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Rainbow [258]
1 year ago
15

Consider the standard galvanic cell based on the following half-reactions The electrodes in this cell are and . Does the cell po

tential increase, decrease, or remain the same when the following changes occur to the standard cell? You may need to use the following data: Half-Reaction 0.80 0.34 a is added to the silver half-cell compartment (assume no volume change). increases decreases remains the same b is added to the silver half-cell compartment. [Hint: reacts with to form .] increases decreases remains the same c is added to the copper half-cell compartment. [Hint: reacts with to form .] increases decreases remains the same
Chemistry
1 answer:
Free_Kalibri [48]1 year ago
6 0

Question: The question is incomplete and can't be comprehended. See the complete question below and the answer.

Consider a galvanic cell based upon the following half reactions: Ag+ + e- → Ag 0.80 V Cu2+ + 2 e- → Cu 0.34 V

How would the following changes alter the potential of the cell?

a) Adding Cu2+ ions to the copper half reaction (assuming no volume change).

b) Adding equal amounts of water to both half reactions.

c) Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

d) Adding Ag+ ions to the silver half reaction (assume no volume change)

Explanation:

Nernst equation relates the reduction potential of an electrochemical reaction (half-cell or full cell reaction) to the standard electrode potential, temperature, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation.

Reaction under consideration:

Ag+ + e- → Ag 0.80 V

Cu+2 + 2 e- → Cu 0.34 V

Clearly, Ag reduction potential is high and this indicates that it gets reduced readily which leaves Cu to oxidize. Cu+2 ions are products of reaction and Ag+ ions are reactant ions.

Nernst equation : Ecell = E°cell­ – (2.303 RT / n F) log Q    

where                            

             Ecell = actual cell potential

             E°cell­ ­​ = standard cell potential

             R = the universal gas constant = 8.314472(15) J K−1 mol−1

             T = the temperature in kelvins

              n = the number of moles of electrons transferred                                    

 F = the Faraday constant, the number of coulombs per mole of electrons:

  (F = 9.64853399(24)×104 C mol−1)

 Q = [product ion]y / [reactant ion] x

Accordingly when applied to above reaction one will get the following

= E°cell­ – (2.303 x RT / 6 F) log [Cu+2] / [Ag+]

Now the given variables can be studied according to Le Chatelier's principle which states when any system at equilibrium is subjected to change in its concentration, temperature, volume, or pressure, then the system readjusts itself to (partially) counteract the effect of the applied change and a new equilibrium is established.

a)        Adding Cu2+ ions to the copper half reaction (assuming no volume change).

Addition of Cu+2 ions increases its concentration and consequently increases the Q value which results in reduction of Ecell. In other words the addition of Cu+2 ions favors the backward reaction to maintain the equilibrium of reaction and hence the forward reaction rate decreases.

b)       Adding equal amounts of water to both half reactions.

Addition of water increases the dilution of the electrochemical cell. For weak electrolytes such as Ag+/ Cu+2 with increase in dilution, the degree of dissociation increases and as a result molar conductance increases.

c)        Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

Based on Le Chatelier's principle when Cu+2 ions amount is decreased by its continuous removal from  the system the forward reaction is favored. As the Cu+2 ions is removed the system attempts to generate more Cu+2 ions to counter the affect of its removal.

d)       Adding Ag+ ions to the silver half reaction (assume no volume change)

Addition of reactant ions, i.e. Ag+ ions, will favour the forward reaction, which results in more product formation.

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mastering chem If the experiment below is run for 60 s, 0.16 mol A remain. Which of the following statements is or are true? At
Dvinal [7]

Answer: All of the statements are true.

Explanation:

(a) Considering the system mentioned in the equation:-

The sum of total moles in the flask will always be equal to 1 which leads to confirmation of this statement as for 60 secs= 0.16 mol A and 0.84 mol B

(b) 0<t< 20s,  mole A got reduced from 1 mole to 0.54 moles while at 40s to 60s A got decreased from 0.30 moles to 0.16 moles.

0 to 20s is 0.46 (1 - 0.54 = 0.46)mol whereas,

40 to 60s is 0.14 (0.30-.16 = 0.14) mol

(0.46 > 0.14) mol leading this statement to be true as well.

(c) Average rate from t1 = 40 to t2 = 60 s is given by:

\delta moles/\delta time  = 0.30-0.16/60-40 = 0.007 Mol/s which is true as well

7 0
2 years ago
What are characteristics of absorption spectra? Check all that apply.
Amanda [17]
The absorption spectrum is the result of absorption of light radiation by a material (solid or liquid) as a function of wavelength or frequency. The figure is shown below:

7 0
2 years ago
Ethene is a useful substance that can form polymers. It has a melting point of 169°C and a boiling point of 104°C. At which temp
blondinia [14]

Answer:

-169°C to -104°C

Explanation:

Ethene, also known as ethylene exists in solid, liquid and gaseous states. Ethene is an aliens with condensed structural formula C2H4. Athens is a colourless gas. It is flammable and is also a sweet smelling gas in its pure form. It is the monomer in the production of polyethylene which is of great importance in the plastic industry. In agriculture, it is used to induce the ripening of fruits. It can be hydrated in order to produce ethanol.

The liquid range of ethene refers to the temperatures at which ethene is found in the liquid state of matter. It is actually the difference between the melting point and the boiling points of ethene. Hence the liquid range of ethene is -169°C to -104°C

4 0
1 year ago
Which compound would be expected to show intense IR absorption at 2710 and 1705 cm-1? (Ph = phenyl group)
Serhud [2]

Answer:

B. PhCHO

Explanation:

Every organic group shows a characteristic IR absorption at certain wavelength . With the help of these absorption spectra we can identify the group present on organic molecules .

The wave number of 2710 cm⁻¹ is absorbed by aldehyde bond stretching .

The wave number of 1705 cm⁻¹ is shown by conjugated aldehyde . So the most likely compound among given compounds is PhCHO .

7 0
2 years ago
Dissolving NaOH(s) in water is exothermic. Two calorimetry experiments are set up. Experiment 1: 2 g of NaOH are dissolved in 10
vlada-n [284]

Answer:

a. both temperature changes will be the same

Explanation:

When sodium hydroxide (NaOH) is dissolved in water, a determined amount is released to the solution following the equation:

Q = m×C×ΔT

<em>Where Q is the heat released, m is the mass of the solution, C is the specific heat and ΔH is change in temperature.</em>

Specific heat of both solutions is the same (Because the solutions are in fact the same). Specific heat = C.

m is mass of solutions: 102g for experiment 1 and 204g for experiment 2.

And Q is the heat released: If 2g release X heat, 4g release 2X.

Thus, ΔT in the experiments is:

Experiment 1:

X / 102C = ΔT

Experiment 2:

2X / 204C = ΔT

X / 102C = ΔT

That means,

<h3>a. both temperature changes will be the same</h3>

5 0
1 year ago
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