EMF of the following cell is 0.6 volt. Ag(s)|A g B r (s)|KBr(0.01 M)|.|.AgNO3(0.001 M)|.Ag(.s.) Ks p of AgBr is expressed as 1× 10- x, x is [T a k e (2.303 R T/F) = 0.06 V]

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Tardigrade Admin · Accepted Answer

Ag arrow Ag+(L)+ e- Ag+(0.001 M)+e- arrow Ag Ag+(.0.001 M.) arrow Ag+(.L.) 0.6=0-(0.06/1)log(Ks p/10- 2 × 10- 3) 10- 10=(Ks p/10- 5)⇒ Ks p=10- 15

Q. EMF of the following cell is $0.6$ volt.

$A g (s) ∣ A g B r (s) ∣ K B r (0.01 M) ∣ ∣ A g N O_{3} (0.001 M) ∣ A g (s)$

$K_{s p}$ of $A g B r$ is expressed as $1 \times 1 0^{- x}, x$ is

$[T ak e \frac{2.303 RT}{F} = 0.06 V]$