A) -0.476 V
B) -0.404 V
C) + 0.404 V
D) + 0.772 V
Correct Answer: D
Solution :
\[\Delta {{G}^{o}}=-nF{{E}^{o}}\] \[F{{e}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Fe\] ... (i) \[\Delta {{G}^{o}}=-2\times F\times (-0.440\,\,V)\] = 0.880 F \[F{{e}^{3+}}+3{{e}^{-}}\xrightarrow{{}}Fe\] ... (ii) \[\Delta {{G}^{o}}=-3\times F\times (-0.036)=0.108\,F\] On subtracting Eq. (i) from Eq. (ii) \[F{{e}^{3+}}+{{e}^{-}}\xrightarrow{{}}F{{e}^{2+}}\] \[\Delta {{G}^{o}}=0.108\,F-0.880\,F=-0.772\,F\] \[{{E}^{o}}_{cell}=\frac{-\Delta {{G}^{o}}}{nF}\] \[=-\frac{(-0.772F)}{1\times F}=+0.772\,V\]
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