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(Ag ) e==Ag, Cu 2 2e =Cu, Eo= 0.50 V; Eo = 0.34 V A 100 ml solution is 1080 mg with respect to Agt and 635 mg with respect to Cu2. If 0.1mg Ag left in the solution is considered to be the complete deposition of Ag , the cathode potential, so that no copper is deposited during the process is?
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(Ag ) e==Ag, Cu 2 2e =Cu, Eo= 0.50 V; Eo = 0.34 V A 100 ml solution...
Given:
Ag + e → Ag, Eo = 0.50 V
Cu2 + 2e → Cu, Eo = 0.34 V
100 ml solution containing 1080 mg Agt and 635 mg Cu2
0.1 mg Ag is left in the solution after complete deposition of Ag
To Find:
Cathode potential to prevent deposition of Cu during the process
Solution:
Step 1: Calculate the number of moles of Agt and Cu2 in the solution
Moles of Agt = (1080 mg)/(107.87 g/mol) = 10.01 × 10^-3 mol
Moles of Cu2 = (635 mg)/(63.55 g/mol) = 9.99 × 10^-3 mol
Step 2: Calculate the initial potential of the cell
Eo(cell) = Eo(reduction at cathode) - Eo(reduction at anode)
Eo(cell) = 0.34 - 0.50 = -0.16 V
Step 3: Calculate the concentration of Ag+ and Cu+ ions
Ag+ ions = (10.01 × 10^-3 mol)/(0.1 L) = 0.100 M
Cu2+ ions = (9.99 × 10^-3 mol)/(0.1 L) = 0.100 M
Step 4: Calculate the concentration of Ag+ and Cu+ ions after deposition of 0.1 mg Ag
Moles of Ag deposited = (0.1 mg)/(107.87 g/mol) = 0.93 × 10^-6 mol
Ag+ ions after deposition = (10.01 × 10^-3 mol - 0.93 × 10^-6 mol)/(0.1 L) = 0.099 M
Cu2+ ions after deposition = (9.99 × 10^-3 mol)/(0.1 L) = 0.100 M
Step 5: Calculate the new potential of the cell after deposition of 0.1 mg Ag
E(cell) = Eo(cell) - (0.0592 V/n) * log(Q)
where n = number of electrons transferred in the half-reaction
Q = (Ag+ ions after deposition)/(Cu2+ ions after deposition)
E(cell) = -0.16 - (0.0592 V/2) * log(0.099/0.100)
E(cell) = -0.157 V
Step 6: Calculate the cathode potential to prevent deposition of Cu during the process
Cathode potential = E(cell) + Eo(reduction at anode)
Cathode potential = -0.157 + 0.50 = 0.343 V
Answer:
The cathode potential to prevent deposition of Cu during the process is 0.343 V.
Ag + e → Ag, Eo = 0.50 V
Cu2 + 2e → Cu, Eo = 0.34 V
100 ml solution containing 1080 mg Agt and 635 mg Cu2
0.1 mg Ag is left in the solution after complete deposition of Ag
To Find:
Cathode potential to prevent deposition of Cu during the process
Solution:
Step 1: Calculate the number of moles of Agt and Cu2 in the solution
Moles of Agt = (1080 mg)/(107.87 g/mol) = 10.01 × 10^-3 mol
Moles of Cu2 = (635 mg)/(63.55 g/mol) = 9.99 × 10^-3 mol
Step 2: Calculate the initial potential of the cell
Eo(cell) = Eo(reduction at cathode) - Eo(reduction at anode)
Eo(cell) = 0.34 - 0.50 = -0.16 V
Step 3: Calculate the concentration of Ag+ and Cu+ ions
Ag+ ions = (10.01 × 10^-3 mol)/(0.1 L) = 0.100 M
Cu2+ ions = (9.99 × 10^-3 mol)/(0.1 L) = 0.100 M
Step 4: Calculate the concentration of Ag+ and Cu+ ions after deposition of 0.1 mg Ag
Moles of Ag deposited = (0.1 mg)/(107.87 g/mol) = 0.93 × 10^-6 mol
Ag+ ions after deposition = (10.01 × 10^-3 mol - 0.93 × 10^-6 mol)/(0.1 L) = 0.099 M
Cu2+ ions after deposition = (9.99 × 10^-3 mol)/(0.1 L) = 0.100 M
Step 5: Calculate the new potential of the cell after deposition of 0.1 mg Ag
E(cell) = Eo(cell) - (0.0592 V/n) * log(Q)
where n = number of electrons transferred in the half-reaction
Q = (Ag+ ions after deposition)/(Cu2+ ions after deposition)
E(cell) = -0.16 - (0.0592 V/2) * log(0.099/0.100)
E(cell) = -0.157 V
Step 6: Calculate the cathode potential to prevent deposition of Cu during the process
Cathode potential = E(cell) + Eo(reduction at anode)
Cathode potential = -0.157 + 0.50 = 0.343 V
Answer:
The cathode potential to prevent deposition of Cu during the process is 0.343 V.
