The cell shown below generates a potential of 0.643 V at 300K. Find (log K_sp) for AgBr:

$A g | A g B r (s) | N a B r (0.1 M) | | A g N O_{3} (0.1 M) | A g$ $[T a k e \frac{2.303 R T}{F} = 0.06]$

Option 1 -

-30.2

Option 2 -

-15.6

Option 3 -

-12.71

Option 4 -

-8.9

5 Views | Posted a month ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 3

Detailed Solution:

Anode Ag + Br^- -> AgBr + e^-

Cathode Ag⁺ + e^- ® Ag

${E^{o}}_{c e l l} = {({E^{o}}_{R P})}_{C} - {({E^{o}}_{R P})}_{A}$

$\begin{array}{l} E_{c e l l} = - 0.06 l o g K_{s p} - 0.06 l o g 100 \\ 0.643 = - 0.06 l o g K_{s p} - 0.12 \\ l o g K_{s p} = - \frac{0.763}{0.06} = - 12.71 \end{array}$

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The cell shown below generates a potential of 0.643 V at 300K. Find (log K_sp) for AgBr:

$A g | A g B r (s) | N a B r (0.1 M) | | A g N O_{3} (0.1 M) | A g$ $[T a k e \frac{2.303 R T}{F} = 0.06]$

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The cell shown below generates a potential of 0.643 V at 300K. Find (log Ksp) for AgBr: A g | A g B r ( s ) | N a B r ( 0 . 1 M ) | | A g N O 3 ( 0 . 1 M ) | A g [ T a k e 2 . 3 0 3 R T F = 0 . 0 6 ]

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The cell shown below generates a potential of 0.643 V at 300K. Find (log K_sp) for AgBr:

$A g | A g B r (s) | N a B r (0.1 M) | | A g N O_{3} (0.1 M) | A g$ $[T a k e \frac{2.303 R T}{F} = 0.06]$