Question 1

Consider an elementary reaction $M _ { s o l } ^ { 3 + }+ 3 e ^ { - }  \underset { k _ { - 1 } } { \stackrel { k _ { 1 } } { \rightleftarrows } } M  $

where M is a metal and 
 $$M _ { s o l } ^ { 3 + }$$ 
 is metal ion in solution. F denotes faraday constant. The net faradaic current density due to metal deposition is given by
A) $$F k _ { 1 } \left[ M _ { s o l } ^ { 3 + } \right]$$
B) $$F \left\{ k _ { 1 } \left[ M _ { s a l } ^ { 3 + } \right] - k _ { - 1 } \right\}$$
C) $$3 F \left( k _ { 1 } \left[ M _ { s o l } ^ { 3 + } \right] + k _ { - 1 } \right)$$
D) $$3 F \left\{ k _ { 1 } \left[ M _ { s a l } ^ { 3 + } \right] - k _ { - 1 } \right\}$$

Accepted Answer

The net faradaic current density for metal deposition in this reaction is given by the difference between the forward and reverse reaction rates, multiplied by the charge transferred per reaction (3 electrons in this case), and the Faraday constant. Hence, option D correctly represents this with the term $3 F \left\{ k _ { 1 } \left[ M _ { s o l } ^ { 3 + } \right] - k _ { - 1 } \right\}$, accounting for the three electrons transferred per metal ion reduced and the difference between the forward and reverse reaction rates.

Question 2

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 standard notation, the fractional surface coverage ($\theta$ ) of the adsorbed intermediate $$M _ { a t } ^ { + }$$ 
 is described by
A) $$\Gamma \frac { d \theta } { d t } = k _ { 1 } ( 1 - \theta ) - k _ { 2 } \theta$$
B) $$\Gamma \frac { d \theta } { d t } = k _ { 1 } ( 1 - \theta ) - k _ { - 1 } \theta - k _ { 2 } \theta + k _ { - 2 } ( 1 - \theta ) C _ { M _ { 2a d } ^ { 2 + } }$$
C) $$\Gamma \frac { d \theta } { d t } = k _ { 1 } ( 1 - \theta ) - k _ { - 1 } \theta$$
D) $$\Gamma \frac { d \theta } { d t } = k _ { 1 } ( 1 - \theta ) - k _ { - 1 } \theta - k _ { 2 } \theta + k _ { - 2 } C _ { M _ { s a } ^ { 2+ } }$$

Accepted Answer

The correct expression for the fractional surface coverage of the adsorbed intermediate $M_{ads}^+$ takes into account both the forward and reverse reactions for adsorption and desorption, as well as the formation and dissolution of $M_{ads}^+$ into the solution. Option B correctly includes all these processes: adsorption ($k_1(1-\theta)$), desorption ($-k_{-1}\theta$), formation ($-k_2\theta$), and dissolution ($+k_{-2}(1-\theta)C_{M_{2ad}^{2+}}$), where $\theta$ is the surface coverage, and $C_{M_{2ad}^{2+}}$ is the concentration of a species in the solution, which might be a typo or misinterpretation in the context of this question.

Question 3

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 steady-state fractional surface coverage of the intermediate is given by
A) $$\theta _ { ss } = \frac { k _ { 1  { dc} } } { k _ { 1 \dot {dc } } + k _ { 2 } }$$
B) $$\theta _ { s s } = \frac { k _ { 1 d c } } { k _ { 1 \dot { dc } } + k _ { - 1  \dot { dc } } + k _ { 2 } + k _ { - 2 } }$$
C) $\theta_{ss}=\frac{k_{1 d c}+k_{-2} C_{M^{+2}_{sol}}}{k_{1 d c}+k_{-1 d c}+k_{2}+k_{-2} C_{M^{+2}_{sol}} }$ 
D) $$\theta _ { ss } = \frac { k _ { 1 \dot { dc } } - k _ { - 1 \dot { dc } } } { k _ { 1 \dot { dc } } + k _ { - 1 \dot { dc } } + k _ { 2 } + k _ { - 2 } C _ { M _ { 2a i } ^ { 2+ } } }$$

Accepted Answer

The steady-state fractional surface coverage of the intermediate is given by:$$	heta_{ss}=\frac{k_{1 d c}+k_{-2} C_{M^{+2}_{sol}}}{k_{1 d c}+k_{-1 d c}+k_{2}+k_{-2} C_{M^{+2}_{sol}} }$$

Question 4

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) $$i _ { F } = F \left[ k _ { 1 } ( 1 - \theta ) - k _ { - 1 } \theta - k _ { 2 } \theta \right]$$

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.

