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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.
- \[\ce{Ag^{\oplus}_{ (aq)} + e^{\ominus} -> Ag_{(s)}E^\circ = 0.8 V}\]
- \[\ce{I2_{(s)} + 2e^\ominus -> 2I^{\ominus}_{(aq)} E^\circ = 0.53V}\]
- \[\ce{Pb^{2\oplus}_{(aq)} + 2e^{\ominus} -> Pb_{(s)} E^\circ = -0.13 V}\]
- \[\ce{Fe^{2\oplus} + 2e^{\ominus} -> Fe_{(s)} E^\circ = -0.44 V}\]
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:
Calculate emf of the cell:
Zn(s) |Zn2+ (0.2 M)||H+ (1.6 M)| H2(g, 1.8 atm)| Pt at 25 °C.
Calculate emf of the cell at 25°C.
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`E_(Zn)^0` = −0.76 V, `E_(Cr)^0` = −0.74 V
Answer the following:
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Calculate `"E"_"cell"^circ` of the following galvanic cell:
Mg(s) / Mg2+(1 M) // Ag+ (1 M) / Ag(s) if `"E"_"Mg"^circ` = – 2.37 V and `"E"_"Ag"^circ` = 0.8 V. Write cell reactions involved in the above cell. Also mention if cell reaction is spontaneous or not.
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\[\ce{Mg_{(s)} | Mg^{2+}_{( aq)} || Ag^+_{( aq)} | Ag_{(s)}}\]
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Given: `"E"_("Cr"^(3+)//"Cr")^0` = −0.74 V,
`"E"_("Fe"^(2+)//"Fe")^0` = −0.44 V
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(E0 = −0.257 V)
What is the ΔG0 for the following reaction?
\[\ce{Al_{(s)} + Fe^{3+}_{( aq)} -> Al^{3+}_{( aq)} + Fe_{(s)}}\]; \[\ce{E^0_{cell}}\] = +2.43
The standard potential of the cell in the following reaction is ______.
\[\ce{Cd_{(s)} + Cu^{2+}_{(1M)} -> Cd^{2+}_{(1M)} + Cu_{(s)}}\]
`("E"_("Cd")^circ = - 0.403V, "E"_("Cu")^circ = 0.334V)`
Answer the following in one or two sentences.
What is the standard cell potential for the reaction?
\[\ce{2Al(s) + 3Ni^{2⊕}(1M) -> 2Al^{3⊕}(1 M) + 3Ni(s)}\]
if \[\ce{E{^{\circ}_{Ni}}}\] = −0.25 V and \[\ce{E{^{\circ}_{Al}}}\] = −1.66 V?
Which element from the following has the highest negative standard reduction potential?
The standard potential of the electrode Zn2+(0.02M) |Zn(s) is − 0.76 V. Calculate the electrode potential of the zinc electrode.
Calculate the emf of the following cell at 25°C.
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E0zn = − 0.76 V, E0cu = 0.36 V.
Write the four applications of emf series.
Standard potential (E°) of
\[\mathrm{Zn_{(aq)}^{+2}+2e^{-}\longrightarrow Zn_{(s)}~is~-0.76~V}\]
What is standard potential of reaction
\[2\mathrm{Zn}_{(\mathrm{s})}\longrightarrow2\mathrm{Zn}_{(\mathrm{aq})}^{+2}+4\mathrm{e}^{-}\] ?
If \[\mathrm{E^o\left(Cd_{(aq)}^{+2}|Cd_{(s)}\right)=-0.40~V}\]. What is potential for \[\mathrm{Cd}_{(s)}\xrightarrow{}\mathrm{Cd}_{(aq)}^{+2}\left(0.01\mathrm{M}\right)+2\mathrm{e}^{-}\] at 298 K?
Which from following formulae is used to obtain value of E°cell for a reaction taking place in Dry cell?
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?
The cell potential for the following cell notation is approximately:
M(s) | M3+ (aq, 0.01 M) || N2+ (aq, 0.1 M) | N(s)
`E_(M^(3+)//M)^0` = 0.6 V and `E_(N^(2+)//N)^0` = 0.1 V
