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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
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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:
- `"E"_"cell"^circ = "E"_"cathode"^circ - "E"_"anode"^circ`
- `"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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