JEE Mains 30-Jan-2023 Shift 1 Solved Paper
© examsiri.com
Question : 51 of 90
Marks:
+1,
-0
Consider the cell
Pt ( s ) | H 2 ( g , 1 atm ) | H + ( aq , 1 M )
Fe 3 + ( aq ) , Fe 2 + ( aq ) ∣ Pt ( s )
When the potential of the cell is0.712 V 298 K [ Fe 2 + ] ∕ [ Fe 3 + ]
(Nearest integer)
Given:Fe 3 + + e − = Fe 2 + , E ∘ Fe 3 + , Fe 2 + ∣ Pt = 0.771
= 0.06 V
When the potential of the cell is
(Nearest integer)
Given:
[30-Jan-2023 Shift 1]
- Your Answer:
Go to Question: