) (18.0 g/mole) = 0.118 g.

It is lower than the mass of water in the flask (2.2 g).

<em><u>3) For C₂H₅OH:</u></em>

Pressure of C₂H₅OH at 308 K is 102.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of C₂H₅OH = (1.0 atm x 102.0 mmHg) / (760.0 mmHg) = 0.13421 atm.

We must check if more 0.3 g of C₂H₅OH is evaporated,

n = PV/RT = (0.13421 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.01594 mole.

m = n x molar mass = (0.01594 mole) (46.07 g/mole) = 0.7344 g.

<em>It is more than the amount in the flask (0.3 g), so the pressure should be less than 0.13421 atm.</em>

We have n = mass / molar mass = (0.30 g) / (46.07 g/mole) = 0.00651 mole.

So, P of C₂H₅OH = nRT / V = (0.00651 mole) (0.082 L.atm/mole.K) (308.0 K) / (3.0 L) = 0.055 atm.

  • <em>So, </em><em>total pressure</em><em> = </em><em>P of N₂ + P of H₂O + P of C₂H₅OH</em><em> = 0.5 atm + 0.0553 atm + 0.055 atm = </em><em>0.6103 atm</em><em>.</em>
3 0
3 years ago
differentiate between the characteristics, composition, and location of comets, asteroids, and meteoroids.
Masja [62]

Answer:

Asteroids

At first glance, asteroids may seem like run-of-the-mill space rocks, but these ancient solar system remnants come in all shapes, sizes and flavors

Comets

For millennia, the sight of a comet elicited fear and awe. Ancient astronomers believed comets foretold the death of princes and the outcomes of wars. Modern astronomers know comets are the ice-clad leftovers from the material that formed our solar system billions of years ago.

Meteoroids

Meteoroids are the true space rocks of the solar system. No larger than a meter in size (3.3 feet) and sometimes the size of a grain of dust, they are too small to be considered asteroids or comets, but many are the broken pieces of either. Some meteoroids originate from the ejected debris caused by impacts on planets or moons.

5 0
3 years ago
How many moles are in 98.3 grams of aluminum hydroxide, Al(OH)3?
Musya8 [376]
We have to calculate the molar mass of  AL(OH)₃
Atomic mass (Al)=27 amu
Atomic mass (O)=16 amu
Atomic mass (H)=1 amu

molecular weight= 27 amu+3(16 amu + 1 amu) =78 amu.

Therefore, the molar mass of Al (OH)₃ is 78 g/ mol

Now, we calculate the number of moles in 98.3 g of aluminum hydroxide.

78 g-------------------1 mol
98.3 g-----------------  x

x=(98.3 g * 1 mol) / 78 g=1.26 moles.

Answer: 1.26 moles.
6 0
3 years ago
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