A) - 0.26 V
B) 0.26 V
C) 0.339 V
D) - 0.339 V
Correct Answer: B
Solution :
Reduction potential of \[F{{e}^{2}}^{+}\,/Fe\] is higher Cathode, \[{{\operatorname{Fe}}^{2}}^{+}+2{{e}^{-}}\,\,\xrightarrow{{}}\,\,Fe\] ... (i) Anode, \[\operatorname{Cr}\,\xrightarrow{{}}\,\,C{{r}^{3\,+}} + 3e\] ... (ii) Eq. (i) \[\times \] 3 - Eq. (ii) \[\times \] 2 \[2Cr\,\,+\,\,3\,F{{e}^{2\,+}}\,\,\xrightarrow{{}}\,\,3Fe+2C{{r}^{3\,+}}\] \[E{{{}^\circ }_{Cell}}={{E}_{cathode}}-{{E}_{anode}}\] \[E{{{}^\circ }_{cell}}=-0.42-(-0.72)\] \[=\,\,\,-\,0.30V\] \[{{E}_{cell}}=E_{cell}^{{}^\circ }-\frac{0.0591}{n}\log \,\,\frac{[C{{r}^{3+}}]}{{{[F{{e}^{2+}}]}^{3}}}\] \[=\,\,-0.30\,\,-\,\,\frac{0.0591}{6}\log \,\,\frac{{{(0.1)}^{2}}}{{{(0.01)}^{3}}}\] \[=\, 0.26 V\]
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