Question
A button cell used in watches functions as following: If half cell potentials are: The cell potential will be:
0.84 V
1.34 V
1.10 V
0.42 V
In the given button cell reaction, Zinc () is oxidised to and silver oxide () is reduced to silver (). Therefore, the zinc electrode acts as the anode and the silver electrode acts as the cathode. The standard cell potential () is calculated using the formula: Given the standard reduction potentials: Substituting these values into the formula: The standard cell potential is .
This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Electrochemistry. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.
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A cell reaction becomes spontaneous when:
In a typical fuel cell, the reactants (R) and products (P) are:
The cell reaction of an electrochemical cell is $Cu^{2+}(C_1) + Zn \rightarrow Cu + Zn^{2+}(C_2)$. The change in free energy will be the function of:
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The number of Faradays (F) required to produce $20 \text{ g}$ of calcium from molten $\text{CaCl}_2$ (Atomic mass of Ca = $40 \text{ g mol}^{-1}$) is:
For a cell involving one electron $E^\ominus_{\text{cell}} = 0.59 \text{ V}$ at $298 \text{ K}$. The equilibrium constant for the cell reaction is: [Given that $\frac{2.303 RT}{F} = 0.059 \text{ V}$ at $T = 298 \text{ K}$]
The value of $E^{\circ}_{cell}$ for the following reaction is: $Cu^{2+} + Sn^{2+} \rightarrow Cu + Sn^{4+}$ (Given, equilibrium constant is $10^6$)
The three cells with their $E^{\circ}_{cell}$ values are given below: 1. $Fe|Fe^{2+}||Fe^{3+}|Fe \ ; \ 0.404\text{ V}$ 2. $Fe|Fe^{2+}||Fe^{3+}, Fe^{2+}|Pt \ ; \ 1.211\text{ V}$ 3. $Fe|Fe^{3+}||Fe^{3+}, Fe^{2+}|Pt \ ; \ 0.807\text{ V}$ The standard Gibbs free energy change values for three cells are, respectively: (F represents the charge on $1\text{ mole}$ of electrons.)
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