| \[C{{d}_{(s)}}+P{{b}^{2+}}_{(aq)}\xrightarrow{{}}C{{d}^{2+}}_{(aq)}+P{{b}_{(s)}}\] | |
| Given | \[E_{{\scriptstyle{}^{C{{d}^{2+}}}/{}_{Cd}}}^{o}=-0.403\,V\] |
| \[E_{{\scriptstyle{}^{P{{b}^{2+}}}/{}_{Pb}}}^{o}=-0.125\,V\] |
A) \[-5.346\text{ }kJ\]
B) \[53.46\text{ }kJ\]
C) \[-10.69\text{ }kJ\]
D) \[-106.92\text{ }kJ\]
Correct Answer: B
Solution :
\[C{{d}_{(g)}}+P{{b}^{2+}}_{(aq)}\xrightarrow{{}}C{{d}^{2+}}_{(aq)}+P{{b}_{(s)}}\] \[n=2\] \[\Delta G=nFE\] \[=-2\times 96500\times E_{cell}^{o}\] \[=-2\times 96500\times ({{E}^{o}}_{cathode}-{{E}^{o}}_{anode})\] \[=-2\times 96500\,(-0.126+0.403)\] \[=-53461J=-53.46\,kJ\]
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