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Calculate the standard cell potentials of galvanic cell in which the following reactions take place:
 Fe2+ (aq) + Ag+ (aq) → Fe3+(aq) + Ag(s)
Calculate the ΔrG° and equilibrium constant of the reaction.


The galvanic cell of the given reaction is represented as
Fe2+(aq) | Fe3+(aq) || Ag| Ag(s)
The formula of standard cell potential is
Eocell = Eo right  – Eoleft
Eocell = 0.80 – 0.77
Eocell =  + 0.03 V
In balanced reaction there are 1 electron are transferring so that n = 1
Faraday constant, F = 96500 C mol−1
Eocell = + 0.03 V
Use formula
rGθ = – nFEocell
Plug the value we get   
Then, = −1 × 96500 C mol−1 × 0.03 V
= −2895 CV mol−1
= −2895J mol−1
= −2.895 kJ mol−1
Again,
Use second formula of ∆rGθ
rGθ = −2.303RT log kC
log KC = (∆rGθ) /( – 2.303RT)
plug the values we get

                                         
               
log kc = nE°cell0.0591 = 1×0.030.0591 = 0.5076      kc = antilog 0.5076 = 3.218.
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Calculate the standard cell potentials of galvanic cell in which the following reactions take place:
 2Cr(s) + 3Cd2+ (aq) → 2Cr3+ (aq) + 3Cd
(Calculate the ΔrG° and equilibrium constant of the reaction.)


E°Cd2+/Cd = -0.40 V,  E°Cr3+/Cr = -0.74 VE°Fe3+/Fe = 0.77 V, E°Ag+/Ag = + 0.80 V

Cr(s) Cr3+(aq)  Cd2+(aq)   Cd(s)                 (anode is on left and cathode on right)   E0cell  = E0right - E0left               = -0.40 - (0.74) = +0.34 V
 Gr0 (or  G°cell = -nFE°cell
  (Here n = 6 (as 6e are involved in overall cell reaction i.e., 
                   2Cr+3Cd2+3Cr3++3Cd)
                              = -6×96500 C × 0.34= -196860 J mol-1= - 196.86 kJ mol-1
  E0cell = 0.0591nlog Kclog kc = 0.34×60.0591 = 34.5177
or      kc = antilog 34.5177 =3.294 × 1034
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The conductivity of  0.20 M solution of KCl at 298 K is 0.0248 s cm–1. Calculate its molar conductivity. 


We have given that
Molarity of solution, M = 0.20
conductivity, i.e., specific conducitivity = k = 0.248 s cm
–1 = 2.48 x 10–2 ohm–1cm–1

Molar conductivity,   λm = 1000 kM ohm-1 cm2 mol-1      = 1000×2.48×10-20.20 = 124.0 s cm2 mol-1

Thus, molar conductivity,λm = 124.0 s cmmol–1
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Given the standard electrode potentials:
K+/K = – 2.93 V Ag+/Ag = 0.80V
Hg2+/Hg = 0.79 V Mg2+/Mg = –2.37 V
Cr3+/Cr = –0.74 V.
Arrange these metals in their increasing order of reducing power.


answer:
Ag < Hg < Cr < Mg < K.
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Arrange the following metals in the order in which they displace each other from the solution of their salts Al, Cu, Fe, Mg and Zn.

Answer:
Mg> Al> Zn> Fe> Cu is decreasing order of their reactivity.
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