A) - 0.339 V
B) - 0.26 V
C) 0.26 V
D) 0.339 V
Correct Answer: C
Solution :
\[2Cr(s)+3F{{e}^{2+}}(aq)\xrightarrow{{}}2C{{r}^{3+}}(aq)+3Fe(s)\] \[{{E}_{cell}}=E_{F{{e}^{2+}}\left| Fe \right.}^{o}-E_{C{{r}^{3+}}\left| Cr \right.}^{o}-\frac{0.0059}{6}\log \frac{{{[C{{r}^{3+}}]}^{2}}}{{{[F{{e}^{2+}}]}^{3}}}\] \[=-0.42-(-0.72)-\frac{0.059}{6}\log \frac{{{(0.1)}^{2}}}{{{(0.01)}^{3}}}=0.26\,V\]
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