# 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. - Chemistry

Sum

Calculate emf of the cell:

Zn(s) |Zn2+ (0.2 M)||H+ (1.6 M)| H2(g, 1.8 atm)| Pt at 25 °C.

#### 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)

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"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.

Concept: Electrode Potential and Cell Potential
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#### APPEARS IN

Balbharati Chemistry 12th Standard HSC for Maharashtra State Board
Chapter 5 Electrochemistry
Exercises | Q 4.05 | Page 119