The half reactions are:
(i) Fe3+ + e → Fe2+, E° = 0.76 V
(ii) Ag++ e → Ag,E° = 0.80 V
Calculate Kc for the following reaction at 25° C:
     Ag++Fe3+    Fe3++Ag                                    (F = 96500 C mol-1)


E°cell = E°Ag+/Ag-E°Fe3+/Fe2+           = + 0.80 V - 0.76 V = + 0.04 V   + nE°F = 2.303 RT log k      log K = 0.040.0591 0.6779          K =  4.763.
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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?


The emf of an electrode when dipped in different concentrated solution is given by Nernst equation.

ECu2+/Cu = E°Cu2+/Cu+0.0592×log Cu2+

Substituting the given values, we get

 ECu2+/Cu = 0.34 V + 0.0592 log 0.1 V                 = + 0.34 V - [0.0295 log 10] V                 = + 0.34 V - 0.0295 V = + 0.3105 V.

When concentration of Cu2+ ion in the solution decreases the emf of the electrode decreases. In this case it has decreased from 0.34 V to 0.3105 V.
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Zinc rod is dipped in 0.1 M solution of ZnSO4. The salt is 95% dissociated at this dilution at 298 K. Calculate the electrode potential. [Given that: E°Zn2+/Zn = – 0.76 V.]

Concentration of Zn2+(aq)
             = 0.1 + 95100=0.095
           Zn2+(aq) + 2e-  Zn
According to Nernst equation,
            E = E°+0.0591nlogZn2+(aq)[Zn]
             = -0.76 + 0.0591nlog0.0951= 0.76+0.02953 × (-1.0223)= -0.76 - 0.03021 = -0.79 V.

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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)

we have given the emf of the cell reaction :

3Sn4++6e-3Sn2+                   At cathode2Cr  2Cr3++6e-                         At anode     n = 6,  E°cell = 0.89 V,     F = 96500 C mol-1


ΔG = –nE°F
= – 6 x 0.89 V x 96500 C mol–1
= – 515310 J = – 515.310 kJ.
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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.

We have given that
Λm = 138.9 S cm2,  M = 1.5, K = ?Λm = 1000 KM   K = Λm×M1000           = 138.9 S cm2 mol-1×1.5 mol Lt-11000 cm3      K = 208.351000 S cm-1      K = 2.0835 × 10-1 S cm-1       K = 0.20835 S cm-1.
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