Question 1: In lab the K sp of a slightly solution salt was determined to be 1.5

Question

Question 1:

In lab the Ksp of a slightly solution salt was determined to be 1.5 10?10 at 25 C and 3.6 10?7 at 80 C. What is the ?H and ?S of this slightly soluble salt?

Question 2:

In Part A of this lab a student measured a cell voltage of 2.04 V. The cathode used was copper, Ered = 0.34 V, the same as in the lab you are doing. What is the Ered of anode of this cell?

Question 3:

In Part B of the lab you will be doing electrolysis. What will happen to masses of the electrodes at the anode and cathode?

Explanation / Answer

Question (1)

The relationship between Delta G and Ksp can be written as

Delta G = - R * T * ln Ksp

Let's find delta G at the given temperatures

we have 2 different temperatures, T = 298 K and T = 353 K

Delta G ( at 25 C) = - 8.314 J/mol K * 298 K * ln ( 1.5 x 10^-10)

Delta G ( at 25 C) = 56044 J

Delta G ( at 25 C) = 56.04 kJ -----------------(1)

Delta G ( at 80 C) = - 8.314 J/mol K * 353 K * ln ( 3.6 x 10^-7)

Delta G ( at 80 C) = 43545 J

Delta G ( at 80 C) = 43.54 kJ ------------------(2)

The relationship between Delta G, Delta H and Delta S can be written as

Delta G = Delta H - T*Delta S

Using the Delta G values we calculated above, let's write 2 equations

56.04 kJ = Delta H - 298 K * Delta S ....................( 3)

43.54 kJ = Delta H - 353 K * Delta S .....................(4)

_________________________________________________ ( Subtract eq. 4 from eq. 3)

12.5 kJ = 0 - 298 K Delta S - ( - 353 K Delta S)

12.5 kJ = 55 K * Delta S

Delta S = 12.5 kJ / 55 K

Delta S = 12500 J/55 K

Delta S = 227.3 J/K

Delta S = 227.3 J/mol K

Let's substitute this value in equation (3)

56.04 kJ = Delta H - 298 K * Delta S

56040 J = Delta H - 298 K * 227.3 J/K

56040 J = Delta H - 67727.3 J

Delta H = 123767.3 J

Delta H = 123.8 kJ

Delta H = 123.8 kJ/mol

_________________________________________________________________________________________________

Question (2)

E cell is calculated as

E cell = E cathode - E anode

They have given us

E cell= 2.04 V and E cathode = 0.34 V , Substituting these values, we get

2.04 V = 0.34 V - E anode

2.04 V - 0.34 V = - E anode

E anode = -1.7 V

___________________________________________________________________________________________________

Question (3)

Oxidation occurs at anode, which can be represented as

A (s) --------------> A^+ (aq) + e- where A is a metal

From the above reaction we can say that metal which is in solid state gets oxidized and goes into the solution in the form of aq.metal ions

This suggests that anode will reduce in size and its mass will decrease.

On the other hand, reduction occurs at the cathode which is represented as

B^+ (aq) + e- ----------> B (s)

from the above reaction,we can see that aqueous ions of metal B are reduced and get converted to solid atoms which get deposited at the cathode. Therefore cathode increases in size and its mass also increases

Mass of anode reduces and mass of cathode increases

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