A) \[-\,5.346\text{ }kJ\]
B) \[-\,53.46\text{ }kJ\]
C) \[-\,106.92\text{ }kJ\]
D) \[-\,10.69\text{ }kJ\]
Correct Answer: B
Solution :
\[Cd(s)+P{{b}^{2+}}(aq.)\xrightarrow{{}}C{{d}^{2+}}(aq.)+Pb(s)\] From the reaction, it is clear that transfer of \[2{{e}^{-}}\] is taking place i.e., \[n=2\] Free energy change \[(\Delta G)=-n\,.\,F\,.\,E\] \[=-\,2\times 96500\times (E_{cell}^{0})\] \[=-\,2\times 96500\times (E_{cathode}^{\text{o}}-E_{anode}^{\text{o}})\] \[=-193000\,(-0.126+0.403)\] \[=-\,193000\times 0.277\] \[=-\,53461\,joules\] \[=-\,53.46\,kJ\]You need to login to perform this action.
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