Silver can be electroplated at the cathode of an electrolysis cell by the half-reaction. Ag+(aq)+e−→Ag(s) Part A Silver can be e
1 answer:
Answer: Mass of silver deposited at the cathode is 37.1g
Explanation: According to Faraday Law of Electrolysis, the mass of substance deposited at the electrode (cathode or anode) is directly proportional to quantity of electricity passed through the electrolyte
Faraday has found that to liberate one gm eq. of substance from an electrolyte, 96500C of electricity is required.
+e− ==> Ag(s)
Given that
Current (I) = 8.5A
Time (t) = 65 *60 = 3900s
Quantity of electricity passed = 8.5*3900 =33150C
Molar mass of Ag= 108g
96500C will liberate 108g
33150C will liberate Xg
Xg= (108*33150)/96500
=37.1g
Therefore the mass of Ag deposited at the cathode is 37.1g.
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Answer:
= 25.05°C
Explanation:
Given:
the value of ΔHcomb (heat of combustion) for dimethylphthalate (C10H10O4) is = 4685 kJ/mol.
mass = 0.905g of dimethylphthalate
molar mass = 194.18g dimethylphthalate
number of moles of dimethylphthalate = ???
= 21.5°C
= 6.15 kJ/°C
= ???
since we have our molar mass and mass of dimethylphthalate ;we can determine the number of moles as;
0.905g of dimethylphthalate ×
number of moles of dimethylphthalate = 0.000466 moles
Heat released = moles of dimethylphthalate × heat of combustion
= 0.000466 moles × 4685 kJ
= 21.84 kJ
∴ Heat absorbed by the calorimeter =
21.84 kJ =6.15 kJ/°C
21.84 KJ =
21.84 KJ = - 132.225 kJ
21.84 KJ + 132.225 kJ =
154.065 kJ =
=
=25.05°C