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Question
Calculate emf of the cell at 25°C.
Zn(s) | Zn2+ (0.08 M) || Cr3+ (0.1 M) | Cr(s)
E°Zn = −076V, E°Cr = −0.74V.
Solution
Given: `"E"_"Zn"^circ` = - 0.76 V, `"E"_"Cr"^circ` = - 0.74 V
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"_((0.08 "M"))^(2+) + 2"e"^(-)] xx 3` (oxidation at anode)
`["Cr"_((0.1 "M"))^(3+) + 3"e"^(-) -> "Cr"_(("s"))] xx 2` (reduction at cathode)
___________________________________________________
`3"Zn"_(("s")) + 2"Cr"_((0.1 "M"))^(3+) -> 3"Zn"_((0.08 "M"))^(2+) + "Cr"_(("s"))` (overall reaction)
Using formula (i),
`"E"_"cell"^circ = "E"_"cathode"^circ - "E"_"anode"^circ`
`"E"_"cell" = "E"_("Cr")^circ - "E"_("Zn")^circ`
= - 0.74 V - (- 0.76 V) = 0.02 V
Using formula (ii),
The cell potential is given by
`"E"_"cell" = "E"_"cell"^circ - (0.0592 "V")/"n" log_10 [["Product"]]/[["Reactant"]]`
= 0.02 - `(0.0592 "V")/6 log_10 (0.08)^3/(0.1)^2`
`= 0.02 - 9.867 × 10^-3 log_10 (5.12 × 10^-4)/(1 × 10^-2)`
`= 0.02 - 9.867 × 10^-3 × bar 2. 7093`
= 0.02 - 9.867 × 10-3 × (- 1.2907)
= 0.02 + 0.01273
= 0.03273 V
The emf of the cell is 0.03273 V.
Notes
Students can refer to the provided solutions based on their preferred marks.
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