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.

Answer 1

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.