A) +0.29 41V
B) + 0.5212 V
C) + 0.1308V
D) -0.2606 V
Correct Answer: A
Solution :
Key Idea : \[{{E}_{cell}}=E_{cell}^{o}-\frac{0.059}{n}\log \frac{[product]}{[reac\tan t]}\] Given \[{{E}^{o}}_{C{{r}^{3+}}/Cr}=-0.74\,\,V\] \[{{E}^{o}}_{F{{e}^{2+}}/Fe}=-0.44\,\,V\] \[Cr/C{{r}^{3+}}\,(0.1\,M)||F{{e}^{2+}}\,\,(0.1\,M)/Fe\] \[\therefore \] \[C{{r}^{3+}}/Cr\]is anode and \[F{{e}^{2+}}/Fe\]is cathode. \[E_{cell}^{o}=E_{C}^{o}-E_{A}^{o}\] \[=(-0.44)-(0.74)\] \[=-0.444+0.74\] \[=0.30\,\,V\] Cell reaction is \[2Cr+3F{{e}^{2+}}\xrightarrow{{}}2C{{r}^{3+}}+3Fe\] number of electrons in cell reaction = 6 \[E_{cell}^{o}=E_{cell}^{o}-\frac{0.059}{n}\log \left[ \frac{product}{reac\tan t} \right]\] \[=+0.30\,V-\frac{0.059}{6}\log \left[ \frac{{{(C{{r}^{3+}})}^{2}}}{{{(F{{e}^{2+}})}^{3}}} \right]\] \[=0.30-\frac{0.059}{6}\log \left[ \frac{{{(0.1)}^{2}}}{{{(0.01)}^{3}}} \right]\] \[=0.30-\frac{0.059}{6}\log \,\,{{10}^{4}}\] \[=0.30-\frac{0.059}{6}\times 0.60\] \[=0.30-5.9\times {{10}^{-3}}\] \[=0.2941\,V\]
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