A) \[{{E}_{3}}={{E}_{1}}+{{E}_{2}}\]
B) \[{{E}_{3}}=\frac{({{n}_{1}}{{E}_{1}}+{{n}_{2}}{{E}_{2}})}{{{n}_{3}}}\]
C) \[{{E}_{3}}=\frac{({{E}_{1}}+{{E}_{2}})}{{{n}_{3}}}\]
D) \[{{E}_{3}}=n{{E}_{1}}+n{{E}_{2}}\]
Correct Answer: B
Solution :
\[{{E}_{cell}}\]is not the intensive property, cannot be additive, instead \[\Delta G\] is extensive property \[\Delta {{G}_{3}}=\Delta {{G}_{1}}+\Delta {{G}_{2}}\] \[{{n}_{3}}F{{E}_{3}}=-{{n}_{1}}F{{E}_{1}}-{{n}_{2}}F{{E}_{2}}\] \[{{E}_{3}}=({{n}_{1}}{{E}_{1}}+{{n}_{2}}{{E}_{2}})/{{n}_{3}}\]You need to login to perform this action.
You will be redirected in
3 sec