NAD+ + H+ + 2e’ <====> NADH -0.32
1/2 O2 + 2H+ + 2e’ <====> H2O 0.816
Standard reduction potential = 0.816 V – (-0.032 V) = 1.14 V (exergonic reaction)
C6H12O6 (glucose) + 6 O2 -> 6 CO2 + 6 H2O
Standard reduction potential = -1.24 (exergonic reaction)
6CO2+12H2O--->C6H12O6+6O2+6H2O
Standard reduction potential = +1.24 (endergonic reaction)
FADH2 → FAD
Eo' = +0.15 V (endergonic reaction)
CoQH2 + 2 cytochrome c1 (Fe3+) + 2 H+ → CoQ + 2 cytochrome c1 (Fe2+) + 4 H+
Eo' = +1.00 V (endergonic reaction)
Using the table of reduction potentials (Eo) for the half-reactions shown, match the given redox reactions...
E Volts) 6 co2/glucose 0.43] -0.5 I-0.42] -0.4 NAD /NADH 0.32] -03 -0.2 -0.1 Fumarate/succinate 0.03] 0.0 [0.10] +0.1 2 H /H2 I- 0.27] FADIFADH2 0.18] S°/H2S CoQ/CoQH2 +0.2 Cyt C (Fe )Cyt C (Fe+) [0.25] +0.3- +0.4 +0.5 +0.6 No,INO2 [0.421 Fe3+/Fe2+ 4 0,H20 (0.82108 10.77) +07
I need this answered and explained!
Redox couple Eo (V) CO2/glucose (-0.43) 24 e 2H /H2 (-0.42) 2e CO2/methanol(-0.38) 6 e NAD /NADH (-0.32) 2 e -0.50 --0.40 -0.30 CO2/acetate (-0.28) 8 e E-0.20 S /H2S (-0.28)2 e SO22 /H2S(-0.22) 8 e Pyruvate/lactate (-0.19) 2 e SO62-S2032(+0.024) 2 e -0.10 0.0 Fumarate/succinate (+0.03) 2 e +0.10 Cytochrome boxred (+0.035) 1 e +0.20 Fe3 /Fe2 (+0.2) 1 e, (pH 7) Ubiquinone ox/red (+0.11) 2 e Cytochrome Cox/red (+0.25) 1 e Cytochrome...
Two half-reactions are shown with their standard cell potentials. If a galvanic cell is constructed using them, which electrode would be the anode, and what would the cell potential be? Fe2+ (aq) + 2e → Fe(s); E=-0.44 V Ag+(aq) + 1e → Ag(s); E=0.80 V O a. Iron, 1.24 V O b. Silver, 0.36 V O c. Iron, 0.36 V O d. Iron, -0.36 V O e. Silver, -0.36 V
Using the standard reduction potentials listed, calculate the
equilibrium constant for each of the following reactions at 298
K.
A) Fe(s)+Ni2+(aq)→Fe2+(aq)+Ni(s)
Express your answer using two significant figures.
B) Co(s)+2H+(aq)→Co2+(aq)+H2(g)
Express your answer using two significant figures.
C) 10Br−(aq)+2MnO−4(aq)+16H+(aq)→2Mn2+(aq)+8H2O(l)+5Br2(l)
Express your answer using two significant figure.
E°(V) -0.83 +0.88 +1.78 +0.79 Half-Reaction E°(V) Half-Reaction Ag+ (aq) + - Ag(s) +0.80 2 H20(1) + 2 e — H2(8) + 2 OH+ (aq) AgBr(s) + - Ag(s) + Br" (aq) +0.10 HO2...