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Electrochemistry

Short Answer Type

281. How long will it take to deposit electrolytically 127 gm of copper on the cathode of a copper voltameter by a current of 50A? Given ECE of copper = 0.0003294 g/c.

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282. An electric current of 0.5 ampere was passed through acidulated water for one hour. Calculate the volume of hydrogen at STP produced. 1 coulomb of electricity deposits 0.00001 gm of hydrogen.

648 Views

Long Answer Type

283.

How much charge is required for the following reduction of
(i) 1 mol of Al³⁺ to Al
(ii) 1 mol of Cu²⁺ to Cu
(iii) 1 mol of MnO₄^–to Mn²⁺

326 Views

Short Answer Type

284. A solution of CuSO₄ is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode? (at mass Cu = 63.5).

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285. A solution of CuSO₄ is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?

1025 Views

286. Represent the cell in which the following reaction takes place:
Mg(s) + 2Ag⁺ (0.0001 M) → Mg²⁺(0.130 m) + 2Ag(s)
Calculate its E_(cell), if E°_cell = 3.17 V.

831 Views

Long Answer Type

287.

A Cell is prepared by dipping a copper rod in 0.01 M copper sulphate solution, and zinc rod in 0.02 M ZnSO₄ solution. The standard reduction potentials of copper and zinc are + 0.34 V and – 0.76 V respectively.
(a) What will be the cell reaction?
(b) How will the cell be represented?
(c) What will be the emf. of the cell?

384 Views

288. Calculate the standard electrode potential of the Mg²⁺/ Mg electrode for a cell in which the cell reaction is
Mg(s) + 2Ag⁺(aq) → Mg²⁺(aq) + 2Ag(s)
Given that, [Mg²⁺] = 0.1 M, [Ag⁺] = 0.01 M
E°_Ag/Ag = + 0.80, E_cell = + 2.90 V

The cell is
Mg(s) | Mg²⁺ (aq) (0.1 M) || Ag⁺ (aq) (0.01 M) | Ag(s)
and the E_cell is given by

$E_{cell} = E °_{cell} - \frac{0.059}{2} \log \frac{[{Mg}^{2 +}]}{{[{Ag}^{+}]}^{2}} 2.90 = E °_{cell} - \frac{0.059}{2} \log [\frac{(0.1)}{(0.01)^{2}}] 2.90 = E °_{cell} - \frac{0.059}{2} \times 3$

   $2.90 = E °_{cell} - 0.0885 E °_{cell} = 2.988 V$

But    $E °_{cell} = E °_{{Ag}^{+} / Ag} - E °_{{Mg}^{2 +} / Mg}$
   $E °_{{Mg}^{2 +} / Mg} = + 0.80 - 2.988 E °_{{Mg}^{2 +} / Mg} = - 2.188 V$

991 Views

289.

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^–

486 Views