Question 5

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 value of $$\frac { d \theta } { d E }$$ 
is given by
A) $\frac{\left(b_{1} k_{1 ak}\left(1-\theta_{s s}\right)-b_{-1} k_{-1ak} \theta_{ss}\right)}{\left(k_{1 dak}+k_{-1 ak}+k_{2}+k_{-2} C_{M_{2a}^{2}}+j \omega \Gamma\right)}$
B) $$\frac { \left( b _ { 1 } k _ { 1 \dot {dc } } - b _ { - 1 } k _ { - 1 \dot {dc } } \right) } { \left( k _ { 1 \dot { dc} } + k _ { - 1 \dot { dc } } + k _ { 2 } + k _ { - 2 } + j \omega \Gamma \right) }$$
C) $$\frac { \left( b _ { 1 } k _ { 1 \dot { d } } - b _ { - 1 } k _ { - 1 \dot { \alpha } } \right) } { \left( k _ { 1 \dot { \alpha } } + k _ { - 1 \dot { k } } + k _ { 2 } + j \Gamma \Gamma \right) }$$
D) $$\frac { \left( b _ { 1 } k _ { 1 \dot { dc } } \right) } { \left( k _ { 1 \dot { dc} } + k _ { - 1 \dot { dc } } + k _ { 2 } + j \omega \Gamma \right) }$$

Accepted Answer

The correct answer is A because it is the only option that includes all the necessary terms in the numerator and denominator. The numerator includes the terms that represent the rates of the forward and reverse reactions of the first step, while the denominator includes the terms that represent the rates of all the reactions in the mechanism.

Question 6

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 impedance (Z_F) is given by A) $$\left( Z _ { F } \right) ^ { - 1 } = F \left( b _ { 1 } k _ { 1 \dot { dc } } \left( 1 - \theta _ { s s } \right) - b _ { - 1 } k _ { - 1 \dot { \alpha } } \theta _ { ss } + \left( k _ { 1 \dot { dc } } + k _ { - 1 \dot { dc } } + k _ { 2 } \right) \frac { d \theta } { d E } \right)$$ B) $$\left( Z _ { F } \right) ^ { - 1 } = F \left( b _ { 1 } k _ { 1 d \dot { c } } \left( 1 - \theta _ { s s } \right) - b _ { - 1 } k _ { - 1 \dot { } } \theta _ { s s } - \left( k _ { 1 \dot { dc } } + k _ { - 1 \dot {dc } } + k _ { 2 } \right) \frac { d \theta } { d E } \right)$$ C) $$\left( Z _ { F } \right) ^ { - 1 } = F \left( b _ { 1 } k _ { 1 \dot { dc } } \left( 1 - \theta _ { s s } \right) - b _ { - 1 } k _ { - 1 \dot {dc } } \theta _ { s s } - \left( k _ { 1 d \dot {c } } + k _ { - 1 \dot { dc } } \right) \frac { d \theta } { d E } \right)$$ D) $$\left( Z _ { F } \right) ^ { - 1 } = F \left( b _ { 1 } k _ { 1 d \dot { c} } \left( 1 - \theta _ { s s } \right) - b _ { - 1 } k _ { - 1 \dot { dc } } \theta _ { ss } + \left( k _ { 1 d \dot { c} } + k _ { - 1 d \dot {c } } \right) \frac { d \theta } { d E } \right)$$

Accepted Answer

The answer of Consider a reaction given by \(\begin{array} {...

Question 7

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 above reaction, the rate constants are given by $$k _ { i } = k _ { i 0 } e ^ { b ^ { \prime } E }$$ where E is measured wrt equilibrium potential. Given that k₁₀ = 10^-7 mol cm^-2 s^-1, b₁ = 20 V^-1, b_-1 = -18 V^-1 , k₂ = 10^-5 mol cm^-2 s^-1, k_-2 = 10⁰ cm s^-1 and the concentration of $$M _ { s o l } ^ { 2 + }$$ as 25 mM, determine (i) k_-10 = ___________ mol cm^-2 s^-1 . At E = 0.3 V vs. "Equilibrium. potential", (ii) value of k_1dc = ___________mol cm^-2 s^-1, (iii) k_-1dc = ___________mol cm^-2 s^-1, (iv) steady state fractional surface coverage of the intermediate is _______ Remember that 1 M = 1 mol/lit = 10^-3 mol/ cm³.

