A) 0.399 V
B) 0.28 IV
C) 0.222V
D) 0.176V
Correct Answer: B
Solution :
It can be calculated with the help of Nernst equation. \[E={{E}^{o}}+\frac{0.059}{n}\log [{{M}^{n+}}]\] Where, E = emf of cell \[{{E}^{o}}=\]Standard emf o fcell n = number of \[{{e}^{-}}\]transferred \[{{\text{M}}^{\text{n+}}}\text{=}\]Concentration \[{{E}^{o}}=0.34\,\,V\] \[n=2\] \[[C{{u}^{2+}}]={{10}^{-2}}M\] So, \[E=0.34+\frac{0.059}{2}\log {{10}^{-2}}\] \[=0.34+\frac{0.059}{2}\times -2\] \[=0.34-0.059\] \[=+\,0.281\,V\]You need to login to perform this action.
You will be redirected in
3 sec