JEE Exam  >  JEE Questions  >  A cell Cu|Cu++||Ag+ |Ag initially contains 2M... Start Learning for Free
A cell Cu|Cu++||Ag+ |Ag initially contains 2M Ag+ and 2M Cu++ ions in 1 L electrolyte .The change in cell potential after the passage of 10 amp current for 4825 seconds during usage of cell is Answer: -0.009V?
Verified Answer
A cell Cu|Cu++||Ag+ |Ag initially contains 2M Ag+ and 2M Cu++ ions in ...
Ans.

This question is part of UPSC exam. View all JEE courses
Most Upvoted Answer
A cell Cu|Cu++||Ag+ |Ag initially contains 2M Ag+ and 2M Cu++ ions in ...
Cell Potential Calculation:

To determine the change in cell potential after the passage of current, we need to consider the individual half-cell reactions and their standard cell potentials.

Half-Cell Reactions:

The half-cell reactions involved in this cell are:

Cu2+ + 2e- → Cu (reduction)
Ag → Ag+ + e- (oxidation)

Standard Cell Potentials:

The standard cell potential, E°, for the cell can be calculated using the Nernst equation:

E°cell = E°reduction + E°oxidation

The standard reduction potentials for Cu2+ + 2e- → Cu and Ag → Ag+ + e- are +0.34V and -0.80V, respectively.

Therefore, E°cell = +0.34V + (-0.80V) = -0.46V

Calculating the Change in Cell Potential:

To calculate the change in cell potential after the passage of current, we need to consider the effect of the current on the concentrations of the ions involved in the half-cell reactions.

The passage of current leads to the migration of Cu2+ ions towards the cathode (reduction half-cell) and Ag+ ions towards the anode (oxidation half-cell).

Calculating the Change in Concentration:

The passage of current for a given time can be calculated using Faraday's law:

Q = nF

Where Q is the charge passed (Coulombs), n is the number of electrons transferred in the reaction, and F is the Faraday constant (96,485 C/mol).

Given that the current is 10A and the time is 4825 seconds, the charge passed is:

Q = 10A * 4825s = 48,250 C

Since the number of electrons transferred in the reduction half-cell is 2 (from the balanced equation), the number of moles of Cu2+ ions reduced is:

n = Q / (2F)

n = 48,250 C / (2 * 96,485 C/mol) = 0.25 mol

Similarly, the number of moles of Ag+ ions oxidized is also 0.25 mol.

Calculating the New Concentrations:

The initial concentration of Cu2+ and Ag+ ions is 2M each. After the passage of current, the new concentration can be calculated using the volume of the electrolyte and the number of moles.

Assuming the volume of the electrolyte remains constant at 1L, the new concentration can be calculated as follows:

New concentration = (Number of moles) / (Volume)

New concentration = 0.25 mol / 1 L = 0.25 M

Calculating the New Cell Potential:

Using the Nernst equation, we can calculate the new cell potential, Ecell, with the new concentrations:

Ecell = E°cell - (RT / nF) * ln(Q)

Where R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, n is the number of electrons transferred, F is the Faraday constant, and Q is the reaction quotient.

Assuming the temperature is 298K, the new cell potential can be calculated as:

Ecell = -0.46
Explore Courses for JEE exam

Similar JEE Doubts

A fuel cell is a cell that is continuously supplied with an oxidant and a reductant so that it can deliver a current indefinitely.Fuel cells offer the possibility of achieving high thermodynamic efficiency in the conversion of Gibbs energy into mechanical work. Internal combustion engines at best convert only the fraction (T2 T1)/T2 of the heat of combustion into mechanical work.While the thermodynamic efficiency of the fuel cell is given by, is theGibbs energy change for the cell reaction and H is the enthalpy change of the cell reaction.A hydrogen-oxygen fuel cell may have an acidic or alkaline electrolyte.Pt|H2(g)|H+(aq.)||H2O( l )|O2(g)|Pt; The above fuel cell is used to produce constant current supply under constant temperature 30 atm constant total pressure condition in a cylinder. If 10 moles H2 and 5 moles of O2 were taken initially. Rate of combustion of O2 is 10 milli moles per minute. The half cell reactions areTo maximize the power per unit mass of an electrochemical cell, the electronic and electrolytic resistances of the cell must be minimized. Since fused salts have lower electrolytic resistances than aqueous solutions, high-temperature electrochemical cells are of special interest for practical applications.Q.The above fuel cell is used completely as an electrolytic cell with Cu voltameter of resistance 26.94 W using Pt electrodes. Initially Cu voltameter contains 1 litre solution of 0.05M CuSO4.[H+] in solution after electrolysis (Assuming no change on volume of solution).

A cell Cu|Cu++||Ag+ |Ag initially contains 2M Ag+ and 2M Cu++ ions in 1 L electrolyte .The change in cell potential after the passage of 10 amp current for 4825 seconds during usage of cell is Answer: -0.009V?
Question Description
A cell Cu|Cu++||Ag+ |Ag initially contains 2M Ag+ and 2M Cu++ ions in 1 L electrolyte .The change in cell potential after the passage of 10 amp current for 4825 seconds during usage of cell is Answer: -0.009V? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A cell Cu|Cu++||Ag+ |Ag initially contains 2M Ag+ and 2M Cu++ ions in 1 L electrolyte .The change in cell potential after the passage of 10 amp current for 4825 seconds during usage of cell is Answer: -0.009V? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cell Cu|Cu++||Ag+ |Ag initially contains 2M Ag+ and 2M Cu++ ions in 1 L electrolyte .The change in cell potential after the passage of 10 amp current for 4825 seconds during usage of cell is Answer: -0.009V?.
Solutions for A cell Cu|Cu++||Ag+ |Ag initially contains 2M Ag+ and 2M Cu++ ions in 1 L electrolyte .The change in cell potential after the passage of 10 amp current for 4825 seconds during usage of cell is Answer: -0.009V? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of A cell Cu|Cu++||Ag+ |Ag initially contains 2M Ag+ and 2M Cu++ ions in 1 L electrolyte .The change in cell potential after the passage of 10 amp current for 4825 seconds during usage of cell is Answer: -0.009V? defined & explained in the simplest way possible. Besides giving the explanation of A cell Cu|Cu++||Ag+ |Ag initially contains 2M Ag+ and 2M Cu++ ions in 1 L electrolyte .The change in cell potential after the passage of 10 amp current for 4825 seconds during usage of cell is Answer: -0.009V?, a detailed solution for A cell Cu|Cu++||Ag+ |Ag initially contains 2M Ag+ and 2M Cu++ ions in 1 L electrolyte .The change in cell potential after the passage of 10 amp current for 4825 seconds during usage of cell is Answer: -0.009V? has been provided alongside types of A cell Cu|Cu++||Ag+ |Ag initially contains 2M Ag+ and 2M Cu++ ions in 1 L electrolyte .The change in cell potential after the passage of 10 amp current for 4825 seconds during usage of cell is Answer: -0.009V? theory, EduRev gives you an ample number of questions to practice A cell Cu|Cu++||Ag+ |Ag initially contains 2M Ag+ and 2M Cu++ ions in 1 L electrolyte .The change in cell potential after the passage of 10 amp current for 4825 seconds during usage of cell is Answer: -0.009V? tests, examples and also practice JEE tests.
Explore Courses for JEE exam

Top Courses for JEE

Explore Courses
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev