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?

Like equilibrium constants, E values are temperature dependent. At 80°C, E for the cell cell cell diagram shown is 0.18 V. Pt | H₂(g) | HCl(aq) || AgCl(s) | Ag(s) The corresponding cell reaction is H₂(g) + 2AgCl(s) + 2Ag(s) + 2H¹ (aq) + 2C1 (aq) Calculate the equilibrium constant for this reaction at 80°C. x 10 (Enter your answer in scientific notation).

Calculate the cell potential for the galvanic cell in which the given reaction occurs at 25 °C, given that 'Ce'[Sn'2] - 0.0652-text M[Sn²+] = 0.0652 M, 'Ce'[Co'3] - 0.0409-text M[Co³+] = 0.0409 M, 'Ce'[Sn'4] - 0.00496 'text' [Sn+] = 0.00496 M, and "C.[Co'2] - 0.01150 'text'[Co²+] = 0.01150 M. Standard reduction potentials can be found in this table. Sn²+(so much) + 2 Co³* (so much) = Sn* "(so much) + 2C0²+ (so much) E = V

Permanganate ion can oxidize sulfite to sulfate in basic is 0.000 V at pH 5 0.00, what is the potential of the same electrode at pH 5 7.00?17.52. Glucose Metabolism The standard potentials for the reduction of nicotinamide adenine dinucleotide (NAD1) and oxaloacetate (reactants in the multistep metabolism of glucose) are as follows: NAD1(aq) 1 2 H1(aq) 1 2 e2 S NADH(aq) 1 H1(aq)E° 5 20.320 V Oxaloacetate(aq) 1 2 H1(aq) 1 2 e2 S malate(aq)E° 5 20.166 V Calculate the standard potential for the following reaction: Oxaloacetate(aq) 1 NADH(aq) 1 H1(aq) S malate(aq) 1 NAD1(aq) Calculate the equilibrium constant for the reaction at 298 K. in the reaction vessel in which PClO 5 2.0 atm and PCl 5 1.00 atm. Calculate [ClO32]2 if, at equilibrium, 2 (g) partial pressure of NO 5 0.00150 atm.17.56. Chlorine dioxide (ClO2) is produced by the following reaction of chlorate (ClO 2) with Cl2 in acid solution: 3 2 ClO 2(aq) 1 2 Cl2(aq) 1 4 H1(aq) S3 2 ClO2(g) 1 Cl2(g) 1 2 H2O(/)…