Class 12th

3.6 The cell in which the following reaction occurs: ( ) ( ) ( ) ( ) 3 2 2Fe 2I 2Fe I aq aqaq 2 s + − + + → + has 0 Ecell = 0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction

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Contributor-Level 10

Given:
2Fe³⁺ (aq) + 2I^- (aq) → 2Fe²⁺ (aq) + I₂ (s)
E^0cell = 0.236V
n = moles of e^- from balanced redox reaction = 2
F = Faraday's constant = 96,485 C/mol
T = 298 K.
Using the formula, we get
? _rG⁰ = – nFE⁰_cell
⇒? _rG⁰ = – 2 * FE⁰_cell
⇒? _rG⁰ = −2 * 96485 C mol^-1 * 0.236 V
⇒? _rG⁰ = −45540 J mol^-1
⇒? _rG⁰ = −45.54 kJ mol^-1
Now,
? _rG⁰ = −2.303RT log K_c
Where, K is the equilibrium constant of the reaction

R is the gas constant; R = 8.314 J-mol-C^-1
⇒ −45540 J mol^-1 = –2.303* (8.314 J-mol-C^-1)* (298 K) * (log K_c)
Solving for K_c we get,
⇒ logK_c = 7.98
Taking antilog both side, we get
⇒ K_c = Antilog (7.98)
⇒ K_c = 9.6 * 1

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8 months ago

3.5 Calculate the emf of the cell in which the following reaction takes place: Ni(s) + 2Ag+ (0.002 M) → Ni2+ (0.160 M) + 2Ag(s) Given that E(cell) V = 1.05 V

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Contributor-Level 10

Given:
[Ag⁺] = 0.002 M
[Ni²⁺] = 0.160 M
n = 2
(n = moles of e- from balanced redox reaction)
E⁰cell= 1.05 V
Now, using the Nernst equation, we get,

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8 months ago

3.4 Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

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Contributor-Level 10

Given:
For hydrogen electrode, pH = 10
n = 1
(n = moles of e- from balanced redox reaction)
On using the formula [H+] = 10– pH

⇒ [H+] = 10 − 10 M
We know,

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8 months ago

3.3 Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.

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Contributor-Level 10

For a substance to oxidise Fe²⁺to Fe³⁺ ion, it must have high reduction potential than Fe³⁺. The reduction potential of Fe³⁺ to Fe²⁺ reaction is 0.77V, the substances which have reduction potentials higher than this value will oxidise Fe²⁺ ions. Comparing the values, from the table:

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8 months ago

3.2 Can you store copper sulphate solutions in a zinc pot?

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Contributor-Level 10

NO, because Zn is very reactive with Cu. It reacts with copper sulphate to form zinc sulphate i.e., Zn displaces Cu and metallic Cu is also formed.
The reaction is given as:
Zn + CuSO₄ ⇒ ZnSO₄ + Cu

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8 months ago

3.1 How would you determine the standard electrode potential of the system Mg²⁺|Mg?

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Contributor-Level 10

To determine the standard electrode potential of the system Mg²⁺|Mg, connect it to the standard hydrogen electrode (SHE). Keep the Mg²⁺|Mg system as cathode and SHE as cathode. This is represented as shown below.
Pt (s) | H₂ (g, 1 bar)| H⁺ (aq, 1 M) |Mg²⁺ (aq, 1M)| Mg
The electrode potential of a cell is given by
E^? = E^?R – E^?L
Where,
E^?R- Potential of the half-cell in the right side of the above representation
E^?L- Potential of the half-cell in the left side of the above representation
It is to be noted that the potential of the standard hydrogen electrode is zero.
Therefore, E^?L = 0
E^? = E^?R – 0
⇒ E^? = E^?R

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8 months ago

4.3 For a reaction, A + B → Product; the rate law is given by, r = k [ A]1/2 [B]2 . What is the order of the reaction?

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Payal Gupta

Contributor-Level 10

4.3 The order of the reaction is sum of the powers of concentration of reactants in the rate law. According to this,

The order of the reaction = 1/2 + 2

= 2 1/2 or 2.5

The order of the reaction is 2.5

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8 months ago

4.2 In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L–1 to 0.4 mol L–1 in 10 minutes. Calculate the rate during this interval?

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Payal Gupta

Contributor-Level 10

4.2 Given-

Initial concentration (A₁) = 0.5M

Final concentration (A₂) = 0.4M

Time = 10 mins.

The formula for average rate of the reaction is,

r_av = -1/2 X Δ{A} / Δt → Equation 1

? {A} = (A₂)-( A₁), the equation 1 is written as,

r_av = -1/2 X A₂- A₁ / Δt

= -1/2 X 0.4-0.5 / 10

= -1/2 X 0.1 / 10

= 0.005 mol L^-1 min^-1

= 5 * 10^-3 mol L^-1 min^-1

The average rate of the reaction is 5 * 10^-3 mol L^-1 min^-1

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8 months ago

4.1 For the reaction R → P, the concentration of a reactant changes from 0.03M to 0.02M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

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Payal Gupta

Contributor-Level 10

4.1 Given-

Initial concentration (R₁) = 0.03M

Final concentration (R₂) = 0.02M

Time = 25 mins.

The formula for average rate of the reaction is,

r_av= - Δ{R} / Δt → Equation 1

? {R} = (R₂)-( R₁), the equation 1 is written as,

r_av= - R₂- R₁/ Δt

= -0.02-0.03 / 25

= 4 X 10^-4 mol L^-1 min^-1

The average rate of reaction in seconds is given by,

= 4 X 10^-4 mol L^-1 / 60 S

(dividing by 60 to convert minutes to seconds)

= 6.6 * 10^-6 mol L^-1 s^-1

The average rate of the reaction in minutes is 4 * 10^-4 mol L^-1 min^-1 and in seconds is 6.6 * 10^-6 mol L^-1 s^-1

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