Question

Will Pu4+ (0.5 M) be reduced to Pu3+ (0.0005 M) by 0.25 M ascorbic acid under 1 M acid conditions at 25 degrees C? (5/5) Pu4+ + e- ⇌ Pu3+ E° = 1.006 V Dehydroascorbic acid + 2H+ + 2e- ⇌ ascorbic acid + H2O E° = 0.390 V

Will Pu4+ (0.5 M) be reduced to Pu3+ (0.0005 M) by 0.25 M ascorbic acid under 1 M acid conditions
at 25 degrees C? (5/5)
Pu4+ + e- ⇌ Pu3+ E° = 1.006 V
Dehydroascorbic acid + 2H+ + 2e- ⇌ ascorbic acid + H2O E° = 0.390 V

Answer 1

Yes, Pu(IV) will be reduced to Pu(III).

Explanation:

Consider the standard electrode potentials:

`sf(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxx)E^(@)" " (V))`

`sf("DHA"+2H^(+)+2erightleftharpoons"AA"color(white)(xxxxxxx)+0.390)`

`sf(Pu^(4+)+erightleftharpoonsPu^(3+)color(white)(xxxxxxxxxx)+1.006)`

The most +ve `sf(E^@)` value will take in the electrons. This means that the Pu(IV)/Pu(III) half cell will be driven left to right and the DHA/AA half cell will be driven right to left.

Under standard conditions the cell reaction will be:

`sf(2Pu^(4+)+ "AA"rarr2Pu^(3+)+2H^(+)+"DHA")`

and

`sf(E_(cell)^(@)=1.006-0.390=+0.616color(white)(x)"V")`

This is confirmed by the fact that :

`sf(DeltaG_(cell)^(@)=-nFE_(cell)^@)`

which must `:.` have a -ve value.

However, we are not under standard conditions. We would need to use The Nernst Equation which at `sf(25^@C)` is:

`sf(E_(cell)=E_(cell)^(@)-0.0591/(z)logQ)`

`sf(Q=([Pu^(3+)]^(2)cancel([H^+]^2)[DHA])/([Pu^(4+)]^2["AA"]))`

We can set up an ICR table based on mol/l. I'll assume `sf([H^+])` is set at IM.

`sf(" "2Pu^(4+)+ "AA"rarrcolor(white)(xxx)2Pu^(3+)+2H^(+)+"DHA")`

`sf(I" "0.5color(white)(xxxx)0.25color(white)(xxxxxxx)0color(white)(xxxxxxxxxx)0)`

`sf(C" "-x" "-x/2color(white)(xxxxxx)+xcolor(white)(xxxxxxx)+x/2)`

Since x = 0.0005 the concentrations become:

`sf([Pu^(4+)]=0.5-0.0005=0.4995color(white)(x)M)`

`sf(["AA"]=0.25-0.0005/2=0.24975color(white)(x)M)`

`sf([Pu^(3+)]=0.0005color(white)(x)M)`

`sf(DHA=0.0005/2=0.00025color(white)(x)M)`

`:.` `sf(Q=(0.0005^2xx0.00025)/(0.4995^2xx0.24975)=1.003xx10^(-9))`

`:.` `sf(E_(cell)=0.616-0.0591/(2)[-0.8998])`

`sf(E_(cell)=0.616+0.266=+0.882color(white)(x)V)`

The +ve value indicates the reduction can happen.

I am not sure about this answer. The wording of the question implies that the concentrations given are initial but no value for `sf([DHA])` is given so you can't get `sf(Q)`. I have therefore assumed that `sf([Pu^(3+)]=0.0005M` is its concentration at some point after the reaction has started, in which case I could have just answered "Yes" to the question and not bothered with the calculation.