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Electrochemistry

NEET Chemistry · 91 questions · Page 4 of 10 · Click an option or "Show Solution" to reveal answer

Q31
A button cell used in watches function as following. Zn (s) + Ag 2 O (s) + H 2 O (l) ⇌\rightleftharpoons 2Ag (s) + Zn 2+ (aq) + 2OH −- (aq) If half cell potentials are Zn 2+ (aq) + 2e −- → \to Zn (s) ; E o = −-0.76 V Ag 2 O (s) + H 2 O (l) + 2e −- → \to 2Ag (s) + 2OH −- (aq), E o = 0.34 V The cell potential will be
A 0.84 V
B 1.34 V
C 1.10 V
D 0.42 V
Correct Answer
Option C
Solution

At anode :Zn 2+ (aq) + 2e −- →\to Zn (s) ; E o = −-0.76 V At cathode : Ag 2 O (s) + H 2 O (l) + 2e −- →\to 2Ag (s) + 2OH −- (aq), E o = 0.34 V E° cell = E° cathode – E° anode = 0.34 – (–0.76) = 1.10 V

Q32
Molar conductivities (Λmo)\left( {\Lambda _m^o} \right) at infinite dilution of NaCl, Hcl and CH 3 COONa are 126.4, 425.9 and 91.0 S cm 2 mol −-1 respectively. (Λmo)\left( {\Lambda _m^o} \right) for CH 3 COOH will be
A 425.5 S cm 2 mol −-1
B 180.5 S cm 2 mol −-1
C 290.8 S cm 2 mol −-1
D 390.5 S cm 2 mol −-1
Correct Answer
Option D
Solution
Λ\Lambda

o CH 3 COOH =

Λ\Lambda

o CH 3 COONa +

Λ\Lambda

o HCl -

Λ\Lambda

o NaCl = = 91 + 425.9 – 126.4 = 390.5

Q33
The Gibb's energy for the decomposition of Al 2 O 3 at 500 o C is as follows 23{2 \over 3} Al 2 O 3 → \to 43{4 \over 3} Al + O 2 Δ\Delta r G = +960 kJ mol −-1 The potential difference needed for the electrolytic reduction of aluminium oxide (Al 2 O 3 ) at 500 o C is at least
A 4.5 V
B 3.0 V
C 2.5 V
D 5.0 V
Correct Answer
Option C
Solution

We know,

Δ\Delta

G o = – nFE o

23{2 \over 3}

Al 2 O 3 →\to

43{4 \over 3}

Al + O 2 Total number of Al atoms in Al 2 O 3 =

23×2=43{2 \over 3} \times 2 = {4 \over 3}

Al 3+ + 3e – →\to Al As 3e – change occur for each Al atom ∴\therefore n =

43×3=4{4 \over 3} \times 3 = 4

E o = -

ΔG∘nF{{\Delta G^\circ } \over {nF}}

= -

960×10004×96500{{960 \times 1000} \over {4 \times 96500}}

= - 2.5 V

Q34
Limiting molar conductivity of NH 4 OH [\left[ {} \right.i.e. Λm(NH4OH)0\Lambda _{m\left( {N{H_4}OH} \right)}^0]\left. {} \right] is equal to
A Λm(NH4OH)0+Λm(NaCl)0−Λm(NaOH)0\Lambda _{m\left( {N{H_4}OH} \right)}^0 + \Lambda _{m\left( {NaCl} \right)}^0 - \Lambda _{m\left( {NaOH} \right)}^0
B Λm(NaOH)0+Λm(NaCl)0−Λm(NH4Cl)0\Lambda _{m\left( {NaOH} \right)}^0 + \Lambda _{m\left( {NaCl} \right)}^0 - \Lambda _{m\left( {N{H_4}Cl} \right)}^0
C Λm(NH4OH)0+Λm(NH4Cl)0−Λm(HCl)0\Lambda _{m\left( {N{H_4}OH} \right)}^0 + \Lambda _{m\left( {N{H_4}Cl} \right)}^0 - \Lambda _{m\left( {HCl} \right)}^0
D Λm(NH4Cl)0+Λm(NaOH)0−Λm(NaCl)0\Lambda _{m\left( {N{H_4}Cl} \right)}^0 + \Lambda _{m\left( {NaOH} \right)}^0 - \Lambda _{m\left( {NaCl} \right)}^0
Correct Answer
Option D
Solution

According to Kohlrausch’s law, the molar conductivity of NH 4 OH

Λm(NH4OH)0\Lambda _{m\left( {N{H_4}OH} \right)}^0

=

Λm(NH4Cl)0+Λm(NaOH)0−Λm(NaCl)0\Lambda _{m\left( {N{H_4}Cl} \right)}^0 + \Lambda _{m\left( {NaOH} \right)}^0 - \Lambda _{m\left( {NaCl} \right)}^0
Q35
A solution contains Fe 2+ , Fe 3+ and I −- ions. This solution was treated with iodine at 35 o C. E o for Fe 3+ /Fe 2+ is + 0.77 V and E o for I 2 /2I −- = 0.536 V. The favourable redox reaction is
A I 2 will be reduced to I −-
B there will be no redox reaction
C I −- will be oxidised to I 2
D Fe 2+ will be oxidised to Fe 3+
Correct Answer
Option C
Solution

Since the reduction potential of Fe 3+ /Fe 2+ is greater than that of I 2 /I – , Fe 3+ will be reduced and I – will be oxidised.

