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Calculate the maximum work and log\; K_{c} for the given reaction at 298\; K :

Ni\; (s)+2\; Ag^{+}(aq)\rightleftharpoons Ni^{2+}\; (aq)+2Ag\; (s)

Given :

E^{\circ}_{Ni^{2+}/Ni}=-0.25\; V,\; \; E^{\circ}_{Ag^{+}/Ag}=+0.80\; V

1\; F=96500\; C\; mol^{-1}

 

 

 
 
 
 
 

Answers (1)

Given :

E^{\circ}_{Ni^{2+}/Ni}=-0.25\; V,\; \; E^{\circ}_{Ag^{+}/Ag}=+0.80\; V

1\; F=96500\; C\; mol^{-1}

Ni\; (s)+2\; Ag^{+}(aq)\rightleftharpoons Ni^{2+}\; (aq)+2Ag\; (s)

ΔGo = -nFEo cell = -2 x 96500 x {0.80- (-0.25)} = -202650 J/ mol

Maximum work = 202650 J/ mol

Eocell = (0.059/2) log Kc 

log Kc = 2\times 1.15\div 0 .0591 = 35.6

 

 

Posted by

Safeer PP

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