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blsea [12.9K]
4 years ago
11

Which statement describes where oxidation and reduction half-reactions occur in an operating electrochemical cell?(1) Oxidation

and reduction both occur at the anode
(2) Oxidation and reduction both occur at the cathode
(3) Oxidation occurs at the anode and reduction occurs at the cathode
(4) Oxidation occurs at the cathode and reduction occurs at the anode
Chemistry
2 answers:
Natalija [7]4 years ago
8 0

Answer : The correct option is, (3) Oxidation occurs at the anode and reduction occurs at the cathode

Explanation :

Electrochemical cell : It is the cell in which the chemical reaction occurs with the production of electric potential difference between the two electrodes.

Or, it is a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy.

In this cell, the oxidation occurs at anode which is a negative electrode and reduction occurs at cathode which is a positive electrode.

Hence, the correct option is, (3) Oxidation occurs at the anode and reduction occurs at the cathode

emmainna [20.7K]4 years ago
3 0
The answer is (3), oxidation occurs at the anode and reduction occurs at the cathode. That's because the oxidation reaction can lose electrons and reduction can gain electrons.
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a 3.50L sample of neon gas has a pressure of .950atm at 20c what would the volume be if the pressure increased to 1.50atm and th
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V_2=2.22L

Explanation:

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In this case, we use the Boyle's law which allows us to understand the volume-pressure behavior as an inversely proportional relationship:

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Thus, solving for the final volume, once the pressure changes, we obtain:

V_2=\frac{P_1V_1}{P_2}=\frac{3.50L*0.950atm}{1.50atm}  \\\\V_2=2.22L

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Which part of the calcium atom in the ground state is represented by the dots in its Lewis electron dot diagram
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5 0
3 years ago
A buffer solution is prepared by placing 5.86 grams of sodium nitrite and 32.6 mL of a 4.90 M nitrous acid solution into a 500.0
N76 [4]

This question is incomplete, the complete question is;

A buffer solution is prepared by placing 5.86 grams of sodium nitrite and 32.6 mL of a 4.90 M nitrous acid solution into a 500.0 mL volumetric flask and diluting to the calibration mark. If 10.97 mL of a 1.63 M solution of potassium hydroxide is added to the buffer, what is the final pH? The Ka for nitrous acid = 4.6 × 10⁻⁴.

Answer:

the final pH is 3.187

Explanation:

Given the data in the question;

Initial moles of HNO2 = 32.6/1000 × 4.90 = 0.15974 mol

Initial moles of NO2- = mass/molar mass = 5.86/68.995 =  0.0849336 mol

Moles of KOH added = 10.97/1000 × 1.63  = 0.0178811 mol

so

HN02 + KOH → NO2- + H2O

moles of HNO2 = 0.15974 - 0.0178811 = 0.1418589 mol

Moles of NO2- = 0.0849336 + 0.0178811  =  0.1028147 mol

Now,

pH = pka + log( [NO2-]/[HNO2])

pH = -log ka + log( moles of NO2- / moles of HNO2 )

we substitute

pH = -log( 4.6 × 10⁻⁴ ) + log( 0.1028147  / 0.1418589  )

pH = -log( 4.6 × 10⁻⁴ ) + log( 0.724767 )

pH =  3.337242 + (-0.1398 )

pH = 3.187

Therefore, the final pH is 3.187

8 0
3 years ago
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