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Calculate the emf of the cell at 298 k pt h2

Web12. All chemical reactions used in galvanic cells are redox reactions. 13. The amount of the product formed by the passage of 1 coulomb of electricity through electrolyte is called. electrochemical equivalent of the substance. 14. The redox reaction involved in galvanic cell is a non- spontaneous process. 15. WebApr 7, 2024 · Hint: First of all we will write Nernst equation. We will write the reaction of anode and cathode separately. Then we will add them both to get the complete cell equation. Then we will try to find the value of concentration from the cell equation and then we will find the value of ${K_a}$ by using the formula $[{H^ + }] = {H^ + } + {A^ - }$

(a) Write the cell reaction and calculate the e.m.f. of the following ...

WebJul 19, 2024 · Calculate the potential of the cell at 298 K : Cd/Cd2+ (0.1M) H+ (0.2M)/Pt, H2 (0.5 atm) asked Dec 11, 2024 in Chemistry by sforrest072 (129k points) electro chemistry concepts; 0 votes. ... Calculate EMF of the cell Pt, H2 (0.1 atm) solution (pH = 4) M /100 KC1 solution (saturated with AgC1) Ag and equilibrium constant. ... WebWrite the Nernst Equation and Emf of the Following Cells at 298 K Exercise Chapter 3 Electrochemistry Chemistry Class. Write the Nernst equation and emf of the following cells at 298 K Chapter 3: Electrochemistry Chemistry Class 12 solutions are developed for assisting understudies with working on their score and increase knowledge of the subjects. sleep evaluation center baytown tx https://globalsecuritycontractors.com

Electrochemical Cell EMF Example Problem - ThoughtCo

Web(a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K : Sn(s) ` Sn^(2+) (0.004M) H^(+) (0.020M) H_(2)(g)` (1 bar ) Pt(s) (Gi... WebDec 10, 2024 · The general Nernst equation correlates the Gibb's Free Energy D G and the EMF of a chemical system known as the galvanic cell. For the reaction. where R, T, Q and F are the gas constant (8.314 J mol -1 K -1 ), temperature (in K), reaction quotient, and Faraday constant (96485 C) respectively. Thus, we have. WebDec 10, 2024 · The electromotive force (EMF) is the maximum potential difference between two electrodes of a galvanic or voltaic cell. This quantity is related to the tendency for an … sleep ethnicity medication use

(a) Write the cell reaction and calculate the e.m.f. of the following ...

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Calculate the emf of the cell at 298 k pt h2

Calculate the emf of the cell Cd Cd^2+ (0.10M)1 H^+ (0.20M) Pt, H2…

WebQuestion From – KS Verma Physical Chemistry Class 12 Chapter 03 Question – 086 ELECTROCHEMISTRY CBSE, RBSE, UP, MP, BIHAR BOARDQUESTION TEXT:-Calculate `EMF`... WebTo use this online calculator for EMF of Due Cell, enter Standard Reduction Potential of Cathode (Ecathode) & Standard Oxidation Potential of Anode (Eanode) and hit the …

Calculate the emf of the cell at 298 k pt h2

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WebOct 31, 2024 · Write the cell reaction and calculate the e.m.f. of the following cell at 298 K : Sn (s) Sn2+ (0.004 M) ... Pt (s) (Given, Eosn^ 2+/sn = - 0.14V) ... Pt (s) (Given, E o sn^ 2+/sn = - 0.14V) cbse; class-12; Share It On Facebook Twitter Email. 1 Answer +1 vote . answered Oct ... WebCalculate the cell EMF in mV for `Pt H_(2)(1atm) HCl(0.01M) AgCl(s) Ag(s)` at 298 K if `triangleG_(1)^(@)` values are at `25^(@)C` `-109.56(kJ)/(mol)` for `...

http://rkt.chem.ox.ac.uk/tutorials/emf/emf.html Webthis dilution at 298 K. Calculate the electrode potential. Given E0zn2+/zn = – 0.76 V. III. SHORT ANSWER (3M) 1. Represent the cell in which the following reaction takes place.The value of E˚ for the cell is 1.260 V. What is the value of Ecell ? 2. A voltaic cell is set up at 25 °C with the following halfcells: Al/Al3+(0.001 M) and Ni/Ni2 ...

WebAssuming that apart from hydrogen and its ion everything is in its standard state, what is (a) the emf of the above cell at pH 9, 1 atm pressure of H 2, and 298 K, (b) the minimum temperature at which hydrogen will reduce nickel ions to nickel at pH 9, p H2 = 1 atm., and (c) the minimum pressure of H 2 required to do the same thing at pH 9 and ... WebNov 2, 2024 · A voltaic cell is setup at 25°C with the half cells Ag^+ (0.001 M) Ag and Cu^2+ (0.10 M) Cu. What should be its cell potential ? asked Nov 2, 2024 in Chemistry by Richa ( 61.0k points)

WebThe cell should be represented as Pt H2 (1 bar), H+ (0.03 M) Br (0.01 M) Br2(l), Pt. ... Write the Nernst equation and emf of the following cells at 298 k. Pt(s) ... (aq) → 2Fe …

WebDec 11, 2024 · The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500. asked Dec 11, 2024 in Chemistry by sforrest072 ( 129k points) electro chemistry concepts sleep every other dayWebCalculate the emf at 298 K for the cell Pt H2 (1 atm) H+ (0.1 M) H+ (0.2 M) H2 (10 atm) Pt. answer is -0.0118 thank u This problem has been solved! You'll get a detailed … sleep essentials mattresses raleigh ncWebAt 298 K, the solubility product constant for Pb(IO3)2 is 2.6 1013, and the standard reduction potential of the Pb2+(aq) to Pb(s) is 0.126 V. (a) Find the standard potential of the half-reaction Pb(IO3)2(s)+2ePb(s)+2IO3(aq) (Hint: The desired half-reaction is the sum of the equations for the solubility product and the reduction of Pb2+. sleep essentials mattress reviewsleep everywhere pillowWeba Write the cell reaction and calculate the e.m.f of the following cell at 298 K :Sns Sn 2+0.004 M H +0.020 M H 2g1 bar Pt s Given : E Sn 2+ / Sn ∘= 0.14 V b Give reasons:i … sleep exerts lasting effectsWebEMF of a Cell. The electromotive force of a cell or EMF of a cell is the maximum potential difference between two electrodes of a cell. It can also be defined as the net voltage … sleep excurtion formWebCalculate the cell EMF in mV for `Pt H_(2)(1atm) HCl(0.01M) AgCl(s) Ag(s)` at 298 K if `triangleG_(1)^(@)` values are at `25^(@)C` `-109.56(kJ)/(mol)` for `... sleep exception form texas