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

Question 1

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Consider the reaction M⟶k1Mads++e−M+Mads+⟶k2Mads++Msol++e−Mads+⟶ksMsol+\begin{array} { l } M \stackrel { k _ { 1 } } { \longrightarrow } M _ { a ds } ^ { + } + e ^ { - } \\M + M _ { a d s } ^ { + } \stackrel { k _ { 2 } } { \longrightarrow } M _ { a d s } ^ { + } + M _ { s o l } ^ { + } + e ^ { - } \\M _ { a d s } ^ { + } \stackrel { k _ { s } } { \longrightarrow } M _ { s o l } ^ { + }\end{array}
-In the same mechanism, the charge transfer resistance (Rt) is given by


A) (Rt) −1=F{b1k1dc(1−θss) −b2k2dc(1−θss) θss}\left( R _ { t } \right) ^ { - 1 } = F \left\{ b _ { 1 } k _ { 1 dc } \left( 1 - \theta _ { ss } \right) - b _ { 2 } k _ { 2 dc } \left( 1 - \theta _ { s s } \right) \theta _ { s s } \right\}
B) (Rt) −1=F{b1k1dc(1−θss) +b2k2dc(1−θss) θss}\left( R _ { t } \right) ^ { - 1 } = F \left\{ b _ { 1 } k _ { 1 dc} \left( 1 - \theta _ { ss } \right) + b _ { 2 } k _ { 2dc } \left( 1 - \theta _ { s s } \right) \theta _ { s s } \right\}
C) (Rt) −1=F{b1k1dc(1−θss) +b2k2dcθss}\left( R _ { t } \right) ^ { - 1 } = F \left\{ b _ { 1 } k _ { 1 d c } \left( 1 - \theta _ { s s } \right) + b _ { 2 } k _ { 2 dc } \theta _ { s s } \right\}
D) (Rt) −1=F(b1k1dc(1−θss) −b2k2dcθss}\left( R _ { t } \right) ^ { - 1 } = F \left( b _ { 1 } k _ { 1 d c } \left( 1 - \theta _ { s s } \right) - b _ { 2 } k _ { 2 dc } \theta _ { s s } \right\}

Correct Answer:

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