Could you show me the oxidation half-reaction and the reduction half-reaction fo
ID: 637296 • Letter: C
Question
Could you show me the oxidation half-reaction and the reduction half-reaction for the following 9 reactions. Also calculate the standard potential of each reaction listed. 1. Mg(s) + ZnSO4 --> Zn(s) + MgSO4 (aq). 2. Cu(s) + ZnSO4(aq) --> Zn(s) + CuSO4. 3. Zn(s) + CuSO4 --> Cu(s) + ZnSO4 . 4. Zn(s) + 2HCl(aq) --> ZnCl2 + H2(g) . 5. Cu(s) + 2HCl(aq) --> CuCl(aq) + H2(g) . 6. 4KI(aq) + 2CuSO4(aq) --> 2CuI(s) + I2(aq) + 2K2SO4 . 7. 2FeCl3(aq) + 6KI(aq) --> 2FeI2(aq) + I2(aq) + 6KCl(aq) . 8. 2FeCl3(aq) + 6KBr(aq) --> 2FeBr2(aq) + Br2(aq) + 6KCl(aq) . 9. I2(aq) + 2Na2S2O3(aq) --> 2NaI(aq) + Na2S4O6(aq) .
Explanation / Answer
ans)
From above data that
Given reactions
1.
Mg(s) + ZnSO4 --> Zn(s) + MgSO4 (aq).
Here the
Oxidation half reaction:
Mg (s) -----------> Mg2+ (aq.) + 2 e-
Reduction Half Reaction :
Zn2+ (aq.) + 2 e- -----------> Zn (s)
Cell potential ,
E0cell = E0Zn2+/Zn - E0Mg2+/Mg
= - 0.76 - ( - 2.38 )
E0cell = 1.62 V
2.
Cu(s) + ZnSO4(aq) --> Zn(s) + CuSO4.
Oxidation half reaction :
Cu (s) -------------> Cu2+ (aq.) + 2 e-
Reduction half reaction :
Zn2+ (aq.) + 2 e- ------------> Zn (s)
E0cell = E0Zn2+/Zn - E0Cu2+/Cu
= - 0.76 - 0.34
E0cell = - 1.10 V
3.
Zn(s) + CuSO4 --> Cu(s) + ZnSO4
Oxidation half reaction :
Zn (s) -----------> Zn2+ (aq.) + 2 e-
Reduction half reaction :
Cu2+ (aq.) + 2 e- ----------> Cu (s)
E0cell = E0Cu2+/Cu - E0Zn2+/Zn
= 0.34 - ( - 0.76 )
E0cell = 1.10 V
4.
Zn(s) + 2HCl(aq) --> ZnCl2 + H2(g) .
Oxidation half reaction :
Zn (s) ------------> Zn2+ (aq.) + 2 e-
Reduction half reaction :
2 H+ (aq.) + 2 e- ------------> H2 (g)
E0cell = E0H+/H2 - E0Zn2+/Zn
= 0.00 - ( - 0.76 )
E0cell= + 0.76 V
5.
Cu(s) + 2HCl(aq) --> CuCl(aq) + H2(g) .
Oxidation half reaction :
Cu (s) -----------> Cu2+ (aq.) + 2 e-
Reduction half reaction :
2 H+ (aq.) + 2 e- ----------> H2 (g)
E0cell = E0H+/H2 - E0Cu2+/Cu
= 0.00 - 0.34
E0cell= - 0.34 V