Calculate the standard free energy change for the reaction occuri

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 Multiple Choice QuestionsShort Answer Type

261. Calculate the equilibrium constant for the reaction:
Zn + Cd2+ (aq) → Zn2+ (aq) + Cd (E°cell = 0.36 V)
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262. Calculate the pH of the following half reactions: Pt. H2 (1 atom) / HCl, E = 0.25 V.

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263. Electrolytic conductivity of 0.20 mol L–1 solution of KCl at 298 k is 2.48 x 10–2 ohm–1cm–1. Calculate its molar conductivity.
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264.

Calculate the molar conductivity at infinite dilution of acetic acid from the following data:
Λm (HCl) = 426 ohm–1 cm2 mol–1, Λm CH3COONa = 91 ohm–1 cm2 mol–1 and Λm(NaCl) = 126 ohm–1 cm2 mol–1.

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265. Calculate the equilibrium constant of reaction at 25°C:
Ni(s) + Cu2+ (aq) → Cu(s) + Ni2+ (aq)
Given : E°Ni2+/Ni = -0.25 V, E°Cu2+/Cu = + 0.34 V, R = 8.314 J K–1 mol–1, F = 96500 C mol–1.
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266. Calculate the standard free energy change for the reaction occuring in the cell:
Zn(s) | Zn2+ (1 M)|| Cu2+ (1M) | Cu(s)
[Given E°Zn2+/Zn = – 0.076 V, E°Cu2+/Cu = + 0.34 V, F = 96500 C mol–1]


Given:
EZn2+/Zn 0= – 0.076 V

ECu2+/Cu0 = + 0.34 V

F = 96500 C mol–1
from the reaction 
n=2

Ecell0 = ECu2+/Cu0 -  EZn2+/Zn 0


Ecell0 =0.34 -(-0.076)
        = 0.416

We know that
G = -nFEcell0

    =  -2 x 96500 x 0.416
     =–802.88 kJ mol
–1

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267.

Consider the cell Zn/Zn2+ (aq) (1.0 M) || Cu2+ (aq) (0.1 M) | Cu The standard reaction potentials are + 0.35 V for 2e + Cu2+ (aq) → Cu and – 0.763 V for 2e + Zn2+ (aq) → Zn
(i) Write down the cell reaction.
(ii) Calculate the emf of the cell.
(iii) Is the cell reaction spontaneous or not?

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268. The standard reaction potential for Cu2+/Cu is + 0.34 V. Calculate the reduction potential at pH = 14 for the above couple. Ksp of Cu(OH)2 is 1.0 x 10–19.
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269.

The standard reduction potential of the reaction at 25°C
2H2O + 2e H2(g) + 2OH is – 0.8277 V
Calculate equilibrium constant for the reaction
2H2O H3O++OH- at 25°C.

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270. The resistance of a conductivity cell containing 0.01 M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001 M KCl solution at 298 K. is 1.46 x 10–6 s cm–1.
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