2Fe 3+ + 2I – →\to 2Fe 2+ + I 2

Q36
Standard electrode potential for Sn 4+ /Sn 2+ couple is + 0.15 V and that for the Cr 3+ /Cr couple is −- 0.74 V. These two couples in their standard state are connected to make a cell. The cell potential will be
A + 1.19 V
B + 0.89 V
C + 0.18 V
D + 1.83 V
Correct Answer
Option B
Solution

E° Sn 4+ /Sn 2+ = 0.15 V E° Cr 3+ /Cr = –0.74 V E° cell = E° cathode – E° anode = 0.15 – (– 0.74) = 0.15 + 0.74 = 0.89 V

Q37
The electrode potentials for Cu 2+ (aq) + e −- → \to Cu + (aq) and Cu + (aq) + e −- → \to Cu (s) are + 0.15 V and + 0.50 V respectively. The value of E o cu 2+ /cu will be
A 0.500 V
B 0.325 V
C 0.650 V
D 0.150 V
Correct Answer
Option B
Solution

Cu 2+ (aq) + e −- →\to Cu + (aq) ; E 1 o = 0.15 V Cu + (aq) + e −- →\to Cu (s) ; E 2 o = 0.50 V Cu 2+ + 2e – →\to Cu ; E o = ?

Δ\Delta

G o =

Δ\Delta

G 1 ° +

Δ\Delta

G 2 ° ⇒\Rightarrow – nFE° = – n 1 FE 1 ° – n 2 FE 2 ° ⇒\Rightarrow E o =

1×0.15+1×0.502{{1 \times 0.15 + 1 \times 0.50} \over 2}

= 0.325 V

Q38
Standard electrode potential of three metals X, Y and Z are −-1.2 V, + 0.5 V and −- 3.0 V respectively. The reducing power of these metals will be
A Y > Z > X
B Y > X > Z
C Z > X > Y
D X > Y > Z
Correct Answer
Option C
Solution

As the electrode potential drops, reducing power increases. So, Z (–3.0 V) > X (–1.2 V) > Y (+ 0.5 V)

Q39
Consider the following relations for emf of an electrochemical cell (i) EMF of cell = (Oxidation potential of anode) −- (Reduction potential of cathode) (ii) EMF of cell = (Oxidation potential of anode) + (Reduction potential of cathode) (iii) EMF of cell = (Reductional potential of anode) + (Reduction potential of cathode) (iv) EMF of cell = (Oxidation potential of anode) −- (Oxidation potential of cathode) Which of the above relations are correct?
A (iii) and (i)
B (i) and (ii)
C (iii) and (iv)
D (ii) and (iv)
Correct Answer
Option D
Solution

EMF of a cell = Reduction potential of cathode – Reduction potential of anode = Reduction potential of cathode + Oxidation potential of anode = Oxidation potential of anode – Oxidation potential of cathode.

Q40
Which of the following expressions correctly represents the equivalent conductance at infinite diluation of Al 2 (SO 4 ) 3 . Given that Λ∘\mathop \Lambda \limits^ \circ Al 3+ and Λ∘\mathop \Lambda \limits^ \circ so42−_4^{2 - } are the equivalent conductances at infinite dilution of the respective ions?
A 2Λ∘2\mathop \Lambda \limits^ \circ Al 3+ + 3Λ∘3\mathop \Lambda \limits^ \circ so42−_4^{2 - }
B Λ∘\mathop \Lambda \limits^ \circ Al 3+ + Λ∘\mathop \Lambda \limits^ \circ so42−_4^{2 - }
C (Λ∘\mathop \Lambda \limits^ \circ Al 3+ + Λ∘\mathop \Lambda \limits^ \circ so42−_4^{2 - } ) × \times 6
D 13{1 \over 3}Λ∘\mathop \Lambda \limits^ \circ Al 3+ + 12{1 \over 2}Λ∘\mathop \Lambda \limits^ \circ so42−_4^{2 - }
Correct Answer
Option B
Solution

Equivalent conductance of an electrolyte at infinite dilution is given by the sum of equivalent conductances of the respective ions at infinite dilution.

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