Accepted Answer

The answer of Consider a reaction given by \(\begin{array} {...

Question 8

When kinetics is rate-limiting, the faradaic impedance of a simple electron transfer reaction can be modeled as
A) a simple resistor
B) a simple capacitor
C) a capacitor in parallel with a resistor
D) none of the above

Accepted Answer

When kinetics is rate-limiting, the faradaic impedance is dominated by the charge transfer resistance, which can be modeled as a simple resistor.

Question 9

Consider the reaction $$\begin{array} { l } P _ { s a } \stackrel { k _ { 1 } } { \longrightarrow } Q _ { s a } + e ^ { - } \ A _ { s i l } ^ { - } \underset { k _ { - 2 } } { \stackrel { k _ { 2 } } { \rightleftarrows } } A _ { \alpha i j } + e ^ { - } \ \end{array}$$ The kinetic parameters are given by kS1U1B110S1U1B0 = 10S1U1P1-5 S1S1P0cm sS1U1P1-1S1S1P0, bS1U1B11S1U1B0 = 20 VS1U1P1-1S1S1P0, kS1U1B120S1U1B0 = 10S1U1P1-6S1S1P0 cm sS1U1P1-1S1S1P0, bS1U1B12S1U1B0 = 20 VS1U1P1-1S1S1P0, kS1U1B1-20S1U1B0 =10S1U1P1-6S1S1P0 mol cmS1U1P1-2S1S1P0 sS1U1P1-1S1S1P0 , bS1U1B1-2S1U1B0 = -18 VS1U1P1-1S1S1P0 and $\varGamma$ = 10S1U1P1-8S1S1P0 mol cmS1U1P1-2S1S1P0. Solution species concentrations are given by [PS1U1B1solS1U1B0] = [QS1U1B1solS1U1B0] =[ $$A _ { s o l } ^ { - }$$ ] = 100 mM. -The equation for fractional surface coverage of A_ads is given by A) $$\frac { k _ { 2 \dot { dc } } } { k _ { 2 k \dot { dc } } + k _ { - 2 \dot { dc } } + k _ { 1 \dot { dc } } }$$ B) $$\frac { k _ { 2 dc } } { k _ { 2dc } + k _ { - 2 \dot {dc } } }$$ C) $\frac{k_{2 dc} C_{\mathcal{A}^-_{\mathrm{sol}}}}{k_{2 dc} C_{A^-_{sol}}+k_{-2 d c}+k_{1 d c} C_{P_{sol}}}$ D) $\frac{k_{2 dc} C_{A^-_{sol}}}{k_{2 dc} C_{A^-_{sol}}+k_{-2dc}}$

Accepted Answer

The fractional surface coverage of $A_{\text{ads}}$ is determined by the balance between its adsorption rate ($k_{2dc} C_{A^-_{sol}}$) and its desorption rate ($k_{-2dc}$). The presence of $k_{1dc}$ in the denominator of options A and C is incorrect because it pertains to a different reaction pathway that does not directly influence the adsorption-desorption equilibrium of $A_{\text{ads}}$. Option B is incorrect due to a typographical error in the subscript notation. Therefore, option D correctly represents the fractional surface coverage of $A_{\text{ads}}$ by considering only the processes directly involved in its adsorption and desorption.

Question 10

Consider the reaction $$\begin{array} { l } P _ { s a } \stackrel { k _ { 1 } } { \longrightarrow } Q _ { s a } + e ^ { - } \ A _ { s i l } ^ { - } \underset { k _ { - 2 } } { \stackrel { k _ { 2 } } { \rightleftarrows } } A _ { \alpha i j } + e ^ { - } \ \end{array}$$ The kinetic parameters are given by kS1U1B110S1U1B0 = 10S1U1P1-5 S1S1P0cm sS1U1P1-1S1S1P0, bS1U1B11S1U1B0 = 20 VS1U1P1-1S1S1P0, kS1U1B120S1U1B0 = 10S1U1P1-6S1S1P0 cm sS1U1P1-1S1S1P0, bS1U1B12S1U1B0 = 20 VS1U1P1-1S1S1P0, kS1U1B1-20S1U1B0 =10S1U1P1-6S1S1P0 mol cmS1U1P1-2S1S1P0 sS1U1P1-1S1S1P0 , bS1U1B1-2S1U1B0 = -18 VS1U1P1-1S1S1P0 and $\varGamma$ = 10S1U1P1-8S1S1P0 mol cmS1U1P1-2S1S1P0. Solution species concentrations are given by [PS1U1B1solS1U1B0] = [QS1U1B1solS1U1B0] =[ $$A _ { s o l } ^ { - }$$ ] = 100 mM. -At E_dc = 0.25 V vs. OCP, the value of fractional surface coverage of A_ads is given by, $\theta$_ss = _________

