Solved

Consider the Reaction -In the Same Mechanism, the Charge Transfer Resistance (Rt) Is

Question 10

Multiple Choice

Consider the reaction Psol⇄k1k−1Q−sol+e−Asol−⇄k2k−2Aads+e−\begin{array} { l } P _ { s o l }\underset { k _ { - 1 } } { \stackrel { k _ { 1 } } { \rightleftarrows } }Q _ { - s o l } + e ^ { - } \\A _ { s ol } ^ { - } \underset { k _ { - 2 } } { \stackrel { k _ { 2 } } { \rightleftarrows } } A _ { a d s } + e ^ { - } \\\end{array}

-In the same mechanism, the charge transfer resistance (Rt) is given by


A) (Rt) −1=F{b1k1dc[Psol](1−θss) −b−1k−1dc[Qsol](1−θss) }\left( R _ { t } \right) ^ { - 1 } = F \left\{ b _ { 1 } k _ { 1 { d c} } \left[ P _ { s o l } \right] \left( 1 - \theta _ { s s } \right) - b _ { - 1 } k _ { - 1 { dc } } \left[ Q _ { s o l } \right] \left( 1 - \theta _ { s s } \right) \right\}
B) (Rt) −1=F{b1k1dc[Psol](1−θss) −b−1k−1dc[Qsol](1−θss) +b2k2dc[Asol−](1−θss) −b−2k−2dcθss}\left( R _ { t } \right) ^ { - 1 } = F \left\{ \begin{array} { l } b _ { 1 } k _ { 1 dc } \left[ P _ { so l } \right] \left( 1 - \theta _ { s s } \right) - b _ { - 1 } k _ { - 1 { dc} } \left[ Q _ { s o l } \right] \left( 1 - \theta _ { ss} \right) \\+ b _ { 2 } k _ { 2 { dc } } \left[ A _ { s o l } ^ { - } \right] \left( 1 - \theta _ { s s } \right) - b _ { - 2 } k _ { - 2 { dc } } \theta _ { s s }\end{array} \right\}
C) (Rt) −1=F{b1k1dc[Psol]−b−1k−1dc[Qsol]+b2k2dc[Asol−]−b−2k−2dc}\left( R _ { t } \right) ^ { - 1 } = F \left\{ b _ { 1 } k _ { 1 d c} \left[ P _ { s ol } \right] - b _ { - 1 } k _ { - 1 dc} \left[ Q _ { s o l } \right] + b _ { 2 } k _ { 2 dc } \left[ A _ { s o l } ^ { - } \right] - b _ { - 2 } k _ { - 2 d c } \right\}
D) (Rt) −1=F{b2k2dc[Asol−](1−θss) −b−2k−2dcθss}\left( R _ { t } \right) ^ { - 1 } = F \left\{ b _ { 2 } k _ { 2 { dc} } \left[ A _ { s ol } ^ { - } \right] \left( 1 - \theta _ { s s } \right) - b _ { - 2 } k _ { - 2 { dc} } \theta _ { ss } \right\}

Correct Answer:

verifed

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Related Questions