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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.
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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?.
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