Accepted Answer

The answer of Consider the reaction \[\begin{array} { l }...

Question 11

Consider the reaction $$\begin{array} { l } P _ { s a } \stackrel { k _ { 1 } } { \longrightarrow } Q _ { s a } + e ^ { - } \ A _ { s i l } ^ { - } \underset { k _ { - 2 } } { \stackrel { k _ { 2 } } { \rightleftarrows } } A _ { \alpha i j } + e ^ { - } \ \end{array}$$ The kinetic parameters are given by kS1U1B110S1U1B0 = 10S1U1P1-5 S1S1P0cm sS1U1P1-1S1S1P0, bS1U1B11S1U1B0 = 20 VS1U1P1-1S1S1P0, kS1U1B120S1U1B0 = 10S1U1P1-6S1S1P0 cm sS1U1P1-1S1S1P0, bS1U1B12S1U1B0 = 20 VS1U1P1-1S1S1P0, kS1U1B1-20S1U1B0 =10S1U1P1-6S1S1P0 mol cmS1U1P1-2S1S1P0 sS1U1P1-1S1S1P0 , bS1U1B1-2S1U1B0 = -18 VS1U1P1-1S1S1P0 and $\varGamma$ = 10S1U1P1-8S1S1P0 mol cmS1U1P1-2S1S1P0. Solution species concentrations are given by [PS1U1B1solS1U1B0] = [QS1U1B1solS1U1B0] =[ $$A _ { s o l } ^ { - }$$ ] = 100 mM. -At the same potential (E_dc = 0.25 V vs. OCP), the total Faradaic current is given by, i_F_-_dc = ______________ mA cm^-2

Accepted Answer

The answer of Consider the reaction \[\begin{array} { l }...

Question 12

The polarization plot of an electrochemical system is given below. -If an impedance spectrum is acquired at E_dc = 0.7 V vs. OCP, then the polarization resistance is expected to be. A) negative B) positive C) can be either negative or positive D) zero.

Accepted Answer

The answer of The polarization plot of an electrochemical system...

Question 13

The polarization plot of an electrochemical system is given below. -In the same system, if an impedance spectrum is acquired at E_dc = 1.2 V vs. OCP, then the polarization resistance is expected to be. A) negative B) positive C) can be either negative or positive D) zero.

Accepted Answer

Polarization resistance is a measure of the resistance that an electrochemical system presents to the flow of charge during an electrochemical reaction. It is inherently a positive quantity because resistance cannot be negative. Negative resistance would imply that the system supplies energy rather than dissipates it, which contradicts the passive nature of polarization resistance in electrochemical systems.

Question 14

Consider the reaction $$\begin{array} { l } M \underset { k _ { - 1 } } { \stackrel { k _ { 1 } } { \rightleftarrows } } M _ { a ds } ^ { + } + e ^ { - } \ M _ { a ds } ^ { + } + M \stackrel { k _ {2 } } { \longrightarrow } M _ { ads} ^ { + } +M _ { sol } ^ { + } + e ^ { - } \end{array}$$ The kinetic parameters are given by kS1U1B110S1U1B0 = 10S1U1P1-S1S1P0S1U1P19S1S1P0S1U1P1 S1S1P0mol cmS1U1P1-2S1S1P0 sS1U1P1-1S1S1P0, bS1U1B11S1U1B0 = 10 VS1U1P1-1S1S1P0, kS1U1B1-1S1U1B0S1U1B10S1U1B0 = 10S1U1P1-S1S1P0S1U1P18S1S1P0 mol cmS1U1P1-S1S1P0S1U1P12S1S1P0 sS1U1P1-1S1S1P0, bS1U1B1-1S1U1B0 = -10 VS1U1P1-1S1S1P0, kS1U1B120S1U1B0 = 10S1U1P1-8S1S1P0 mol cmS1U1P1-2S1S1P0 sS1U1P1-1S1S1P0, bS1U1B12S1U1B0 = 10 VS1U1P1-1S1S1P0 and$\varGamma$ = 10S1U1P1-8S1S1P0 mol cmS1U1P1-2S1S1P0. -At E_dc = 0.1 V vs. OCP, the fractional surface coverage of M_ads is given by, $\theta$_ss = _________

