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Question
Answer the following:
Calculate emf of the cell:
Zn(s) |Zn2+ (0.2 M)||H+ (1.6 M)| H2(g, 1.8 atm)| Pt at 25 °C.
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Solution
Given: [Zn2+] = 0.2 M, [H+] = 1.6 M, `"P"_("H"_2)` = 1.8 atm
To find: Emf of the cell `("E"_"cell")`
Formulae:
1) `"E"_"cell"^circ = "E"_"cathode"^circ - "E"_"anode"^circ`
2) `"E"_"cell" = "E"_"cell"^circ - (0.0592 "V")/"n" log_10 [["Product"]]/[["Reactant"]]`
Calculation:
`"Zn"_(("s")) -> "Zn"_((0.2 "M"))^(2+) + 2"e"^(-)` (oxidation at anode)
`2"H"_((1.6 "M"))^+ + 2"e"^(-) -> "H"_(2(1.8 "atm"))` (reduction at cathode)
___________________________________________________
`"Zn"_(("s")) + 2"H"_((1.6 "M"))^+ -> "Zn"_((0.2 "M"))^(2+) + "H"_(2(1.8 "atm"))` (overall reaction)
`"E"_("H"_2)^circ = 0.0 "V" and "E"_("Zn")^circ` = - 0.763 V
Using formula (i),
`"E"_"cell"^circ = "E"_"cathode"^circ - "E"_"anode"^circ`
`"E"_"cell" = "E"_("H"_2)^circ - "E"_("Zn")^circ`
= 0.0 V - (- 0.763 V) = 0.769 V
Using formula (ii),
The cell potential is given by
`"E"_"cell" = "E"_"cell"^circ - (0.0592 "V")/2 log_10 [["Product"]]/[["Reactant"]]`
= 0.763 - `(0.0592 "V")/2 log_10 ((0.2)(1.8))/(1.6)^2`
= 0.763 + 0.0252 = 0.7882 V
The emf of the cell is 0.7882 V.
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