I have the answers I just need an explaination as to how we got to the answer an

Question

I have the answers I just need an explaination as to how we got to the answer and why. Answer: D 4) In a galvanic cell, the half-reaction Mn04-(ag) +8 H (a)+5eMn2+(ag) + 4 H20 +4 H200) is A) an oxidation half-reaction and occurs at the anode. B) an oxidation half-reaction and occurs at the cathode. C) a reduction half-reaction and occurs at the anode. D) a reduction half-reaction and occurs at the cathode. Answer: D 5) What species is oxidized in the reaction: CuS04a)+ Felo) A) CuSO4 (aq) B) Fe (s) C) FeSO4 (a?) D) Cu (s) FeS04(a)Cu(s)y Answer: B t Cl20g)+2e2 C-(ag) is the reduction half-reaction for the overall reaction 6) Given that Cl2( 2Ag(s) + Cl2(g) 2AgCI(s) what is the oxidation half reaction? A) Ag(s) Ag(aq) + e B) Ag(s) + Cl-(a?)AgCI(s) +e

Explanation / Answer

MnO4^- (aq) + 8H^+ (aq) + 5e^- ---------> Mn^2+ (aq0 + 4H2O(l)A chemical reaction gain of electrons are takes place is called reduction.
it is reduction reaction. The reduction
reaction takes place at cathode.
D. a reduction half -reaction and occurs at the cathode.

5. CuSo4(aq) + Fe(s) ----------> FeSo4 + Cu
Cu^2+ (aq) + Fe(s) ---------> Fe^2+ (aq) + Cu
Fe oxidation number increases from 0 to +2 . It is oxidation reaction.So Fe is oxidised.
B. Fe(s)

6.B. Ag(s) + Cl^- (aq) --------> AgCl(s) + e^-

7. CrO7^2- (aq) + 14H^+ (aq) + 6e^- ---------> 2Cr^3+ (aq) +7H2O(l)
     3Pb(s) --------------------------> Pb^2+ (aq) + 6e^-
   ------------------------------------------------------------------------
   CrO7^2- (aq) + 14H^+ (aq) + 3Pb(s) ---------> Pb^2+ (aq) +2Cr^3+ (aq) +7H2O(l)
The ration of Cr2O7^2- and Pb(s) = 1 :3
C.3
8. Galvanic cell is combination of oxidation half reaction and reduction hlaf reaction.
Galvanic cell writen on the paper left side oxidation half cell and reduction half cell at right side.
Mg electrode is oxidation half cell
Mg(s) ----------> Mg^2+ (aq) + 2e^-
A.Mg(s) ----------> Mg^2+ (aq) + 2e^-
9. 2Al(s) ---------> 2Al^3+ (aq) + 6e^-
3Cu^2+ +6e^- ------> 3Cu(s)
--------------------------------------------
2Al(s) + 3Cu^3+ (aq) -------> 2Al^3+ (aq) + 3Cu(s) 6 electrons are involve in the reaction.
   n= 6
D. 6
10. n = 2
DG^0 = -nE0cell*F
           = -2*1.56*96500 = -301080J = -301KJ
C.-301KJ
        

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---