Accepted Answer

The answer of Consider the reaction 
 \[\begin{array} { l...

Question 15

Consider the reaction 
 $$\begin{array} { l } 
M \underset { k _ { - 1 } } { \stackrel { k _ { 1 } } { \rightleftarrows } } M _ { a ds } ^ { + } + e ^ { - } \
M _ { a ds } ^ { + } + M  \stackrel { k _ {2 } } { \longrightarrow } M _ { ads} ^ { + }  +M _ { sol } ^ { + } + e ^ { - }
\end{array}$$
 
The kinetic parameters are given by kS1U1B110S1U1B0 = 10S1U1P1-S1S1P0S1U1P19S1S1P0S1U1P1 S1S1P0mol cmS1U1P1-2S1S1P0 sS1U1P1-1S1S1P0, bS1U1B11S1U1B0 = 10 VS1U1P1-1S1S1P0, kS1U1B1-1S1U1B0S1U1B10S1U1B0 = 10S1U1P1-S1S1P0S1U1P18S1S1P0 mol cmS1U1P1-S1S1P0S1U1P12S1S1P0 sS1U1P1-1S1S1P0, bS1U1B1-1S1U1B0 = -10 VS1U1P1-1S1S1P0, kS1U1B120S1U1B0 = 10S1U1P1-8S1S1P0 mol cmS1U1P1-2S1S1P0 sS1U1P1-1S1S1P0, bS1U1B12S1U1B0 = 10 VS1U1P1-1S1S1P0 and$\varGamma$ = 10S1U1P1-8S1S1P0 mol cmS1U1P1-2S1S1P0.
-To model a system with the above mechanism, the minimum number of dc potentials at which EIS data should be taken to avoid an infinite number of solutions is __________

Accepted Answer

The answer of Consider the reaction 
 \[\begin{array} { l...

Question 16

In a kinetic limited electrochemical reaction, the number of loops appearing in the complex plane plot of EIS can be related to the
A) number of steps in the mechanism
B) the number of reacting species
C) the number of product species
D) the number of adsorbed intermediates

Accepted Answer

The answer of In a kinetic limited electrochemical reaction, the...

Question 17

Consider the reaction 
 $$\begin{array} { l } 
M \stackrel { k _ { 1 } } { \longrightarrow } M _ { a ds} ^ { + } + e ^ { - } \
M _ { a ds } ^ { + } \stackrel { k _ { 2 } } { \longrightarrow } M _ { s o l } ^ { + }
\end{array}$$ . Frumkin isotherm is used to describe the adsorption. The kinetic parameter values are kS1U1B110S1U1B0 = 10S1U1P1-9 S1S1P0mol cmS1U1P1-2S1S1P0 sS1U1P1-1S1S1P0, bS1U1B11S1U1B0 = 10 VS1U1P1-1S1S1P0, g = 15, $\beta$S1U1B11S1U1B0 = -0.5 , kS1U1B120S1U1B0 = 10S1U1P1-8S1S1P0 mol cmS1U1P1-2S1S1P0 sS1U1P1-1S1S1P0, $\beta$S1U1B12S1U1B0 = +0.5 and $\varGamma$ = 10S1U1P1-8S1S1P0 mol cmS1U1P1-2S1S1P0.
-The maximum number of 'loops' that can appear in complex plane plots is _____________

Accepted Answer

The answer of Consider the reaction 
 \[\begin{array} { l...

