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andrew11 [14]
3 years ago
13

What is produced at each electrode in the electrolysis of agf(aq)?

Chemistry
1 answer:
Rzqust [24]3 years ago
8 0
AgF consists of Ag+ and F- ions, which are fully dissociated in aqueous solution. When solving electrolysis problems, it is important to remember that water itself may also be a subject to electrolysis. Therefore, determining which species is oxidized and which species is reduced depends on selecting the processes that are the most energetically favorable. The most preferred reduction reaction will be  Ag+ + e- = Ag (Emf=0.7996 V) which will occur at the cathode, on the other hand, the most favorable oxidation reaction will be
2H2O = O2 +4H+ + 4e- (Emf = -1.3 V) that will occur at the anode. Thus, the product at the anode is oxygen gas and at the cathode electrode is silver metal.
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How is enthalpy used to predict whether a reaction is endothermic or exothermic?
Dahasolnce [82]
If the enthalpy change is negative, then the reaction is exothermic because heat energy was lost to the surroundings.

If the enthalpy change was positive, then the reaction was endothermic because heat energy was gained from the surroundings!

Hope this helps!! x
8 0
3 years ago
Limiting reactants don’t know how to do
Assoli18 [71]

Answer:

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Explanation:

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6 0
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What is the main cause of increased erosion
anyanavicka [17]
The main cause of increased erosion is natural elements like rain, wind and ice. Erosion is the process by which the surface of the Earth gets worn down. 
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Does a reaction occur when aqueous solutions of copper(II) sulfate and lead(II) nitrate are combined?
Dafna11 [192]

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6 0
3 years ago
The following physical constants are for water, H2O.
Delicious77 [7]

Answer:

Q\approx6.4~kJ

Explanation:

Quantity of heat required by 10 gram of ice initially warm it from -5°C to 0°C:

Q_1=m.C_s.\Delta T

here;

mass, m = 10 g

specific heat capacity of ice, C_s=2.09~J.g^{-1}.^{\circ}C^{-1}

change in temperature, \Delta T=(5-0)=5^{o}C

Q_1=10\times2.09\times 5

Q_1=104.5~J

Amount of heat required to melt the ice at 0°C:

Q_2=m.\Delta H_{fus}

where, \Delta H_{fus}=6020~J/mol

we know that no. of moles is = (wt. in gram) \div (molecular mass)

Q_2=\frac{10}{18} \times 6020

Q_2=3344.44~J

Now, the heat required to bring the water to 70°C from 0°C:

Q_3=m.C_L.\Delta T

specific heat of water, C_L=4.18~J/g/^oC

change in temperature, \Delta T=(70-0)=70^oC

Q_3=10\times 4.18\times 70

Q_3=2926~J

Therefore the total heat required to warm 10.0 grams of ice at -5.0°C to a temperature of 70.0°C:

Q=Q_1+Q_2+Q_3

Q=104.5+3344.44+2926

Q=6374.94~J

Q\approx6.4~kJ

8 0
3 years ago
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