A) \[{{P}_{1}}={{P}_{2}}\]
B) \[{{P}_{1}}<{{P}_{2}}\]
C) \[{{P}_{1}}>{{P}_{2}}\]
D) None of these
Correct Answer: B
Solution :
[b] The cell reaction is \[C{{l}_{2}}\left( g \right)\left( {{P}_{2}}\text{ }atm \right)\xrightarrow{{}}C{{l}_{2}}\left( aq \right)\left( {{P}_{1}}\,atm \right)\] \[{{E}_{cell}}=-\frac{0.0592}{2}\log \frac{{{P}_{1}}}{{{P}_{2}}}=\frac{0.0592}{2}\log \frac{{{P}_{2}}}{{{P}_{1}}}\] \[{{E}_{cell}}\] will be positive when \[{{P}_{2}}>{{P}_{1}}.\]
You need to login to perform this action.
You will be redirected in
3 sec