Question 18

Consider the reaction 
 $$\begin{array} { l } 
M \stackrel { k _ { 1 } } { \longrightarrow } M _ { a ds} ^ { + } + e ^ { - } \
M _ { a ds } ^ { + } \stackrel { k _ { 2 } } { \longrightarrow } M _ { s o l } ^ { + }
\end{array}$$ . Frumkin isotherm is used to describe the adsorption. The kinetic parameter values are kS1U1B110S1U1B0 = 10S1U1P1-9 S1S1P0mol cmS1U1P1-2S1S1P0 sS1U1P1-1S1S1P0, bS1U1B11S1U1B0 = 10 VS1U1P1-1S1S1P0, g = 15, $\beta$S1U1B11S1U1B0 = -0.5 , kS1U1B120S1U1B0 = 10S1U1P1-8S1S1P0 mol cmS1U1P1-2S1S1P0 sS1U1P1-1S1S1P0, $\beta$S1U1B12S1U1B0 = +0.5 and $\varGamma$ = 10S1U1P1-8S1S1P0 mol cmS1U1P1-2S1S1P0.
-In the above question, the total number of electrical elements required to model the reaction (i.e., only the Faradaic part, without considering the double-layer capacitor or solution resistance) is ____

Accepted Answer

The answer of Consider the reaction 
 \[\begin{array} { l...

Question 19

An impedance spectrum of an electrochemical reaction could be modeled well by the equivalent circuit given below. The solution resistance was negligible, mass transfer was rapid, and there was not film formation on the surface.
  If a mechanism is proposed to explain the results, what is the minimum number of adsorbed intermediate species necessary to model the data? __________

Accepted Answer

The answer of An impedance spectrum of an electrochemical reaction...

Question 20

Choose the mechanism(s), which, for appropriate values of kinetic parameters and dc potentials, can yield negative differential impedance.
A) $$\begin{array} { l } 
M \rightarrow M _ { a ds } ^ { + } + e ^ { - } \
\
M _ { a ds } ^ { + } \rightarrow M _ { a d s } ^ { 2 + } + e ^ { - } \
\
M _ { a d s } ^ { + } \rightarrow M _ { s o l } ^ { 2 + }
\end{array}$$
B) $$M \rightarrow M _ { s o l } ^ { + } + e ^ { - }$$
C) $$\begin{array} { l } 
M \rightarrow M _ { a d s } ^ { + } + e ^ { - } \
\
M _ { a ds } ^ { + } \rightarrow M _ { \text {sol } } ^ { 2 + } + e ^ { - }
\end{array}$$
D) $$\begin{array} { l } 
M \rightarrow M _ { a ds } ^ { + } + e ^ { - } \
\
M _ { a d s } ^ { + } + M ^ {  } \rightarrow M _ { a d s } ^ { + } + M _ { s o l } ^ { + }
\end{array}$$

Accepted Answer

The answer of Choose the mechanism(s), which, for appropriate values...

Quiz 5: Mechanistic Analysis

When kinetics is rate-limiting, the faradaic impedance of a simple electron transfer reaction can be modeled as

The polarization plot of an electrochemical system is given below. -If an impedance spectrum is acquired at E_dc = 0.7 V vs. OCP, then the polarization resistance is expected to be.

The polarization plot of an electrochemical system is given below. -In the same system, if an impedance spectrum is acquired at E_dc = 1.2 V vs. OCP, then the polarization resistance is expected to be.

In a kinetic limited electrochemical reaction, the number of loops appearing in the complex plane plot of EIS can be related to the

Choose the mechanism(s) , which, for appropriate values of kinetic parameters and dc potentials, can yield negative differential impedance.

Quiz 5: Mechanistic Analysis

When kinetics is rate-limiting, the faradaic impedance of a simple electron transfer reaction can be modeled as

The polarization plot of an electrochemical system is given below. -If an impedance spectrum is acquired at Edc = 0.7 V vs. OCP, then the polarization resistance is expected to be.

The polarization plot of an electrochemical system is given below. -In the same system, if an impedance spectrum is acquired at Edc = 1.2 V vs. OCP, then the polarization resistance is expected to be.

In a kinetic limited electrochemical reaction, the number of loops appearing in the complex plane plot of EIS can be related to the

Choose the mechanism(s) , which, for appropriate values of kinetic parameters and dc potentials, can yield negative differential impedance.

The polarization plot of an electrochemical system is given below. -If an impedance spectrum is acquired at E_dc = 0.7 V vs. OCP, then the polarization resistance is expected to be.

The polarization plot of an electrochemical system is given below. -In the same system, if an impedance spectrum is acquired at E_dc = 1.2 V vs. OCP, then the polarization resistance is expected to be.