A) \[Zn|Z{{n}^{2+}}(1M)||C{{u}^{2+}}(1M)|Cu\]
B) \[Zn|Z{{n}^{2+}}(1M)||A{{g}^{+}}(1M)|Ag\]
C) \[Cu|C{{u}^{2+}}(1M)||A{{g}^{+}}(1M)|Ag\]
D) \[Zn|Z{{n}^{2+}}(1M)||C{{o}^{2+}}(1M)|Co\]
Correct Answer: B
Solution :
\[{{E}_{cell}}=E_{cell}^{o}-\frac{0.0591}{n}\log \frac{[oxidised\,species]}{[reduced\text{ }species]}\] Since, concentration of all the ions is equal, higher the value of\[E_{cell}^{o},\]higher is the\[{{E}_{cell}}\] \[E_{cell}^{o}={{E}_{cathode}}-{{E}_{anode}}\] \[=0.337-(-0.762)=1.09\] \[E_{cell}^{o}=0.799-(-0.762)=1.56V\] \[E_{cell}^{o}=0.799-0.337=0.46\,V\] \[E_{cell}^{o}=-0.227-(-0.762)=0.53\,V\] Thus, emf is maximum for the cell given in option .
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