Answer :
(a) Reaction at anode (oxidation) :
(b) Reaction at cathode (reduction) :
(c) 
(d) Yes, we have have enough information to calculate the cell voltage under standard conditions.
Explanation :
The half reaction will be:
Reaction at anode (oxidation) :

Reaction at cathode (reduction) :

To balance the electrons we are multiplying oxidation reaction by 4 and then adding both the reaction, we get:
Part (a):
Reaction at anode (oxidation) :

Part (b):
Reaction at cathode (reduction) :

Part (c):
The balanced cell reaction will be,

Part (d):
Now we have to calculate the standard electrode potential of the cell.


For a reaction to be spontaneous, the standard electrode potential must be positive.
So, we have have enough information to calculate the cell voltage under standard conditions.
If i am not mistaken ,
density = mass / volume
density = 7.87 g/cm3
volume = 2.5 cm3
thus ,
mass = density x volume
7.87 g/cm3 x 2.5 cm3 = 19.675 g
Molar mass NaOH =23+16+1=40 g/mol
<span> 0.100 M= 0.100 mol/L
</span>500 ml=0.500 L
0.500L*0.100 mol/L=0.0500 mol NaOH we need to prepare 500 ml solution
0.0500 mol NaOH*40g/1mol=2 g NaOH we need to prepare 500 ml solution
we need 2 g NaOH, dissolve it in small amount of water, and dilute it with water up to 500 mL
In carbohydrates the C:H:O is 1:2:1