\[{{M}^{x+}}_{(aq)}/\,{{M}_{(s)}}\] | \[A{{u}^{+}}_{(aq)}\]\[/\,A{{u}_{(s)}}\] | \[A{{u}^{+}}_{(aq)}/\]\[\,A{{g}_{(s)}}\] | \[F{{e}^{3+}}_{(aq)}/\]\[F{{e}^{2+}}_{(aq)}\] | \[F{{e}^{2+}}_{(aq)}/\]\[F{{e}_{(s)}}\] |
\[{{E}^{o}}_{{{M}^{x+}}/M}(V)\] | 1.40 | 0.80 | 0.77 | -0.44 |
A) \[A{{u}^{3+}}/Au\]
B) \[F{{e}^{3+}}/F{{e}^{2+}}\]
C) \[A{{g}^{+}}/Ag\]
D) \[F{{e}^{2+}}/Fe\]
Correct Answer: C
Solution :
\[{{E}^{o}}_{cell}={{E}^{o}}_{cathode}-{{E}^{o}}_{anode}\] (i) For \[A{{u}^{3+}}/Au\]: \[{{E}^{o}}_{cell}=1.40-(-0.76)=2.16V;\]\[\frac{2.16}{3}=0.72V\] (ii) For\[A{{g}^{+}}/Ag\]: \[{{E}^{o}}_{cell}=0.80-(-0.76)=1.56V\] (iii) For \[F{{e}^{3+}}/F{{e}^{2+}}:\] \[{{E}^{o}}_{cell}=0.77-(-0.76)=1.53V\] (iv) For \[F{{e}^{2+}}/Fe:\] \[{{E}^{o}}_{cell}=-0.44-(-0.76)=0.32V;\] \[\frac{0.32}{2}=0.16V\]\[{{E}^{o}}_{cell}\]is maximum for\[{{E}^{o}}_{A{{g}^{+}}_{(ag)}/A{{g}_{(s)}}.}\]You need to login to perform this action.
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