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At 298K the standard free energy of formation of H2O(l) is –237.20kJ/mole while that of its ionisation into H+ iion and hydroxyl ions is 80 kJ/mole, then the emf of the following cell at 298 K will be
H2(g,1 bar) | H+ (1M) || OH–(1M) | O2 (g, 1bar)
Cell reaction
Cathode: H2O(l) + O2(g) + 2e– 2OH–(aq.)
Anode: H2(g) 2H+ (aq.) + 2e–
H2O(l) + O2(g) 2H+ (aq.) + 2OH–(aq.)
Also we have
H2(g)(l) + O2(g) H2O(l)ΔG°f = –237.2 kj/mol
H2O(l) H+ (aq.) + OH–(aq.)ΔG° = 80 kj/mol
Hence for cell reaction
ΔG° = –77.20 kj/mole
So, E = – = 0.40 V
Which of the following cell can produce more electric work.
Ecell = –
For Ecell to be highest [H+]a should be lower and [H+]c should be higher and that why anode compartment should be more basic and cathodic compartment should be acidic.
At what does the following cell have its reaction at equilibrium?
Ag(s) | Ag2CO3(s) | Na2CO3 (aq) || KBr(aq) | AgBr(s) | Ag(s)
KSP = 8 × 10–12 for Ag2CO3 and KSP = 4 × 10–13 for AgBr
anode : Ag(s) Ag+ (aq) + 1e–
Cathode : Ag+ (aq) + 1e– Ag
Net :
0 = 0 + ⇒ =
⇒ ⇒ =
It is observed that the voltage of a galvanic cell using the reaction M(s) + xH+ Mx+ + H2 varies linearly with the log of the square root of the hydrogen pressure and the cube root of the Mx+ concentration. The value of x is
Acetic acid has Ka = 1.8 × 10–5 while formic acid had Ka = 2.1 × 10–4. What would be the magnitude of the emf of the cell
Pt(H2) Pt(H2) at 25°C
Reaction is Hc+
∴ E = = 0.0629 V
Consider the cell Ag(s)|AgBr(s)|Br–(aq)||AgCl(s)|Cl–(aq)|Ag(s) at 25°C. The solubility product constants of AgBr & AgCl are respectively 5 × 10–13 & 1 × 10–10. For what ratio of the concentrations of Br– & Cl–ions would the emf of the cell be zero ?
= log KSPAgBr = – 0.7257
and = log KSPAgCl = – 0.59
Now cell reaction is
Ag + Br– AgBr + 1e–
AgCl + 1e– AgBr + Cl–
Br–AgCl Cl– + AgBr
0 = (0.7257–0.59) +
⇒ = 0.005
Value of for SrCl2 in water at 25°C from the following data:
Conc. (mol/lt) 0.25 1
(Ω–1 cm2 mol–1) 260 250
260 = – 0.5 b…
250 = – b … (2)
On solving & (2), we get
= 270 Ω– cm2 mol–1
Calculate the useful work of the reaction Ag(s) + 1/2Cl2(g) AgCl(s)
Given
If = 1 atmand T = 298 K
AgCl(s) + e Ag(s) + Cl–E° = 22 V
1/2Cl2 + e Cl– E° = 1.36 V
We get
Ag(s) + Cl2(g) AgCl(s)E°cell = 1.14 V
∴ ΔG = –nEF° = – (96500) (1.4) = –110 KJ/mol
Which of these ions Cu+, Co3+, Fe2+ is stable in aqueous medium.
Given :E°Cu2+/Cu+ = 0.15 volt ; E°Cu+/Cu = 0.53 V ;E°Co3+/Co2+ = 1.82 V ;
E°Fe3+/Fe2+ = 0.77 V ;E°Fe2+Fe = –0.44 V ;E°O2,H+/H2O = 1.23 V
if reduction potential of metal ion is greater then O2/H2O couple, the ion is stable in water. So Co3+ is stable in water
Select the correct statement if –
E°Mg2+/Mg = –2.4V, E°Sn4+/Sn2+ = 0.1 V, E°MnO4–,H+/Mn2+ = 1.5 V, E° I2/I–= 0.5 V
Here,
[Hint: Reverse of (2) & (3) is spontaneous; weakest Oxidizing Agent here is Mg2+]
The temperature coefficient of a standard Cd–cell is –5.0 10–5 Vk–1 whose emf at 25°C is 1.018 V. During the cell operation, the temperature will –
∵ ΔH = ΔG + T ΔS
= –nFEcell + nFT
= –199.35 KJ/mol [n = 2 for Cd2+ +2e–→ Cd & putting other values]
NowΔH < O ⇒ Exothermic reaction
⇒ Heat will be released increasing the temperature.
A cell Ag | Ag+ || Cu++ | Cu initially contains 2M Ag+ and 2M Cu++ ions. The charger in cell potential after the passage of 10 amp current for 4825 sec is:
Q = 10 × 4825 = 4825 = 48250 C
no. of mole = = 0.5
Ag + Cu++ Ag+ + Cu
2.00 2.00
2 – 0.25 2 + 0.50
Ecell –E°cell – log
E1 = E°ell – log
E2 = E°ell – log
ΔE = E2 –E1 = [log 1.41 – log 1.88]
= [0.1492 –0.2742] = – × 0.125 = –.00738 V
For the cell (at 298 K)
Ag(s) | AgCl(s) | Cl–(aq) || AgNO3(aq) | Ag(s)
Which of the following is correct –
Conductivity is high due to [H+]
During an electrolysis of conc. H2SO4, perdisulphuric acid (H2S2O8) and O2 from in equimolar amount. The amount of H2 that will form simultaneously will be (2H2SO4 → H2S2O8 + 2H+ + 2e–)
Anode
Cathode {2H2O → H2 + 2OH––2e–} × 3.
Net: 2H2SO4 + 8H2O → H2S2O8 + O2 + 3H2 + 6H+ + 6OH–
Hence ratio of and is 1: 3.
In producing chlorine by electrolysis 100 kW power at 125V is being consumed.How much chlorine per minute is liberated (ECE of chlorine is 0.367)
Mass of the substance deposited at the cathode
=
=0.367
=