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Calculate the voltage of the cell Sn(s) / Sn2+(0.02 M) // Ag+ (0.01 M) / Ag(s) at 25 °C. Given: ESn∘ = - 0.136, EAg∘ = 0.800 V

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Question

Calculate the voltage of the cell Sn(s) / Sn2+(0.02 M) // Ag+ (0.01 M) / Ag(s) at 25 °C.

Given: `"E"_"Sn"^circ` = - 0.136, `"E"_"Ag"^circ` = 0.800 V

Numerical
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Solution

Given: `"E"_"Sn"^circ` = - 0.136, `"E"_"Ag"^circ` = 0.800 V

To find: Voltage of the cell (Ecell)

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: First we write the cell reaction.

\[\ce{Sn_{(s)} -> Sn^2+ (0.02 M) + 2e-}\] (oxidation at anode)
[Ag+ (0.01 M) + e → Ag(s)] × 2 (reduction at cathode)
Overall reaction:
\[\ce{Sn_{(s)} + 2Ag^+ (0.01 M) -> Sn^2+ (0.02 M) + 2Ag_{(s)}}\]

Using formula (1),

`"E"_"cell"^circ = "E"_"Ag"^circ - "E"_"Sn"^circ` = 0.800 V - (- 0.136 V) = 0.936 V

Using formula (2),

The cell potential is given by

`"E"_"cell" = "E"_"cell"^circ - (0.0592  "V")/"n" log_10  (["Sn"^(2+)])/(["Ag"^+]^2)`

∴ `"E"_"cell" = 0.936  "V" - (0.0592  "V")/2 log_10  0.02/(0.01)^2`

`= 0.936  "V" - (0.0592  "V")/2 log_10  200`

`= 0.936  "V" - (0.0592  "V")/2 xx 2.301`

Calculation using log table:

0.0592 × 2.303

= Antilog10 [log10 0.0592 + log10 2.303]

= Antilog10 `[bar(2).7723 + 0.3623]`

= Antilog10 `[bar(1).1346]`

= 0.1363

= 0.936 V – `(0.1363  "V")/2`  (Using log table)

= 0.936 V – 0.0681 V

= 0.8679 V

The voltage of cell is 0.8679 V.

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Chapter 5: Electrochemistry - Short answer questions (Type- II)

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SCERT Maharashtra Chemistry [English] Standard 12 Maharashtra State Board
Chapter 5 Electrochemistry
Short answer questions (Type- II) | Q 6

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