A) 0.42 V
B) - 0.42 V
C) - 0.30 V
D) -1.10 V
Correct Answer: C
Solution :
Half-cell reactions are (i) \[F{{e}^{2\,+}}+2e\xrightarrow{{}}\,Fe\]\[E_{F{{e}^{2+}}/Fe}^{{}^\circ }=-\,0.44\,V\] (ii) \[\operatorname{Sn} \xrightarrow{{}}\,\,S{{n}^{2+}}\,+\, 2e\]\[E_{Sn/S{{n}^{2+}}}^{{}^\circ }=+\,0.14\,V\] Adding Eqs. (i) and (ii), \[E{}^\circ =E_{F{{e}^{2+}}/Fe}^{{}^\circ }+E_{Sn/S{{n}^{2+}}}^{{}^\circ }\] \[=\,\,-0.44 +0.14=-\,0.30V\]You need to login to perform this action.
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