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Vista previa 2 fuera de 8 páginas
Examen

electrochemistry

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Vista previa 2 fuera de 8 páginas

Exam of 8 pages for the course Jee advanced at Jee advanced (Questions)

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SINGLE CHOICE CORRECT QUESTIONS

1. The emf of the cell, Ni | Ni2+ (1.0 M) || Ag+ (1.0M) | Ag [E° for Ni2+ / Ni = – 0.25 volt, E° for Ag+/Ag = 0.80
volt] is given by -
(A) – 0.25 + 0.80 = 0.55 volt (B) – 0.25 – (+ 0.80) = –1.05 volt
(C) 0 + 0.80 – (– 0.25) = + 1.05 volt (D) – 0.80 – (– 0.25) = – 0.55 volt

2. Zn | Zn2+ (C1)|| Zn2+ (C2)|Zn. for this cell G is negative if -
(A) C1 = C2 (B) C1 > C2 (C) C2 > C1 (D) None

3. Pt | (H2) | pH = 1 || pH = 2 | (H2)Pt
1 atm 1 atm
The cell reaction for the given cell is :
(A) spontaneous (B) non - spontaneous (C) equilibrium (D) none of these


H2 (Pt ) H3 O (aq) Ag
4. Consider the cell Ag. The measured EMF of the cell is 1.0 V. What is the value of x?
1 atm pH 5.03 x M

E0Ag ,Ag = + 0.8 V. [T = 25°C]
(A) 2 × 10–2 M (B) 2 × 10–3 M (C) 1.5 × 10–3 M (D) 1.5 × 10–2 M

1
5. The standard potential of the reaction H2O + e– H + OH– at 298 K by using Kw (H2O) = 10–14, is :
2 2
(A) – 0.828 V (B) 0.828 V (C) 0 V (D) – 0.5 V

6. Given : Hg22+ + 2 e 2 Hg , E0 = 0.789 V & Hg2+ + 2 e Hg , E0 = 0.854 V, calculate the equilibrium
constant for Hg22+ Hg + Hg . 2+

(A) 3.13 × 10 3 (B) 3.13 × 10 4 (C) 6.26 × 10 3 (D) 6.26 × 10 4

7. What must be the concentration of Ag+ in an aqueous solution containing Cu2+ = 1.0 M so that both the
0
metals can be deposited on the cathode simultaneously. Given that E Cu Cu2 = – 0.34 V and

E 0Ag Ag = 0.812 V, T = 298 K
(A) nearly 10–19 M (B) 10–12 M (C) 10–8 M (D) nearly 10–16 M

0
8. If E Fe 2 / Fe = – 0.441V and E 0Fe 3 / Fe 2 = 0.771V the standard EMF of the reaction Fe + 2Fe +3 3Fee+2 will be:
(A) 0.330V (B) 1.653V (C*) 1.212V (D) 0.111V

9. For the cell (at 298 K)
Ag(s) | AgCl(s) | Cl– (aq) || AgNO3 (aq) | Ag(s)
Which of following is correct ?
(A) The cell emf will be zero when [Ag+]a = [Ag+]c ([Ag+] in anodic compartment = [Ag+] in cathodic
compartment)
(B) The amount of AgCl(s) precipitate in anodic compartment will decrease with the working of the cell.
(C) The concentration of [Ag+] = constant, in anodic compartment during working of cell.

0.059 1
(D) Ecell = E 0Ag | Ag
E 0Cl | AgCl | Ag
log
1 [Cl ]a

, 10. During the electrolysis of 0.1 M CuSO4 solution using copper electrodes, a depletion of [Cu++] occurs near
the cathode with a corresponding excess near the anode, owing to inefficient stirring of the solution. If the local
concentrations of [Cu++] near the anode & cathode are respectively 0.12 M & 0.08 M, calculate the back
e.m.f. developed . Temperature = 298 K .
(A) 22 mV (B) 5.2 mV (C) 29 mV (D) 59 mV


[Cu ]
11. The equilibrium Cu++ (aq) + Cu(s) 2Cu+ established at 20°C corresponds to 2 = 2.02 × 104+ . The
[Cu ]

standard potential, E0Cu ,Cu
= 0.33 volt at this temperature. What is the standard potential, E 0 ?
Cu / Cu
(A) – 0.457 V (B) – 0.125 V (C) – 0.66 V (D) – 0.250 V

12. Given : Eº(Cu2+ | Cu) = 0.337 V and Eº (Sn2+ | Sn) = – 0.136 V. Which of the following statements is correct?
(A) Cu2+ ions can be reduced by H2(g) (B) Cu can be oxidized by H+
2+
(C) Sn ions can be reduced by H2(g) (D) Cu can reduce Sn2+

13. The standard reduction potentials E° of the following systems are :
System E° (volts)
(i) MnO4¯ + 8H+ + 5e– Mn2+ + 4H2O 1.51
(ii) Sn4+ + 2e– Sn2+ 0.15
(iii) Cr2 O7 + 14H + 6e–
2– +
2Cr3+ + 7H2O 1.33
(iv) Ce4+ + e– Ce3+ 1.61
The oxidising power of the various species decreases in the order :
(A) Ce4+ > Cr2 O72– > Sn4+ > MnO4¯ (B) Ce4+ > MnO4¯ > Cr2 O72– > Sn4+
2– 4+ 4+
(C) Cr2O7 > Sn > Ce > MnO4¯ (D) MnO4¯ > Ce4+ > Sn4+ > Cr2 O72–

14. Consider the reaction : (T = 298 K)
Cl2(g) + 2Br¯(aq) 2Cl¯ (aq) + Br2 (aq.)
The emf of the cell, when [Cl¯] = [Br2] = [Br¯] = 0.01M and Cl2 gas is at 1 atm pressure, will be :
(E° for the above reaction is = 0.29 volt)
(A) 0.54 volt (B) 0.35 volt (C) 0.24 volt (D) –0.29 volt


15. 2Ce4+ + Co 2Ce3+ + Co2+ Eºcell = 1.89 V, Eº Co2 / Co = – 0.277 V hence, Eº Ce4 / Ce3 is :
(A) 0.805 V (B) 1.62 V (C) – 0.805 V (D) – 1.61 V


16. MnO4– + 8H+ + 5e– Mn2+ + 4H2O ; Eº = 1.51 V; G10 = – 5 × 1.51 × F

MnO2 + 4H+ + 2e– Mn2+ + 2H2O ; Eº = 1.23 V; G02 = – 2 × 1.23 × F
EºMnO is :
4 | MnO 2

(A) 1.70 V (B) 0.91 V (C) 1.37 V (D) 0.548 V

17. Two weak acid solutions HA1 and HA2 each with the same concentration and having pKa values 3 and 5 are
placed in contact with hydrogen electrode (1 atm, 25ºC) and are interconnected through a salt bridge. The emf
of the cell is :




(A) 0.21 V (B) 0.059 V (C) 0.018 V (D) 0.021 V

Información del documento

Subido en
11 de septiembre de 2026
Número de páginas
8
Escrito en
2026/2027
Tipo
Examen
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