Consider a reaction given by $\begin{array} { l }
M \underset { k _ { - 1 } } { \stackrel { k _ { 1 } } { \rightleftarrows } } M _ { a ds } ^ { + } + e ^ { - } \
M _ { a d s } ^ { + }\underset { k _ { - 2 } } { \stackrel { k _ { 2 } } { \rightleftarrows } } M _ { sol } ^ { + }
\end{array}$ .
-In the same mechanism, the faradaic current is given by
A) $$i _ { F } = F \left[ k _ { 1 } ( 1 - \theta ) \right]$$
B) $$i _ { F } = F \left[ k _ { 1 } ( 1 - \theta ) - k _ { - 1 } \theta \right]$$
C) $$i _ { F } = F \left[ k _ { 1 } ( 1 - \theta ) + k _ { - 1 } \theta \right]$$
D) $$i _ { F } = F \left[ k _ { 1 } ( 1 - \theta ) - k _ { - 1 } \theta - k _ { 2 } \theta \right]$$

Question

Consider a reaction given by $\begin{array} { l } 
M \underset { k _ { - 1 } } { \stackrel { k _ { 1 } } { \rightleftarrows } } M _ { a ds } ^ { + } + e ^ { - } \
M _ { a d s } ^ { + }\underset { k _ { - 2 } } { \stackrel { k _ { 2 } } { \rightleftarrows } } M _ { sol } ^ { + }
\end{array}$ .
-In the same mechanism, the faradaic current is given by
A) $$i _ { F } = F \left[ k _ { 1 } ( 1 - \theta ) \right]$$
B) $$i _ { F } = F \left[ k _ { 1 } ( 1 - \theta ) - k _ { - 1 } \theta \right]$$
C) $$i _ { F } = F \left[ k _ { 1 } ( 1 - \theta ) + k _ { - 1 } \theta \right]$$
D) $$i _ { F } = F \left[ k _ { 1 } ( 1 - \theta ) - k _ { - 1 } \theta - k _ { 2 } \theta \right]$$

Quizplus · Accepted Answer

The faradaic current ($i_F$) in an electrochemical reaction is related to the rates of the forward and reverse reactions at the electrode surface. For the given reaction, $i_F$ is proportional to the rate of the forward reaction $k_1(1-\theta)$, where $\theta$ is the surface coverage, and the rate of the reverse reaction $k_{-1}\theta$. The correct expression subtracts the reverse reaction rate from the forward reaction rate, as the reverse reaction works against the current flow generated by the forward reaction.

Consider a Reaction Given By -In the Same Mechanism, the Faradaic Current Is Given

Consider a Reaction Given By
-In the Same Mechanism, the Faradaic Current Is Given