1-Write the oxidation half­reaction and the reduction half­reaction for each of

Question

1-Write the oxidation half­reaction and the reduction half­reaction for each of the reactions listed.?

2-Calculate the standard potential of each reaction listed.?

1. Mg(s) + ZnSO4(aq) ? Zn(s) + MgSO4(aq)  

2. Cu(s) + ZnSO4(aq) ? Zn(s) + CuSO4(aq)

3. Zn(s) + CuSO4(aq) ? Cu(s) + ZnSO4(aq)

4. Zn(s) + 2 HCl(aq) ? ZnCl2(aq) + H2(g)

5. Cu(s) + 2HCl(aq) ? CuCl2(aq) + H2(g)

6. 4KI(aq) + 2CuSO4(aq) ? 2CuI(s) + I2(aq) + 2K2SO4(aq)

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) + Na2S2O3(aq) ? 2NaI(aq) + Na2S4O6(aq)

1. Mg(s) + ZnSO4(aq) ? Zn(s) + MgSO4(aq)  

2. Cu(s) + ZnSO4(aq) ? Zn(s) + CuSO4(aq)

3. Zn(s) + CuSO4(aq) ? Cu(s) + ZnSO4(aq)

4. Zn(s) + 2 HCl(aq) ? ZnCl2(aq) + H2(g)

5. Cu(s) + 2HCl(aq) ? CuCl2(aq) + H2(g)

6. 4KI(aq) + 2CuSO4(aq) ? 2CuI(s) + I2(aq) + 2K2SO4(aq)

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) + Na2S2O3(aq) ? 2NaI(aq) + Na2S4O6(aq)

Explanation / Answer

1.

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 ) = 1.62 V

2.

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 = - 1.10 V

3.

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 ) = 1.10 V

4.

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 ) = + 76 V

5.

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 = - 0.34 V

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