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Answer the following: Calculate emf of the cell at 25 °C. Zn(s) |Zn2+ (0.08 M)| |Cr3+ (0.1 M)| |Cr3+ (0.1 M)| Cr(s) EZn∘ = - 0.76 V, ECr∘ = - 0.74 V - Chemistry

Sum

Answer the following:

Calculate emf of the cell at 25 °C.

Zn(s) |Zn2+ (0.08 M)| |Cr3+ (0.1 M)| |Cr3+ (0.1 M)| Cr(s)

`"E"_"Zn"^circ` = - 0.76 V, `"E"_"Cr"^circ` = - 0.74 V

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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 + 0.0127 = 0.0327 V

The emf of the cell is 0.0327 V.

  Is there an error in this question or solution?
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APPEARS IN

Balbharati Chemistry 12th Standard HSC for Maharashtra State Board
Chapter 5 Electrochemistry
Exercise | Q 4.06 | Page 119
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