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Electrochemistry

How many grams of silver could be placed out on a shild by electrolysis of a solution containing Λ° ions for a period of 4 hours by a current strength of 8.5 amperes? [F = 96,500 C mol^–1, Molar mass of Ag = 107.8 g]

Molar mass of silver 107.8g
Current = 8.5 A

{Ag}^{+} + e \to Ag (s)

m = Z \times I \times t = \frac{107.8}{96500} \times 8.5 amp \times 4 \times 60 \times 60 m = \frac{131947.2}{965} = 136.73 g .

99 Views

Calculate the resistance offered by 0.5 M CH₃COOH solution when its molar conductivity. is 7.4 Ω^–1 cm² mol^–1 and the cell constant is 0.037 cm^–1.

We have gien the molar conductivity 7.4Ω^–1 cm² mol^–1
cell constant 0.037cm^-1thus by using formula

k = \frac{1}{R} \times (\frac{l}{A})

and therefore

R = \frac{1}{k} \times (\frac{l}{A}) k = Λ_{m} \times C = 7.4 Ω^{- 1} {cm}^{2} {mol}^{- 1} \times 0.5 mol / L = 7.4 Ω^{- 1} \times \frac{0.05 mol}{1000 {cm}^{3}} = 3.7 \times 10^{- 4} Ω^{- 1} {cm}^{- 1} .

∴

R = \frac{1}{k} . \frac{l}{a}

= \frac{1}{3.7 \times 10^{- 4} Ω^{- 1} {cm}^{- 1}} \times 0.037 {cm}^{- 1} = 100 Ω .

132 Views

Write the Nernst equation and calculate the emf of the following cell at 298 K : Cu(s) | Cu²⁺ (0.130 M) || Ag⁺ (1.00 x 10^–4 M) | Ag(s)

E°_Cu²⁺/ cu = + 0.34 V and E°_Ag+/Ag = + 0.80 V.

Cu + 2 {Ag}^{+} \to {Cu}^{2 +} + 2 Ag (s) A p p l y i n g t h e n e r n s t e q u a t i o n E = E ° - \frac{0.059}{2} \log \frac{[{Cu}^{2 +}]}{{[{Ag}^{+}]}^{2}} E ° = E °_{cathode} - E °_{anode} = 0.80 - 0.34 = 0.46 V E = 0.46 - \frac{0.059}{2} \log \frac{0.130}{1.0 \times 10^{- 4}} = 0.46 - \frac{0.059}{2} \times 1.30 \times 10^{7} = 0.46 - 0.21 = 0.25 V .

247 Views

Conductivity of 0.00241 M acetic acid is 7.896 x 10^–5 S cm^–1. Calculate its molar conductivity, if Λ° for acetic acid is 390.5 S cm² mol^–1, what is its dissociation constant?

We have given that
Concentration = 0.00241M
Conductivity =0.00241S Cm^-1
thus by using the formula
we get

\land = K \times \frac{1000}{C} C = 0.00241 M K = 7.896 \times 10^{- 5} S {cm}^{- 1} \land = \frac{7.896 \times 10^{- 5} \times 1000}{0.00241} = 32.76 α = \frac{\land}{\land_{0}} = \frac{32.76}{390.5} = 0.084 K = \frac{{Cα}^{2}}{1 - α} = \frac{0.00241 \times (0.084)^{2}}{(1 - 0.084)} = 1.856 \times 10^{- 5} .

212 Views

Calculate the equilibrium constant for the reaction,
Cu(s) + 2Ag⁺ (aq) → Cu²⁺ (aq) + 2Ag(s)
[E°_cell = 0.46 V]
(T = 298 K, F = 96500 C mol^–1, R = 8.31 J K^–1 mol^–1)

Using the formula we get:

∆ G ° = - nFE °_{cell} = - 2.303 RT \log K \log K = \frac{nFE °_{cell}}{2.303 \times R \times T} = \frac{2 \times 96500 \times 0.46}{2.303 \times 8.31 \times 298} = 15.5 K = 3.69 \times 10^{15} .

1504 Views

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Electrochemistry

How many grams of silver could be placed out on a shild by electrolysis of a solution containing Λ° ions for a period of 4 hours by a current strength of 8.5 amperes? [F = 96,500 C mol–1, Molar mass of Ag = 107.8 g]

How many grams of silver could be placed out on a shild by electrolysis of a solution containing Λ° ions for a period of 4 hours by a current strength of 8.5 amperes? [F = 96,500 C mol^–1, Molar mass of Ag = 107.8 g]