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Determine Ecell at 25∘C for the following voltaic cell.

Pb(s)/Pb2+ (0.10M)//Cu 2+ (2.00M)/Cu(s)
a) What is the maximum amount of work possible from this cell?

User Eactor
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1 Answer

5 votes

Final answer:

The question involves calculating the Ecell of a voltaic cell at 25°C using the Nernst equation and then determining the maximum work possible. Standard reduction potentials are required to find E°cell, which is then used with ion concentrations to calculate Ecell and subsequently the maximum work.

Step-by-step explanation:

The question at hand relates to the electrochemistry subject in high school level chemistry, focusing on calculating the electromotive force (EMF) or Ecell of a voltaic cell and the maximum work that can be obtained from the cell at 25°C. The voltaic cell described consists of a Pb/Pb2+ half-cell and a Cu2+/Cu half-cell, with respective molar concentrations of 0.10 M and 2.00 M. To determine the Ecell, you would apply the Nernst equation:

Ecell = E°cell - (RT/nF) * ln(Q)

where:

  • E°cell is the standard cell potential
  • R is the gas constant (8.314 J/mol·K)
  • T is the temperature in kelvin (298K for 25°C)
  • n is the number of moles of electrons transferred
  • F is Faraday's constant (96485 C/mol)
  • Q is the reaction quotient

To find E°cell, we look up the standard reduction potentials for both half-cells and calculate:

E°cell = E°(Cu2+/Cu) - E°(Pb2+/Pb)

Q can be calculated using the ion concentrations given by:

Q = [Pb2+]2/[Cu2+]

Plugging in the numbers and solving for Ecell will give the voltage of the cell. The maximum work (W) possible from the cell can be calculated using the formula:

W = nFEcell

The correct option for maximum work will be determined using the calculated value of Ecell.

User Chris Bouchard
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