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(Ag ) e==Ag, Cu 2 2e =Cu, Eo= 0.50 V; Eo = 0.34 V A 100 ml solution is 1080 mg with respect to Agt and 635 mg with respect to Cu2. If 0.1mg Ag left in the solution is considered to be the complete deposition of Ag , the cathode potential, so that no copper is deposited during the process is?
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(Ag ) e==Ag, Cu 2 2e =Cu, Eo= 0.50 V; Eo = 0.34 V A 100 ml solution is 1080 mg with respect to Agt and 635 mg with respect to Cu2. If 0.1mg Ag left in the solution is considered to be the complete deposition of Ag , the cathode potential, so that no copper is deposited during the process is? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about (Ag ) e==Ag, Cu 2 2e =Cu, Eo= 0.50 V; Eo = 0.34 V A 100 ml solution is 1080 mg with respect to Agt and 635 mg with respect to Cu2. If 0.1mg Ag left in the solution is considered to be the complete deposition of Ag , the cathode potential, so that no copper is deposited during the process is? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for (Ag ) e==Ag, Cu 2 2e =Cu, Eo= 0.50 V; Eo = 0.34 V A 100 ml solution is 1080 mg with respect to Agt and 635 mg with respect to Cu2. If 0.1mg Ag left in the solution is considered to be the complete deposition of Ag , the cathode potential, so that no copper is deposited during the process is?.
(Ag ) e==Ag, Cu 2 2e =Cu, Eo= 0.50 V; Eo = 0.34 V A 100 ml solution is 1080 mg with respect to Agt and 635 mg with respect to Cu2. If 0.1mg Ag left in the solution is considered to be the complete deposition of Ag , the cathode potential, so that no copper is deposited during the process is? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about (Ag ) e==Ag, Cu 2 2e =Cu, Eo= 0.50 V; Eo = 0.34 V A 100 ml solution is 1080 mg with respect to Agt and 635 mg with respect to Cu2. If 0.1mg Ag left in the solution is considered to be the complete deposition of Ag , the cathode potential, so that no copper is deposited during the process is? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for (Ag ) e==Ag, Cu 2 2e =Cu, Eo= 0.50 V; Eo = 0.34 V A 100 ml solution is 1080 mg with respect to Agt and 635 mg with respect to Cu2. If 0.1mg Ag left in the solution is considered to be the complete deposition of Ag , the cathode potential, so that no copper is deposited during the process is?.
Solutions for (Ag ) e==Ag, Cu 2 2e =Cu, Eo= 0.50 V; Eo = 0.34 V A 100 ml solution is 1080 mg with respect to Agt and 635 mg with respect to Cu2. If 0.1mg Ag left in the solution is considered to be the complete deposition of Ag , the cathode potential, so that no copper is deposited during the process is? in English & in Hindi are available as part of our courses for IIT JAM.
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(Ag ) e==Ag, Cu 2 2e =Cu, Eo= 0.50 V; Eo = 0.34 V A 100 ml solution is 1080 mg with respect to Agt and 635 mg with respect to Cu2. If 0.1mg Ag left in the solution is considered to be the complete deposition of Ag , the cathode potential, so that no copper is deposited during the process is?, a detailed solution for (Ag ) e==Ag, Cu 2 2e =Cu, Eo= 0.50 V; Eo = 0.34 V A 100 ml solution is 1080 mg with respect to Agt and 635 mg with respect to Cu2. If 0.1mg Ag left in the solution is considered to be the complete deposition of Ag , the cathode potential, so that no copper is deposited during the process is? has been provided alongside types of (Ag ) e==Ag, Cu 2 2e =Cu, Eo= 0.50 V; Eo = 0.34 V A 100 ml solution is 1080 mg with respect to Agt and 635 mg with respect to Cu2. If 0.1mg Ag left in the solution is considered to be the complete deposition of Ag , the cathode potential, so that no copper is deposited during the process is? theory, EduRev gives you an
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