A) \[~+\text{ }0.2941\,V\]
B) \[~+\text{ }0.5212\,V\]
C) \[+\text{ }0.1308\,V\]
D) \[-\,0.2606\,V\]
Correct Answer: A
Solution :
Key Idea: \[{{E}_{cell}}=E_{cell}^{o}-\frac{0.059}{n}\log \left[ \frac{\text{product}}{\text{reactant}} \right]\] 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 \[E_{cell}^{o}=E_{C}^{o}-E_{A}^{o}\] \[=(-\,0.44)-(0.74)\] \[=-0.44+0.74\] \[=+\,0.30\,V\] Cell reaction is \[2Cr+3F{{e}^{2+}}\xrightarrow{{}}2C{{r}^{3+}}+3Fe\] number of electrons in cell reaction = 6 \[{{E}^{o}}_{cell}={{E}^{o}}_{cell}-\frac{0.059}{n}\log \left[ \frac{\text{product}}{\text{reactant}} \right]\] \[=+\,0.30\,V-\frac{0.059}{6}\log \left[ \frac{{{(C{{r}^{3+}})}^{2}}}{{{(F{{e}^{2+}})}^{3}}} \right]\] \[=0.30-\frac{0.059}{6}\log \left[ \frac{{{(0.1)}^{2}}}{{{(0.01)}^{3}}} \right]\] \[=0.30-\frac{0.059}{6}\log {{10}^{4}}\] \[=0.30-\frac{0.059}{6}\times 0.60\] \[=0.30-5.9\times {{10}^{-3}}\] \[=0.2941\,V\]
You need to login to perform this action.
You will be redirected in
3 sec
