In the cell reaction `Cu(S) + 2 Ag^(+) (aq) to Cu^(2+) (aq) + 2 Ag(s) , E_("cell")^(@) = 0.46` V By doubling the concentration of `Cu^(2+) , E_("cell"

Question

In the cell reaction `Cu(S) + 2 Ag^(+) (aq) to Cu^(2+) (aq) + 2 Ag(s) , E_("cell")^(@) = 0.46` V By doubling the concentration of `Cu^(2+) , E_("cell"

In the cell reaction
`Cu(S) + 2 Ag^(+) (aq) to Cu^(2+) (aq) + 2 Ag(s) , E_("cell")^(@) = 0.46` V
By doubling the concentration of `Cu^(2+) , E_("cell")^(@)` is
A. doubled
B. halved
C. increases but less than double
D. decreases by a small fraction .

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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

Salim Ahmed Kawsar

Correct Answer - C
`E_("cell") = E_("cell")^(@) - (RT)/(nF) "ln" ([Cu^(2+)])/([Ag^(+)]^(2))` .
Doubling `[Cu^(2+)]` decreases the EMF by a small fraction .

4 views Answered 2023-02-05 15:29