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Class 10 Class 12

Electrochemistry

The charge required for the reduction of 1 mol of I₂ to I^– is:

96500 C
2 x 96500 C
1/2 x 96500 C
4 x 96500 C

2 x 96500 C

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Charge required to liberate 11.5 g sodium is

0.5 F
0.1 F
1.5 F
96500 coulombs

0.5 F

205 Views

Calculate the potential (emf.) of the cell
Cd | Cd²⁺ (0.10 M) || H⁺(0.20 M) | Pt, H₂ (0.5 atm)
(Given : E° _{Cd2+ / Cd} = – 0.403 V, R = 8.314 K^–1 K^–1 mol^–1, F = 96500 C mol^–1)

Cell reaction will be

Cd + 2 H^{+} \to {Cd}^{2 +} + H_{2} (hence n = 2)

E °_{cell} = E °_{right} - E °_{left} E °_{cell} = 0 - (- 0.403 = + 0.403 V)

E_{cell} = E °_{cell} + \frac{2.303}{nF} \log \frac{{[H^{+}]}^{2} [Cd]}{{pH}_{2} [{Cd}^{2 +}]}

= 0.403 + \frac{8.314 \times 298 \times 2.303}{2 \times 96500} \log \frac{(0.2)^{2}}{(0.5) (0.1)} = 0.403 + 0.0295 \log 0.8 = 0.403 - 0.0028

E_{cell} = 0.400 V

Equilibrium Constant from Nernst Equation: If the Daniell cell is short circuited, then we note that the reacton.
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
takes place and as time passes, the concentration of Zn²⁺ keeps on increasing while the concentration of Cu²⁺ keeps decreasing. At the same time, voltage of the cell as read on the voltmeter keeps on decreasing. After some time, we shall note that there is no change in the concentration of Cu²⁺ and Zn²⁺ ions and at the same time, voltmeter gives zero reading. This indicates that equilibrium has been attained. In this situation the Nernst equation may be written as

E_{cell}^{°} = \frac{2.303 RT}{nF} \log \frac{[{Cu}^{2 +}]}{[{Zn}^{2 +}]} E_{cell}^{°} = \frac{2.303 RT}{2 F} \log \frac{[{Cu}^{2 +}]}{[{Zn}^{2 +}]}

But at equilibrium [Zn²⁺] / [Cu²⁺] = K
and the above equation can be written as

E_{cell}^{°} = \frac{0.0591}{2} logK

(E° = 1.1 V)
K = 2 x 10³⁷ at 298
In general E°_cell = 2.303 RT / nF x log K

595 Views

Write the Nernst equation and calculate the emf of the following cell at 298 K:
Pt(s) | Br₂ (l) Br^–(0.01M) || H⁺ (0.03 M) | H₂(g) (1 bar) | Pt(s)
Given E°_Br2/Br– = + 1.08 V
Br₂/ Br^–

The net reaction is 2Br^–(aq) + 2H⁺(aq) → H₂(g) + Br₂(l)
The Nernst equation is

E_{cell} = E °_{cell} - \frac{0.059}{2} \log [\frac{1}{{[H^{+}]}^{2} {[{Br}^{-}]}^{2}}] E °_{cell} = 0 - (+ 1.08) = 1.08

∴

E_{cell} = - 1.08 - \frac{0.059}{2} \log [\frac{1}{[0.03]^{2} [0.01]^{2}}]

E_{cell} = - 1.08 - 0.21 E_{cell} = - 1.29 V .

486 Views

The conductivity of molten NaCl is due to

free ions
free electrons
free molecules
atoms of Na and CI

free ions

146 Views

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Electrochemistry

Write the Nernst equation and calculate the emf of the following cell at 298 K:
Pt(s) | Br₂ (l) Br^–(0.01M) || H⁺ (0.03 M) | H₂(g) (1 bar) | Pt(s)
Given E°_Br2/Br– = + 1.08 V
Br₂/ Br^–

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Electrochemistry

Write the Nernst equation and calculate the emf of the following cell at 298 K:Pt(s) | Br2 (l) Br– (0.01M) || H+ (0.03 M) | H2(g) (1 bar) | Pt(s)Given E°Br2/Br– = + 1.08 VBr2 / Br–

Write the Nernst equation and calculate the emf of the following cell at 298 K:
Pt(s) | Br₂ (l) Br^–(0.01M) || H⁺ (0.03 M) | H₂(g) (1 bar) | Pt(s)
Given E°_Br2/Br– = + 1.08 V
Br₂/ Br^–