Before You Begin

Question #44800

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following galvanic cell operating at $298\text{ K}$ :

$\ce{Cu(s) | Cu^{2+}(aq, 0.1 M) || Ag+(aq, 0.2 M) | Ag(s)}$

As the cell discharges, the instantaneous rate of disappearance of $\ce{Ag+(aq)}$ is $4 \times 10^{-3}\text{ M s}^{-1}$ . Assuming the volume of the half-cells remains constant and activity coefficients are unity, what is the rate of change of the cell potential, $\frac{dE_{cell}}{dt}$ , at this instant?

Options

A.
$-0.03 \frac{RT}{F} \text{ V s}^{-1}$
B.
$-0.04 \frac{RT}{F} \text{ V s}^{-1}$
C.
$-0.02 \frac{RT}{F} \text{ V s}^{-1}$
D.
$-0.06 \frac{RT}{F} \text{ V s}^{-1}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Stoichiometric Coefficient Division|Nernst Equation Concentration Dependence

Why This Question Is Tricky

Common Mistakes

This problem tests the synthesis of Chemical Kinetics and Electrochemistry.

The primary trap is the 'Stoichiometric Coefficient Division' in kinetics: students often fail to divide the rate of disappearance of $\ce{Ag+}$ by its stoichiometric coefficient (2) to find the rate of appearance of $\ce{Cu^{2+}}$ .

A secondary 'Thermodynamic Trap' involves the Nernst equation's reaction quotient ( $Q$ ); students may forget the exponent for $[\ce{Ag+}]^2$ or fail to correctly apply the chain rule when differentiating the logarithmic term, conflating the thermodynamic dependence with the kinetic rate expression.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Chemical Kinetics - Class 12
Concept	Stoichiometric Coefficient Division
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free