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#cell potential

40 public questions tagged with this topic.

Why does the emf of a cell depend on its chemical composition rather than its size?

**Drift velocity** v_d = I/(n e A), I current (A), n number density of conduction electrons (m⁻³) ≈8.5×10²⁸ m⁻³ for copper, e =1.6×10⁻¹⁹ C, A cross-sectional area (m²). Typical v_d ≈10⁻⁴ m/s for 1 A in mm² wire, slow despite fast signal propagation due to electric field establishment. Emf is the potential difference generated by chemical reactions between electrodes and electrolyte, determined by the materials’ electrochemical properties, not physical dimensions like size. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and P = I²R,

Ref: NCERT > Physics Book > Current Electricity > Electric Current, Drift Velocity and Mobility

What is the EMF of a ll with Zn(s) | Zn²⁺(0.05 M) || H⁺(0.1 M) | H₂(g, 1 bar) | Pt(s) at 298 K, given E_{Zn^{2+/Z

Given: What is the EMF of a ll with Zn(s) | Zn²⁺(0.05 M) || H⁺(0.1 M) | H₂(g, 1 bar) | Pt(s) at 298 K, given E_{Zn^{2+/Zn⁰ = -0.76 V and E_{H^{+/H_2⁰ = 0 V ? These values define the system as per NCERT data. Formula: E_{ll⁰ = 0 - (-0.76) = 0.76 V, n = 2. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: E_{ll = E_{ll⁰ - 0.059/2 log frac[Zn^{2+][H^{+]² . Q = 0.05/(0.1)² = 5, log Q = 0.699 . E_{ll = 0.76 - 0.059/2 × 0.699 = 0.76 - 0.0206 = 0.7394 V approx 0.74 V . Result: The computed value matches the expected outcome and confi

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

What is the EMF of a ll with Ni(s) | Ni²⁺(0.005 M) || Ag⁺(0.001 M) | Ag(s) at 298 K, given E_{Ni^{2+/Ni⁰ = -0.25 V and E

Given: What is the EMF of a ll with Ni(s) | Ni²⁺(0.005 M) || Ag⁺(0.001 M) | Ag(s) at 298 K, given E_{Ni^{2+/Ni⁰ = -0.25 V and E_{Ag^{+/Ag⁰ = 0.80 V ? These values define the system as per NCERT data. Formula: E_{ll⁰ = 0.80 - (-0.25) = 1.05 V, n = 2. This is standard NCERT relation. Substitution & Calculation: E_{ll = E_{ll⁰ - 0.059/2 log frac[Ni^{2+][Ag^{+]² . Q = 0.005/(0.001)² = 5000, log Q = 3.699 . E_{ll = 1.05 - 0.059/2 × 3.699 = 1.05 - 0.109 = 0.941 V approx 0.94 V . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.

Totipotency refers to the ability of a cell to:

Totipotency denotes maximal developmental potential to generate organism in entirety including embryonic lineages forming all fetal tissues plus extraembryonic tissues such as trophectoderm forming placenta and primitive endoderm forming yolk sac membranes. In mammals zygote and early blastomeres up to 2-cell stage retain totipotency as demonstrated by blastomere isolation producing viable twins and tetraploid complementation. Pluripotency limited to embryonic tissues excluding trophectoderm seen in inner cell mass embryonic stem cells forming teratoma. Unipotent limited to single lineage like

Ref: Condic, Totipotency and Pluripotency, Stem Cell Reviews 2014, Chapter on Zygote Potential.

A cell operates with Ecell = 1.50 V at 298 K when [Anode] = 0.001 M and [Cathode] = 0.1 M . If E°cell = 1.44 V , what is

Ecell = E°cell - (0.059/n) log Q , 1.50 = 1.44 - (0.059/2) log Q . 0.06 = -0.0295 log Q , log Q = -(0.06/0.0295) = -2.034, Q = 10⁻².⁰³⁴ ≈ 0.0092 . Since Q = ([Anode]/[Cathode]) = (0.001/0.1) = 0.01 , matches approximately.

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement