| \[F{{e}^{2+}}(aq)+A{{g}^{+}}(aq)\to F{{e}^{3+}}(aq)+Ag(s)\] |
| Given that |
| \[E_{A{{g}^{+}}/Ag}^{o}=xV\] |
| \[E_{F{{e}^{2+}}/Fe}^{o}=yV\] |
| \[E_{F{{e}^{3+}}/Fe}^{o}=zV\] |
A) \[x+2y-3z\]
B) \[x-z\]
C) \[x-y\]
D) \[x+y-z\]
Correct Answer: A
Solution :
\[F{{e}^{+2}}(aq)+A{{g}^{+}}(aq)\to F{{e}^{+3}}(aq)+Ag(s)\] Cell reaction anode :\[F{{e}^{+2}}(aq)\to F{{e}^{+3}}(aq)+{{e}^{\Theta }};\] \[E_{F{{e}^{+2}}/F{{e}^{+3}}}^{o}=mV\] cathode :\[A{{g}^{+}}(aq)+{{e}^{\Theta }}\to Ag(s);\] \[E_{A{{g}^{+}}/Ag}^{o}=xV\] \[\Rightarrow \]cell standard potential \[=(m+x)V\] \[\therefore \]to find ?m?; \[F{{e}^{+2}}+2{{e}^{\Theta }}\to Fe;\] \[E_{1}^{o}=yV\Rightarrow \Delta _{1}^{o}G=-(2Fy)\] \[F{{e}^{+3}}+3{{e}^{\Theta }}\to Fe;\] \[E_{2}^{o}=zV\Rightarrow \Delta _{2}^{o}G=-(3Fz)\] \[F{{e}^{+2}}(aq)\to F{{e}^{+3}}(aq)+{{e}^{\Theta }};\] \[E_{3}^{0}=mV\Rightarrow \Delta _{3}^{o}G=-(1Fm)\] \[\Delta _{3}^{o}G=\Delta G_{1}^{o}-\Delta G_{2}^{o}=(-2Fy+3Fz)=-Fm\] \[\Rightarrow \]\[m=(2y-3z)\] \[\Rightarrow \]\[E_{cell}^{o}=(x+2y-3z)V\]
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