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Electrochemistry

Short Answer Type

191. A current of 1.50 amp passed through an electrolytic cell containing AgNO₃solution with inert electrodes. The weight of Ag deposited was 1.50 g. How long did the current flow?

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192. Write the reactions taking place at the anode and cathode in the above cell.
When AgNO₃ is electrolysis in aqueous solution?

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193. Give reactions taking place at the two electrodes if these are made up of Ag.

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Long Answer Type

194. The following chemical reaction is occuring in an electro-chemical cell
Mg(s) + 2Ag⁺(0.001 M) → Mg²⁺(0.10 M) + Ag(s)
The E° electrode values are
Mg²/Mg = – 2.36 V
Ag⁺/Ag = 0.81
V For the cell calculate/write:
(a) (i) E° value for the electrode 2Ag⁺/2Ag
(ii) Standard cell potential E°_Cell(b) Cell potential (E)_cell(c) (i) Symbolic representation of the above cell
(ii) Will the above cell reaction be spontaneous?

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Short Answer Type

195. Calculate the equilibrium constant for the reaction:
Cd²⁺(aq) + Zn(s) → Zn²⁺(aq) + Cd(s)
if E°_Cd²⁺/Cd = – 0.403 V.
E°_zn²⁺/zn = – 0.763 V

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196. When a current of 0.75 A is passed through a CuSO₄ solution for 25 min, .36 g of copper is deposited at the cathode. Calculate the atomic mass of copper.

Time = 25 min = 25 x 60 sec
Current = 0.75 A
${Cu}^{2 +} + 2 e^{-} \to Cu$
Electricity passed = 25 x 60 x 0.75 = 1125 C of electricity deposit copper = 0.369 g. 2 x 96500 C
of electricity will deposit copper.
M= ZxIxT

$= \frac{2 \times 96500 \times 0.369}{1125}$
Atomic mass of copper = 63.3 u.

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197. Tarnished silver contains Ag₂S. Can this tarnish removed by placing silver in an aluminium pan containing an inert electrolytic solution such as NaCl. The standard electrode potential for half reaction.

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198. Calculate the standard free energy change for the following reaction at 25°C.
Au(s) + Ca²⁺(aq.1M) → Au³⁺(aq. 1M) + Ca(s)
E°_Au²⁺/Au = – 0.403 V.
E°_Au²⁺/Cu = – 0.763 V.
Predict whether the reaction will be spontaneous or not at 25°C. Which of the above two half cells will act as an oxidizing agent and which one will be a reducing agent?

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199. The conductivity of 0.001 M acetic acid is 4 x 10^–5 s/cm. Calculate the dissociation constant of acetic acid, if λ°_m for acetic acid is 390.5 s cm² mol.

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