CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

What is the solubility product of a saturated solution of Ag2CrO4in water at 298K if the emf of the cell Ag Is 0.164V At 298K?


Open in App
Solution

Cell potential:

  • In an electrochemical cell, the cell potential,Ecell, is the gap between two half cells.
  • The capacity of electrons to pass by one-half cell to another causes the potential difference.
  • Because the chemical process is a redox reaction, electrons can travel between electrodes.
  • When one chemical is oxidised while another is reduced, a redox reaction occurs.
  • The component loses one or more electrons during oxidation and consequently becomes positively charged.
  • During reduction, on the other hand, the material acquires electrons and becomes negatively charged.
  • This pertains to cell potential measurement because the difference between the potential for the reducing agent to get oxidized and the potential for the oxidising agent to become reduced determines the cell potential.
The Cell Potential - Chemistry LibreTexts

Formula for Cell potential:

  • Ecell=0.0591log[Ag+]RHS[Ag+]LHS

Step 1: Analyzing the given quantities

  • Ecell=0.164[Ag+]RHS=0.1

Step 2: Applying values in the formula

  • 0.164=0.0591log0.1[Ag+]LHS[Ag+]LHS=1.66×10-4M

Step 3: finding solubility product Ksp(Ag2CrO4)

  • [CrO42-]=1.66×10-42Ksp(Ag2CrO4)=[Ag+]2[CrO42-]Ksp(Ag2CrO4)=(1.66×10-4)2((1.66×10-42)Ksp(Ag2CrO4)=2.287×10-12mol3L-3

Thus the solubility product is Ksp(Ag2CrO4)=2.287×10-12mol3L-3


flag
Suggest Corrections
thumbs-up
3
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Electrode Potential and emf
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon