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प्रश्न
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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उत्तर
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.
संबंधित प्रश्न
Consider the half reactions with standard potentials.
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- \[\ce{I2_{(s)} + 2e^\ominus -> 2I^{\ominus}_{(aq)} E^\circ = 0.53V}\]
- \[\ce{Pb^{2\oplus}_{(aq)} + 2e^{\ominus} -> Pb_{(s)} E^\circ = -0.13 V}\]
- \[\ce{Fe^{2\oplus} + 2e^{\ominus} -> Fe_{(s)} E^\circ = -0.44 V}\]
The strongest oxidising and reducing agents respectively are ______.
Answer the following in one or two sentences.
What is standard cell potential for the reaction
\[\ce{3Ni_{(s)} + 2Al^{3+} (1M) → 3Ni^{2+} (1M) + 2Al(s)}\], if `E_"Ni"^circ` = –0.25 V and `"E"_("Al")^circ` = –1.66 V?
Answer the following in one or two sentences.
Under what conditions the cell potential is called standard cell potential?
Answer the following:
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| Element | `"E"^0 ("V")` |
| Al | −1.66 |
| Fe | −0.44 |
| Hg | +0.79 |
| Cu | +0.337 |
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\[\ce{Ni_{(s)} | Ni^{2+}_{( aq)} || Au^{3+}_{( aq)} | Au_{(s)}}\]
if \[\ce{E^0_{Ni}}\] = −0.25 V, \[\ce{E^0_{Au}}\] = 1.50 V.
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Answer the following in one or two sentences.
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\[\ce{2Al(s) + 3Ni^{2⊕}(1M) -> 2Al^{3⊕}(1 M) + 3Ni(s)}\]
if \[\ce{E{^{\circ}_{Ni}}}\] = −0.25 V and \[\ce{E{^{\circ}_{Al}}}\] = −1.66 V?
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\[\ce{Zn + Cu^{2+} -> Zn^{2+} + Cu}\],
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Construct a cell from Ni2+ | Ni and Cu2+ | Cu Cu half cells. Write the cell reaction and calculate `E_("cell")^0`
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Calculate the emf of the following cell at 25°C.
Zn(s)|Zn2+(0.08 M) || Cu2+(0.l M) |Cu(s)
E0zn = − 0.76 V, E0cu = 0.36 V.
Write the value of `(2.303 RT)/F` in the Nernst equation?
Write the four applications of emf series.
Standard potential (E°) of
\[\mathrm{Zn_{(aq)}^{+2}+2e^{-}\longrightarrow Zn_{(s)}~is~-0.76~V}\]
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What is the reduction potential of hydrogen gas electrode when pure hydrogen gas is at 1 atmosphere pressure and platinum electrode is in contact with HCI of pH 2 at 298 K?
