Advertisements
Advertisements
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
Advertisements
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
Choose the most correct option.
The standard potential of the cell in which the following reaction occurs:
H2 (g,1 atm) + Cu2+ (1M) → 2H+ (1M) + Cu(s),
`("E"_"Cu"^circ = 0.34 "V")` is
Answer the following in one or two sentences.
Write Nernst equation. What part of it represents the correction factor for nonstandard state conditions?
Answer the following in one or two sentences.
Under what conditions the cell potential is called standard cell potential?
Answer the following in brief.
Construct a galvanic cell from the electrodes Co3+ | Co and Mn2+/Mn. \[\ce{E{^{\circ}_{Co}}}\] = 1.82 V, \[\ce{E{^{\circ}_{Mn}}}\] = –1.18 V. Calculate \[\ce{E{^{\circ}_{cell}}}\]
Calculate emf of the cell at 25°C.
Zn(s) | Zn2+ (0.08 M) || Cr3+ (0.1 M) | Cr(s)
`E_(Zn)^0` = −0.76 V, `E_(Cr)^0` = −0.74 V
What is cell voltage?
The correct representation of Nernst's equation for half-cell reaction \[\ce{Cu^{2+} (aq) + e^- -> Cu^+(aq)}\] is ______.
Calculate E.M.F. of following cell at 298 K Zn(s) |ZnSO4 (0.01 M)| |CuSO4 (1.0 M)| Cu(s) if \[\ce{E^0_{cell}}\] = 2.0 V.
Identify the strongest reducing agent from the data given below:
| Element | `"E"^0 ("V")` |
| Al | −1.66 |
| Fe | −0.44 |
| Hg | +0.79 |
| Cu | +0.337 |
What is the standard emf of the following cell?
\[\ce{Ni_{(s)} | Ni^{2+}_{( aq)} || Au^{3+}_{( aq)} | Au_{(s)}}\]
if \[\ce{E^0_{Ni}}\] = −0.25 V, \[\ce{E^0_{Au}}\] = 1.50 V.
What is the standard potential of cell, Ni | Ni2+ (1M) || Cu2+ (1 M) | Cu?
If E°Cu = 0.337 V and E°Ni = - 0.236 V.
The standard electrode potential of Zn and Ni are - 0.76 V and - 0.25 V respectively. If the reaction takes place in the cell constructed between these two electrodes is spontaneous. What is the standard emf of the cell?
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?
Construct a cell from Ni2+ | Ni and Cu2+ | Cu Cu half cells. Write the cell reaction and calculate `E_("cell")^0`
`(E_("Ni")^0 = - 0.236 V and E_("Cu")^0 = + 0.337 V)`
Define standard electrode potential.
Write net cell reaction.
Write the value of `(2.303 RT)/F` in the Nernst equation?
Write the four applications of emf series.
Standard potential (E°) of
\[\mathrm{Zn_{(aq)}^{+2}+2e^{-}\longrightarrow Zn_{(s)}~is~-0.76~V}\]
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?
