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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.

The both electrode are same thus there

E_{cell}^{0}

=0
Since the potential of the hydrogen electrode measured against standard hydrogen electrode is a positive value, the electrode is acting as the anode. If the unknown [H⁺] be x, the Nernst equation takes the form

E_{cell} = E °_{cell} - \frac{0.0592}{2} \log \frac{x^{2}}{1^{2}}

Since both electrodes are the same,

E °_{cell} = 0 .

∴

0.29 V = 0 - \frac{0.0592 V}{2} \times 2 \log x

0.29 V = - 0.0592 V \times \log x

Since

- logx = - \log [H^{+}] = pH

0.029 V = 0.0592 V \times pH

pH = \frac{0.29}{0.0592} = 4.9 .

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)

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:
${Ag}^{+} + {Fe}^{3 +} ⇋ {Fe}^{3 +} + Ag (F = 96500 C {mol}^{- 1})$