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please explain and show how you came to conclusion... maybe one of you can help

ID: 893019 • Letter: P

Question

please explain and show how you came to conclusion... maybe one of you can help me to understand this stuff.

1. Calculate the standard change in free energy G   (kJ) at 298 K for the reaction C2H4(G)+3o2(G)à2CO2(G)+2H2O(G)

-1310   /   +1180   /   -691   /   +1310   /   -1180

2. The value of Ecell for the reaction 2Cr3+(aq) + 6Hg(l)à2Cr(s)+3Hg22+(aq) is 1.59 V.

-921   /   -153   /   -767   /   -460   /   -307

3. What is the value of the equilibrium constant for the cell reaction below at 25 deg C Ecell­=0.61V                                    2Cr(s)+3Pb2+(aq)ßà3Pb(s)+2Cr3+(aq)

8.2*10^30   /   6.7*10^61   /   9.9*10^99   /   4.1*10^20   /   3.3*10^51

4. Which one of the following is not a redox reaction?

Na6FeCl8(s)+2Na(l)à8NaCl(s)+Fe(s)

Al(OH)3(aq)+3H+àAl3+(aq)+3H2O(l)

CuCl2(aq)+Ni(s)àCu(s)+NiCl2(aq)

C6H12O6(s)+6O2(g)à6CO2+6H2O(l)

CO­2(g)+H2(g)àCO(g)+ H2O(g)­­­­­­­­

5. A voltaic cell consists of a Zn/ZN2+ electrode and a Fe/Fe2+ electrode. If [ZN2+] = 0.005M and if [Fe2+] = .0500 M , what is Ecell?

.76   /   0.25   /   0.45   /   0.31   /   0.37

6. What is the oxidation state of Fe in FeO42-

+8   /   +5   /   +2   /   +4   /   +6

7. A voltaic cell is prepared using copper and silver. Its cell notation is Cu(s) Cu2+(aq) Ag+(aq) Ag(s)

Which of the following occurs at the cathode?

Cu2+(aq)+2e-àCu(s)

Ag(s)àAg+(aq)+e-

Cu(s)+2Ag+(aq)+àCu2+(aq)+2Ag(s)

Cu(s)àCu2+(aq)+2e-

Ag+(aq)+e-àAg(s)

8. Calculate E­­cell and indicate wheter the overall reaction is spontaneous or nonspontaneous.

I2(s)+2e-ßà2I-(aq) E=0.53V        Cr3+(aq)+3e-ßàCr(s) E= -0.74V Overall reaction: 2Cr(s)+3I2(s)à2Cr3+(aq)+(aq)+6I-(aq)

Ecell= 1.54V , spontaneous

Ecell= -1.27V, nonspontaneous

Ecell= 1.27V, nonspontaneous

Ecell= 1.27V, spontaneous

Ecell= 1.27V, spontaneous

Ecell= -1.27V, spontaneous

­­­­­­­­­

9. When the following redox equation is balanced with the smallest whole number coefficients, the coefficient for zinc will be _____.         Zn(s)+ReO4-(aq)àRe(s)+Zn2+(aq)(acidic soln)

2   /   8   /   16   /   7   /   10

____________________________________________________________________________________________________

All of your help is greatly appreciated… I am beyond lost on this

Explanation / Answer

First to all, you need to put some order in this. I understand if you don't know how to solve these questions, but I need you to help me to help you. Questions 1-4 I don't understand how you write some values, and the meaning of values there. So, My advice, put all those questions in separate post and write it yourself with calm, and putting the values. Also, in the reactions of question 4, try to write the equations well, in that way I can easily identify the reactants and products of the reaction and I can answer it correctly.

With that being said, I'll answer the questions that I manage to understand. The rest of the questions, I'll gladly answer it for you once you put some order in that and post them in new questions.

Question 6.

Oxidation state in FeO42-

let's call "x" to Fe oxidation state.

Oxidation state of O = (-2); numbers of atoms: 4

Formal charge: -2

so:

-2x4 + x = -2 -----> x = 8-2 = +6

The oxidation state would be +6.

Question 7.

Cu pass from 0 (because it's solid) to Cu2+

Cu(s) ----------> Cu2+ + 2e- This means that it's oxidating

The reaction that it would take place in the cathode is the reduction reaction, so,

Cu(s) --------> Cu2+ + 2e- is in the anode.

Therefore, the reaction with the Ag is the one that is taking place in the cathode. The reaction is:

Ag+(aq) -------------> Ag(s)

Question 8.

The overall reaction is: 2Cr(s) + 3I2(s) ---------> 2Cr3+(aq) + 6I-(aq) with this:

Oxidation: 2Cr -----------> 2Cr+3 + 6e-   E° = -0.74 V

Reduction: 3I2 + 6e-  ----------> 6I-   E° = 0.53 V

Ecell = Ered - Eox

Ecell = 0.53 - (-0.74) = 1.27 V. This value indicates that is spontaneus

Question 9. Zn(s) + ReO4-(aq) --------> Zn2+(aq) + Re(s)

7 x (Zn(s) -----------> Zn2+ + 2e-) = 7Zn ---------------> 7Zn2+ + 14e-

2 x (ReO4- + 8H+ + 7e--------------> Re + 4H2O) = 2ReO4- + 16H+ + 14e- ---------------> 2Re+ 8H2O

7Zn(s) + 2ReO4-(aq) + 16H+ --------> 7Zn2+(aq) + 2Re(s) + H2O

So the coefficient for Zinc, would be 7