a See page 896 11 Question (2 points) The standard potentials for the reduction

Question

a See page 896 11 Question (2 points) The standard potentials for the reduction of nicotinamide adenine dinucleotide (NAD+) and oxaloacetate (reactants in the multistep metabolism of glucose) are as follows: NAD+(aq) +2H + (aq) + 2e--> NADH(aq) + H+(aq) E 0.330V Oxaloacetate2-(aq) +2H+(aq) + 2e-(aq)- malate2-(aq) E=-0.166 V 1st attempt Part 1 (1 pt) O Feedback Calculate the standard potential for the following reaction: Oxaloacetate® (aq) + NADH (aq)-H"(aq)- malate?-(aq) + NAD"(aq) d See Periodic Table See Hint 17

Explanation / Answer

For Rxn2, Eo = - 0.166 V

For reverse of Rxn 1 , Eo = + 0.330 V

Rxn 2 + reverse of Rxn1 gives the net reaction.

Hence,

Eocell = - 0.166 V + 0.330 V

= + 0.164 V

Eocell = 0.164 V

no of electrons transferred n = 2

We know that

  Go = -nFEocell                F = Faraday constant = 96485 Coulomb/mol

   Go  = -RT InK,      K is equilibrium constant , R = Universal gas constant = 8.314 J/K/mol

Hence,

-nFEocell =  -RT InK

InK =nFE/RT

K = e^[nFE/RT]

= e^[ (2 x 96485 x 0.164) /(8.314 x 298.15) ]                                 [ Given that Temperature T = 298.15 K]

= 350458

K =  350458

Therefore,

Equilibrium constnat at 298.15 K = 350458

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