A) 0.86 V
B) 0.336 V
C) - 0.339 V
D) 0.26V
Correct Answer: D
Solution :
From the given representation of the cell, \[{{E}_{cell}}\] can be found as follows. \[{{E}_{cell}}=\left( E_{F{{e}^{2+}}/Fe}^{{}^\circ }-E_{C{{r}^{3+}}/Cr}^{{}^\circ } \right)-\frac{0.059}{6}\log \frac{{{\left[ C{{r}^{3+}} \right]}^{2}}}{{{\left[ F{{e}^{2+}} \right]}^{3}}}\] \[\left[ Nernst\,Eqn. \right]\] \[=-\,0.42-(-\,0.72)-\frac{0.059}{6}\,\log \frac{{{(0.1)}^{2}}}{{{(0.01)}^{3}}}\] \[=-\,0.42\,\,+\,\,0.72-\frac{0.059}{6}\,\log \,\frac{0.1\times 0.1}{0.01\times 0.01\times 0.01}\] \[=0.3-\frac{0.059}{6}\log \frac{{{10}^{-2}}}{{{10}^{-6}}}=0.3-\frac{0.059}{6}\times 4\] \[=0.30-0.0393=0.26\,V\]
You need to login to perform this action.
You will be redirected in
3 sec
