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A Balance the following reactions in acidic solution: NO_3^-(aq)+Cr(aq) rightarr

ID: 993733 • Letter: A

Question

A Balance the following reactions in acidic solution: NO_3^-(aq)+Cr(aq) rightarrow No(g)+CI_2(g) AuCI_4(aq) + Fe^+2(aq)rightarrow Au(s) + cr(aq) + Fe^+3(aq) H_2O_2(aq) + SO_4^2^-(aq)rightarrow SO_3^2-(aq) +O_Z(g) MnO_4^+(aq) + H_2O_2(aq) rightarrow Mn^+2(aq)+O_2(g) Question B. Balance the following reactions m basic solution: NO_3^+(aq) + Cr(aq) rightarrow NO(g) + O_2(g) MnO_4^-(aq) + H_2O_2(aq)rightarrow Mn^+2(aq) + O_2(g) Question C. Consider the reaction: Pb(s) + 2 Ag^+(aq)rightarrow Pb^+2(aq) + 2 Ag(s) Sketch this galvanic cell using two brokers and a salt bridge Label the cathode and the anode Which electrode is gaining moss as the cell discharges? Which solution is increasing in concentration is the reaction proceeds ? What is the oxidation holf reaction? The reduction half reaction? f What is the oxidizing agent ? The reducing agent? g. What is being reduced? Being reduced? Question J. Which of the following are spontaneous (delta G

Explanation / Answer

Too many questions and reactions in a same post. I will answer one reaction from part a) one of b) and question 6. The other questions post them in another post (and per separate)

Question A.

a) NO3- + Cl- ---------> NO + Cl2

(4H+ + NO3- + 3e----------> NO + 2H2O)*2
(2Cl- -------> Cl2 + 2e-)*3

8H+ + 2NO3- + 6e----------> 2NO + 4H2O
6Cl- -------> 3Cl2 + 6e-

Final reaction:
8H+ + 2NO3- + 6Cl- -----------> 2NO + 3Cl2 + 4H2O

c) H2O2 + SO42- -----------> SO32- + O2

H2O2 -----------> O2 + 2e- + 2H+
2H+ + SO42- + 2e- ----------> SO32- + H2O

Final reaction:
H2O2 + SO42- ----------> SO32- + O2 + H2O

For question B:
a)  NO3- + Cl- ---------> NO + Cl2

a) NO3- + Cl- ---------> NO + Cl2

2H2O + NO3- + 3e----------> NO + 4OH-)*2
(2Cl- -------> Cl2 + 2e-)*3

4H2O + 2NO3- + 6e----------> 2NO + 8OH-
6Cl- -------> 3Cl2 + 6e-

Final reaction:
4H2O + 2NO3- + 6Cl- -----------> 2NO + 3Cl2 + 8OH-

Question 6.
dG° = -nFE°
E° = -dG°/nF

E° = 47000 / 2*96500
E° = 0.2435 V

Hope this helps

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