For the cell prepared from electrode A and B , Electrode `A : Cr_(2)O_(7)^(2-) | Cr^(3+) , E_("red")^(@) = 1.33 V` and Electrode `B : Fe^(3+) | Fe^(2+

Question

For the cell prepared from electrode A and B , Electrode `A : Cr_(2)O_(7)^(2-) | Cr^(3+) , E_("red")^(@) = 1.33 V` and Electrode `B : Fe^(3+) | Fe^(2+

For the cell prepared from electrode A and B ,
Electrode `A : Cr_(2)O_(7)^(2-) | Cr^(3+) , E_("red")^(@) = 1.33 V` and
Electrode `B : Fe^(3+) | Fe^(2+) , E_("red")^(@) = 0.77V `. Which of the following statement is correct ?
A. The electrons will flow from B to A when connection are made
B. The emf of the cell will be `-0.56`V
C. A will be negative electrode
D. B will be positive electrode

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - A
`E_(RP_(Cr_(2)O_(7)^(2-)|Cr^(3+))^(@) = 1.33 V ` and `E_(RP_(Fe^(3+)|Fe^(2+))^(@) = 0.77 V E_(RP_(Fe^(3+) |Fe^(2+))^(@) ` is less thus will oxidise or electron will flow Fe(B) electrode to Cr(A) electrode . Also Fe electrons will be negative . Also
`E_(cell)^(@) = E_(RP_(Cr))^(@) - E_(RP_(Fe))^(@)`
`= 1.33 - 0.77 = + 0.56` V