🗊Презентация Electrochemistry. Oxidation-reduction equilibrium in water solutions

Electrochemistry Oxidation-reduction equilibrium in water solutions.

oxidation reductant oxidant + ne reduction ΔGo = - 2,3·RT·lgK – chemical work ΔGo = - nF؏o - electrical work ؏= E2 – E1 E – red-ox potential

Criteria for spontaneous red-ox reactions Mn+7 + 5Fe+2 Mn+2 + 5Fe+3 EoMn+7/Mn+2 = 1,51 В; EoFe+3/Fe+2 = 0,77 В EoMn+7/Mn+2 > EoFe+3/Fe+2 Mn+7 + 5ē → Mn+2 oxidant Fe+2 - ē → Fe+3 reductant ΔGo = -nF؏o ΔGo < 0 for spontaneous reaction ؏o = 1,51 – 0,77 = 0,74 V ΔGo = -5·96500·0,74 = -357050 J = -357,050 kJ.

Red-ox potentials of biological systems reductant oxidant + 2e + 2 H+ ΔGo = -2F؏o Oxidant - acceptor of electrons and protons

Electrodes

1 type-electrodes Me Me+n – metal electrode Me Me+n + ne Pt(H2) H+ - hydrogen electrode H2 2H+ + 2e

Hydrogen electrode

2 type-electrodes Hg Hg2Cl2, KCl – calomel electrode Ag AgCl, KCl – silver chloride electrode Ag Ag+ + Cl- In saturated KCl solutions ESC = const.=0.202V Ecal. = const. = 0.244V

Silver chloride and calomel electrodes

Redox electrodes Pt | ox, red

Quinhydrone electrode Pt (QH) H+

Nernst equation for electrode potential ΔGo = - 2,3·RT·lgK – chemical work ΔGo = - nF (E0 – E) - electrical work Eox/red = E0ox/red + (2,3RT/nF) lgK 2,3RT/F = 0.059 at T = 2890K, F= 96500, R=8.31 Eox/red = E0ox/red + (0.059/n) lg[ox]/[red] E0ox/red – standard potential

Hydrogen electrode Pt (H2) | H+ ½ Н2 Н+ + ē Eox/red = E0ox/red + (0.059/n) lgK EН2 = E0Н2 + (0.059/1) lg[H+] E0Н2 = 0 Standard potential of hydrogen electrode EН2 = - 0.059 pH

Nernst equation for biological redox system reductant oxidant + 2e + 2 H+ Eox/red =E0ox/red + (0.059/2) lg[ox][H+]2/[red] Eox/red = E0ox/red – 0.059pH + 0.03 lg[ox]/[red]

Nernst equation for biological redox system at pH = 7 Eox/red = E0ox/red – 0.059pH + 0.03 lg[ox]/[red] E0ox/red – 0.059pH = E0ox/red – 0.059 . 7 E0’ox/red = E0ox/red – 0.41 Standard potential of biological system at pH = 7 Eox/red = E0’ox/red + 0.03 lg[ox]/[red]

Quinhydrone electrode Pt (QH) H+ Eox/red = E0ox/red – 0.059pH + 0.03 lg[ox]/[red] EQH = E0QH – 0.059pH + 0.03lg[quinone]/[hydroquinone] Quinhydrone = 1M quinone + 1Mhydroquinone EQH = E0QH – 0.059pH

Galvanic (electrochemical) cell Zn | ZnSO4(р- р) || CuSO4(р- р) | Cu

electromotive force (EMF) electromotive force (EMF) ؏= E2 – E1 ؏о = ЕoCu – ЕoZn = 0,34 – ( - 0,76) = 1,1B CuSO4 + Zn → ZnSO4 + Cu reductant Zno - 2ē → Zn2+ oxidation oxidant Cu2+ + 2ē → Cuo reduction

Potentiometry

Potentiometry

1. Calculate the potential of quinhydronе electrode in the 0.01M solution of HF. pKHF= 4, E0QH= 0.699V 1. Calculate the potential of quinhydronе electrode in the 0.01M solution of HF. pKHF= 4, E0QH= 0.699V 2. What is the degree of dissociation of a weak acid, if the equilibrium potential of hydrogen electrode immersed into the 0.2M solution of this acid is equal to - 0.18V? 3. Estimate the value of ΔG0’ and determine the direction of spontaneous reaction: NADH + H+ + ethanal NAD+ + ethanol E0’NAD+/NADH= - 0.32V, E0’ethanal/ethanol= - 0.20V.

4. Estimate the value of the equilibrium potential of platinum electrode in the solution of two salts: FeCl2 (0.25M) and FeCl3 (0.25M), E0Fe3+/Fe2+= 0.77V, t= 250C. 4. Estimate the value of the equilibrium potential of platinum electrode in the solution of two salts: FeCl2 (0.25M) and FeCl3 (0.25M), E0Fe3+/Fe2+= 0.77V, t= 250C. 5. Determine the ratio of concentrations [fumarate]/[succinate] if the potential of platinum electrode in this solution is equal to 0.04V, t= 250C, E0’fum/suc= - 0.031V, pH = 7. 6. Diagram the cell consisting of the silver chloride electrode in saturated KCl solution and hydrogen electrode in the 0.01M solution of acetic acid . What is the electromotive force (EMF) of this cell? pKCH COOH = 4.75, ESC=0.202V