NEET Chemistry Questions: Redox Reactions & Electrochemisty

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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) + 12O2(g) + 2e 2OH(aq.)

Anode: H2(g) 2H+ (aq.) + 2e

H2O(l) + 12O2(g) 2H+ (aq.) + 2OH(aq.)

Also we have

H2(g)(l) + 12O2(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 = – 772002×96500 = 0.40 V

Which of the following cell can produce more electric work.





Ecell = – 0.05911log[H+]a[H+]c

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 [Br-][CO32] 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 : Ag(AgBr)+1eAg(Ag2Co3)+

0 = 0 + 0.0591logKSPAgBr[Br]KSPAg2Co3[CO32]KSPAgBr[Br] = KSPAg2Co3[CO32]

4×10138×1012=[Br][CO32][Br][CO32] = 2×107

It is observed that the voltage of a galvanic cell using the reaction M(s) + xH+ Mx+ + x2 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





The linear dependence of voltage on the log of the square root of hydrogen pressure and the cube root of M^(x+) concentration suggests that the balanced reaction involves the transfer of 3 electrons and the formation of 1.5 H2 molecules, which implies x = 3.

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) 0.1M acetic acid+0.1M sodium acetate0.1M formic acid+0.1M sodium formate Pt(H2) at 25°C





Reaction is Hc+ 1e-HA+

∴ E = 0.0591log2.1×1041.8×105 = 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& Clions would the emf of the cell be zero ?





EBr-/AgBr/Ag0 = EAg+/Ag/Ag0+0.0591 log KSPAgBr = EAg+/Ag0 – 0.7257

and ECr-/AgCl/Ag0 = EAg+/Ag0+0.0591 log KSPAgCl = EAg+/Ag0 – 0.59

Now cell reaction is

Ag + Br AgBr + 1e

AgCl + 1e AgBr + Cl

BrAgCl 1e- Cl + AgBr 

0 = (0.7257–0.59) + 0.0591log[Br][Cl] 

[Br][Cl] = 0.005

Value of Λm for SrCl2 in water at 25°C from the following data:

Conc. (mol/lt)            0.25         1

Λm–1 cm2 mol–1)    260        250





Λm=Λm b c

260 = Λm – 0.5 b…

250 = Λm – b … (2)

On solving & (2), we get

Λm = 270 Ω cm2 mol–1

Calculate the useful work of the reaction Ag(s) + 1/2Cl2(g) AgCl(s)

Given E°Cl2/Cl = + 1.36 V, E°AgCl/Ag,Cl = 0.22 V

If PCl2 = 1 atmand T = 298 K





AgCl(s) + e Ag(s) + ClE° = 22 V

1/2Cl2 + e ClE° = 1.36 V

We get

Ag(s) + 12Cl2(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 ;

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 –

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 ETp

= –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 = 4825096500 = 0.5

Ag + 12Cu++ Ag+ + 12Cu

        2.00           2.00

      2 – 0.25        2 + 0.50

Ecell –E°cell0.05911 log [Ag+][Cu++]1/2

E1 = E°ell0.05911 log 2.00(2.00)1/2

E2 = E°ell0.05911log 2.50(1.75)1/2

ΔE = E2 –E1 = 0.05911 log2log2.501.75=0.05911[log 1.41 – log 1.88]

= 0.05911 [0.1492 –0.2742] = – 0.05911 × 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 2H2SO4H2S2O8+ 2H++ 2e-2H2OO2+ 4H++ 4e-

Cathode {2H2O → H2 + 2OH–2e} × 3.

Net: 2H2SO4 + 8H2O → H2S2O8 + O2 + 3H2 + 6H+ + 6OH

Hence ratio of nO2 and nH2 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 0.367×10-6kgC-1)





Mass of the substance deposited at the cathode 

              m=zit

                 =zWVt

                =0.367×10-6×100×103125×60

                =17.6×10-3kg