An air-gap parallel plate capacitor whose plates have an area of A=0,38m² are separated by adistance d=0,61m and are connected to a battery of V₁ = 1,8 V. The space between the parallelplates is partially filled (x=0,26m) with some material of dielectric constant K=4,56. Find theelectric potential energy stored in the capacitor in terms of permitivity of free-space €0. Expressyour answer using 2 decimal places.

Answered: An air-gap parallel plate capacitor…

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter21: Potentiometry

Section: Chapter Questions

Problem 21.23QAP

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Calculate the cell potential (in V) of the following cell with transference at 25°C: Pb(s) PbSO4(s), CuSO4(m = 0.2, y = 0.110) CuSO4(m = 0.02, y = 0.320), PbSO4(s) Pb(s) The transport number of Cu²+ is 0.370. Express your answer in three significant figures.
An electrochemical cell consists of a Pt|H+ (aq,1.00 M)|H₂(g) cathode connected to a Pt|H+ (aq)|H₂(g) anode in which the H+ concentration is that of a buffer consisting of a weak acid, HA(0.181 M), mixed with its conjugate base, A (0.157 M). The measured cell voltage is Eºcell = 0.258 V at 25 °C, with PH₂ = 1.00 atm at both electrodes. Calculate the pH in the buffer solution and the Ka of the weak acid. Submit Answer Retry Entire Group pH = K₂ = 9 more group attempts remaining
The exchange-current density for a Pt|Fe3+,Fe2+ electrode is 2.5 mA cm−2. The standard potential of the electrode is +0.77 V. Derive an expression for the current flowing through an electrode of surface area 1.0 cm2 as a function of the potential applied to the electrode; assume standard conditions.
9:01 A elearning.alsafwa.edu.iq الوقت المتبقي 0:43:09 1 Jlgw غير مجاب علیه بعد الدرجة من 3.00 علم هذا السؤال When a current of 0.4 A flows for 1.5 micro-seconds in a coper wire, estimate the number of electrons crossing the cross-section of the wire No. of electron= 4.8×10-19 electron No. of electron= 1.875×1012 electron No. of electron= 3x10-7 electron No. of electron= 4.8×10-26 electron الصفحة التالية الإعلانات
In the temperature range 0-90 ºC , the potential difference of the following cell:Pt (s) | H2 (g, f = 1) | HCl (aq, m = 0.1) | AgCl (s) | Ag (s)changes as follows with the temperature:Ecell (V) = 0.35510 – 0.3422x10-4t - 3.2347x10-8t2 + 6.314x10-9t3 (t being the temperature in ºC) Write the reaction and calculate ΔG, ΔH and ΔS at 90 ºC.Result: -34388 J/mol, 10.94 J/Kmol, -30.42 KJ/mol
Although the hydrogen electrode may be conceptually the simplest electrode and is the basis for the choice of reference potential in electrochemical systems, it is cumbersome to use. Therefore, several substitutes for it have been devised. One of these alternatives is the quinhydrone electrode (quinhydrone, Q ⋅ QH2, is a complex of quinone, C6H4O2 = Q, and hydroquinone, C6H4O2H2 = QH2), where the concentrations of Q ⋅ QH2 and QH2 are equal to each other. The electrode half-reaction is Q(aq) + 2 H+(aq) + 2 e− → QH2(aq), E⦵ = +0.6994 V. If the cell Hg(s)|Hg2Cl2(s)|HCl(aq)|Q ⋅ QH2|Au(s) is prepared, and the measured cell potential is +0.190 V, what is the pH of the HCl solution?
Consider the following cell at 25°C: 2+ Pb|Pb²+(a = 0.1)|| Hg₂²+(a 0.1)|Hg(1)|Pt The data is as follows: Half cell Pb|Pb²+ Pt|Hg(1)|Hg2+ Equation Pb²+ + 2e- Pb Hg₂2+ + 2e- → 2Hg(1) Calculate the AG value for the cell Select one: O A. -177965 J/mole B. -129329 J/mol C. +88982 J/mol OD. -88982 J/mol E°(V) -0.126 +0.79
Consider the following electrochemical cell: Pt|H,(8,p°)|HCI(aq,b)|Hg.Cl,(s)|Hg() The potential of this cell, in relation to different molalities (b), has been measured with high precision at 25°C. The following results have been obtained: b/mmol.kg E/V I (ionic strength) Y: (DHLL) |Y: (linear fit) 1.6077 3.0769 5.0403 7.6938 10.9474 0.60080 0.56825 0.54366 0.52267 0.50532 (i) Write the half-reactions for the cell, and show the standard potential of each half-reaction. (ii) Determine the standard cell potential through simple calculation.
Consider the following electrochemical cell: Pt | H₂(g,pᵒ) | HCl(aq,b) | Hg₂Cl₂(s) | Hg(1) The potential of this cell, in relation to different molalities (b), has been measured with high precision at 25°C. The following results have been obtained: b/mmol.kg¹ E/V I (ionic strength) Y+ (DHLL) Y+ (linear fit) 1.6077 0.60080 (ii) (iii) (iv) 3.0769 0.56825 5.0403 0.54366 7.6938 0.52267 10.9474 0.50532 (i) Write the half-reactions for the cell, and show the standard potential of each half-reaction. Determine the standard cell potential through simple calculation. Write the cell reaction for this cell. Write the Nernst equation for this cell in terms of molality (b) and the mean activity coefficient (y) of HCI. (v) (vi) (vii) Write the Nernst equation for this cell only in terms of molality (b), where the mean activity coefficient (Y) of HCI does not feature. (viii) Rearrange the equation in (vii) and make a suitable plot, using the data in the table, to determine the standard cell…
The resistances of a series of aqueous NaCl solutions, formed by successive dilution of a sample, were measured in a cell with cell constant (the constant C in the relation κ = C/R) equal to 0.2063 cm-1 and the following values were found at 25oC: c/(mol dm−3) 0.00050 0.0010 0.0050 0.010 0.020 0.050 R/W 3314 1669 342.1 174.1 89.08 37.14 Find the molar conductance of NaCl at 25oC at infinite dilution from the graph.
Consider the following electrochemical cell: Pt | H₂(g,pᵒ) | HCl(aq,b) | Hg₂Cl₂(s) | Hg(1) The potential of this cell, in relation to different molalities (b), has been measured with high precision at 25°C. The following results have been obtained: b/mmol.kg¹ E/V I (ionic strength) Y+ (DHLL) Y+ (linear fit) (1) (ii) (iii) 1.6077 0.60080 3.0769 0.56825 5.0403 0.54366 7.6938 0.52267 10.9474 0.50532 Write the half-reactions for the cell, and show the standard potential of each half-reaction. Determine the standard cell potential through simple calculation. Write the cell reaction for this cell. (iv) Write the Nernst equation for this cell in terms of molality (b) and the mean activity coefficient (y) of HCI.
The potential of electrodes Cu2+ (aq) | Cu(s) is ϕ1 = 0.337 V , and Cu2+ (aq), Cu+ (aq) | Pt is ϕ2 = 0.153 V , respectively. a) Calculate the value of ϕ0 for electrode Cu+(aq) | Cu(s). [Hint: Write half reactions for the above two electrodes, then derive the target electrode.]b) Calculate the equilibrium constant Ka0 for the reaction Cu(s) + Cu2+ (aq) = 2Cu+ (aq) at 298.15K. [Hint: Based on the reaction, design a cell with two electrodes and verify that the two half reactions will give you the target reaction. Then calculate the cell potential.]

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