Electrochemistry Test

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Electrochemistry Test - 9

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Weekly Quiz Competition

Question 1
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The E °in the given diagram is,

A
0.5

B
0.6

C
0.7

D
0.8

Solution

∴  ΔG⁰ = ΔG₁⁰ +ΔG₂⁰ + ΔG₃⁰
-6FE^0. = -4F x 0.54 - 1F x 1.07
Question 2
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What is cell entropy change of the following cell?
Pt(s) | H₂ (g) | CH₃ COOH, HCl || KCl (aq) |Hg₂ Cl₂ | (s) | Hg
P = 1 atm        0.1M               0.1M
Emf of the cell is found to be 0.045 V at 298 K and temperature coefficient is
3.4 x10^–4 VK^–1
Given Ka _(CH3COOH) = 10^–5 M

A
60

B
65.2

C
69.2

D
63.5

Solution

= 65.223J/K/mole
Question 3
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Following cell has EMF 0.7995 V.
Pt | H₂ (1 atm) | HNO₃ (1M) || AgNO₃ (1M) | Ag
If we add enough KCl to the Ag cell so that the final Cl^- is 1M. Now the measured emf of the cell is 0.222 V.
The K_sp of AgCl would be :

A
1 x 10 ^–9.8

B
1x10 ^–19.6

C
2x10^–10

D
2.64x10^–14

Solution
Question 4
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The solubility of [Co(NH₃ )₄ Cl₂ ] CIO₄ _________ if the = 50, = 70, and the measured resistance was 33.5 Ω in a cell with cell constant of 0.20 is ____.

A
59.7 mmol/L

B
49.7 mmol/L

C
39.7 mmol/L

D
29.7 mmol/L

Solution

The correct answer is option B
Given,
λCo(NH3)4Cl2+=50    λClo-4 =70
λ∞= λCo(NH3)4Cl2+ + λClo-4
λ∞= 50       + 70
λ∞=120
(x) Cell constant = 1/A
0.02 = l/A
Resistance(R) =33.5 Ω
K =c.x        (x = is cell constant)

S =49.7 mol/L
Question 5
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We have taken a saturated solution of AgBr.K_sp of AgBr is 12 x 10^–14 . If 10 ^–7 mole of AgNO₃ are added to 1 litre of this solution then the conductivity of this solution in terms of 10 ^–7 Sm ^–1 units will be
[given Sm² mol^-1 Sm² mol^-1 , 5 x 10^-3 Sm² mol^-1 ]

A
39

B
55

C
15

D
41

Solution

The solubility of Agbr in presence of 10^-7 molar AgNo³ is 3 x 10^-7 M. therefore [Br] 3 x 10^-4 M³ , [Ag⁺ ] = 4 x 10^-4 m³ and [No^3- ]= 10^-4 m³
Therefore
Question 6
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At 298K the standard free energy of formation of H₂ O(L) is –237.20kJ/mole while that of its ionisation into H⁺ ion 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)

A
0.40 V

B
0.81 V

C
1.23 V

D
–0.40 V

Solution

Cell reaction
also we have
Hence for cell reaction
Question 7
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Which of the following cell can produce more electric work.

A
pt,H₂ |NH₄ Cl||0.1MCH₃ COOH|H₂ ,pt

B
pt,H₂ |0.1MHCl||0.1MNaOH|H₂ ,pt

C
pt,H₂ |0.1MHCl||0.1MCH₃ COOK|H₂ ,pt

D
pt,H₂ |0.1MCH₃ COOK||0.1MHCl|H₂ ,pt

Solution

For E_cell 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.
Question 8
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A hydrogen electrodes is immersed in a solution with pH = 0 (HCl). By how much will the potential (reduction) change if an equivalent amount of NaOH is added to the solution. (Take P_H2 = 1 atm) T = 298 K.

A
increase by 0.41V

B
increase by 50mV

C
decrease by 0.41V

D
decrease by 59mV

Solution

pH changes from 0 to 7
∴[H⁺ ] changes from 1 to 10^-7 M
Accordingly E_red. Decrease by 0.059 log 10^-7 i.e 0.059 x (-7) = -0.41volt
Question 9
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At what does the following cell have its reaction at equilibrium?
Ag(s) | Ag₂ CO₃ (s) | Na₂ CO₃ (aq) || KBr(aq) | AgBr(s) | Ag(s)
K_SP = 8 x 10 ^–12 for Ag₂ CO₃ and K_SP = 4 x 10 ^–13 for AgBr

A
X10 ^–7

B
X10 ^–7

C

D
X10^–7

Solution
Question 10
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Calculate the EMF of the cell at 298 K
Pt|H₂ (1atm)|NaOH(xM),NaCl(xM)|AgCl(s)|Ag
If E °_cl-/AgCl/Ag = + 0.222 V

A
1.048 V

B
–0.04 V

C
–0.604

D
Emf depends on x and cannot be determined unless value of x is given

Solution