Advertisements
Advertisements
प्रश्न
Calculate emf of the following cell at 25°C:
\[\ce{Sn/Sn^2+ (0.001 M) || H+ (0.01 M) | H2_{(g)} (1 bar) | Pt_{(s)}}\]
Given: \[\ce{E^\circ(Sn^2+/sn) = -0.14 V, E^\circ H+/H2 = 0.00 V (log 10 = 1)}\]
Advertisements
उत्तर
-
Anode: Sn → Sn2+ + 2e− (oxidation)
-
Cathode: 2H+ + 2e− → H₂ (reduction)
\[\ce{E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.00V - (-0.14 V) = 0.14 V}\]
`E_(cell) = E_(cell)^\circ - 0.0591/n log Q`
Where:
-
n = 2 (electrons exchanged)
-
Q = reaction quotient
\[\ce{Sn(s) + 2H+ -> Sn^2+ + H 2 (g)}\]
`Q = ([Sn^(2+)][H2])/[H^+]^2 = (0.001xx1)/(0.01)^2 = 0.001/0.0001 = 10`
`E_(cell) = 0.14 - 0.0591/2 log(10) = 0.14 - 0.0591/2 xx 1 = 0.14 - 0.02955`
= 0.11045 V
Ecell = 0.110 V
APPEARS IN
संबंधित प्रश्न
Draw a neat and well labelled diagram of primary reference electrode.
Arrange the following reducing agents in the order of increasing strength under standard state conditions. Justify the answer
|
Element |
Al(s) |
Cu(s) |
Cl(aq) |
Ni(s) |
|
Eo |
-1.66V |
0.34V |
1.36V |
-0.26V |
Calculate e.m.f of the following cell at 298 K:
2Cr(s) + 3Fe2+ (0.1M) → 2Cr3+ (0.01M) + 3 Fe(s)
Given: E°(Cr3+ | Cr) = – 0.74 VE° (Fe2+ | Fe) = – 0.44 V
Calculate e.m.f. and ∆G for the following cell:
Mg (s) |Mg2+ (0.001M) || Cu2+ (0.0001M) | Cu (s)
`"Given :" E_((Mg^(2+)"/"Mg))^0=−2.37 V, E_((Cu^(2+)"/"Cu))^0=+0.34 V.`
Calculate the emf of the following cell at 25°C :
Standard electrode potential is measured taking the concentrations of all the species involved in a half-cell is ____________.
The difference between the electrode potentials of two electrodes when no current is drawn through the cell is called ______.
What does the negative sign in the expression `"E"^Θ ("Zn"^(2+))//("Zn")` = − 0.76 V mean?
Consider the figure and answer the following question.
If cell ‘A’ has ECell = 0.5V and cell ‘B’ has ECell = 1.1V then what will be the reactions at anode and cathode?
Which is the correct order of second ionization potential of C, N, O and F in the following?
