Answer : The balanced reduction half-reaction is:

Explanation :
Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.
Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.
Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.
The given balanced redox reaction is :

The half oxidation-reduction reactions are:
Oxidation reaction : 
Reduction reaction : 
In order to balance the electrons, we multiply the oxidation reaction by 2 and reduction reaction by 3 and then added both equation, we get the balanced redox reaction.
Oxidation reaction : 
Reduction reaction : 
The balanced redox reaction will be:

Thus, the balanced reduction half-reaction is:

The phase's composition is as follows: 27
What is the alloy's composition?
Both have mass fractions of w a = w b = 0.5.
A-B alloy composition; C o = 57 wt% B - 43 wt% A
C b = 87 wt% B - 13 wt% A is the new phase composition.
Using the Lever rule, we can calculate the mole fraction (x i) or mass fraction (w i) of each phase of a binary equilibrium phase as follows:
W a = W b = 0.5
This provides us with;
0.5 = (C b - C o)/(C b - C a)
(87 - 57)/(87 - C a) = 0.5
30/0.5 = 87 - C a
60 = 87 - C a
C a = 87 - 60
C a = 27
To learn more about composition please click on below link
brainly.com/question/13808296
#SPJ4
Answer:
E. 8.08 x 10⁴.
Explanation:
Hello,
In this case, for the reaction:

We can compute the Gibbs free energy of reaction via:

Since both the entropy and enthalpy of reaction are given at 298 K (standard temperature), therefore:

Then, as the equilibrium constant is computed as:

We obtain:

For which the answer is E. 8.08 x 10⁴.
Best regards,