Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Electrochemistry

Short Answer Type

151. The potential of a hydrogen electrode in a solution of unknown [H⁺] is 0.29 V at 298 k measured against a standard hydrogen electrode. Calculate the pH of the solution.

124 Views

152. The measured resistance of a conductance cell containing 7.5 x 10^–3 M solution of KCl at 25° C was 1005 ohms. Calculate (a) specific conductance (b) molar conductance of the solution cell constant = 1.25 cm^–1.

695 Views

153. Estimate the minimum potential difference needed to reduce Al₂O₃ at 500° C. The free energy change for the decomposition reaction.

2 {Al}_{2} O_{3} \to \frac{4}{3} Al + O_{2} is ∆ G = + 960 kJ . (F = 96500 C {mol}^{- 1})

1099 Views

154. Two students use same stock-solution of ZnSO₄ and solution of CuSO₄. The emf of one cell is 0.03 V higher than the other. The conc. of CuSO₄ in the cell with higher emf value is 0.5 M. Find out the concentration of CuSO₄ in the other cell (2.303 RT/F = 0.06).

468 Views

Long Answer Type

155. The measured resistance of a conductance cell containing 7.5 x 10^–3 M solution of KCl at 25° was 1005 ohms. Calculate (a) specific conductance (b) Molar conductance of the solution. Cell constant = 1.25 cm^–1.

135 Views

Short Answer Type

156. Molar conductance of a 1.5 M solution of an electrolyte is found to be 138.9 S cm²mol^–1. What would be the specific conductance of this solution.

267 Views

157. The emf (E°_cell) of the cell reaction : 3Sn⁴⁺ + 2Cr → 3Sn²⁺ + 2Cr³⁺ is 0.89 V. Calculate ΔG° for the reaction. (F = 96,500 (mol^–1 and VC ≡ J)

356 Views

Long Answer Type

158. Calculate emf of the following cell at 298 K
Mg(s) | Mg²⁺ (0.001 M) || Cu²⁺ (0.0001 M) | Cu(s)
(Given : E°_Cu2/Cu = + 0.34 V; E° _Mg²⁺/Mg = – 2.37 V)

Mg left parenthesis straight s right parenthesis space rightwards arrow space Mg to the power of 2 plus end exponent space plus space 2 straight e to the power of minus

206 Views

Short Answer Type

159. If E° for copper electrode is + 0.34 V how will you calculate its emf value when the solution in contact with it is 0.1 M in copper ions? How does emf for copper electrode change when concentration of Cu²⁺ ions in the solution is decreased?

1187 Views

160.

The half reactions are:
(i) Fe³⁺ + e^– → Fe²⁺, E° = 0.76 V
(ii) Ag⁺+ e^– → Ag,E° = 0.80 V
Calculate K_c for the following reaction at 25° C:
${Ag}^{+} + {Fe}^{3 +} ⇋ {Fe}^{3 +} + Ag (F = 96500 C {mol}^{- 1})$