Answer:
Explanation:
1. the 1/2 reaction that occurs at the cathode
3Cl2(g) +6e^- -------------> 6Cl^- (aq)
2 the 1/2 reaction that occurs at the anode
2MnO2(s) + 8OH^-(aq) ----------> 2MnO4^- (aq) + 4H2O(l) +6e^-
2MnO2(s) + 8OH^-(aq) ----------> 2MnO4^- (aq) + 4H2O(l) +6e^-
E0 = -0.59v
3Cl2(g) +6e^- -------------> 6Cl^- (aq)
E0 = 1.39v
3Cl2 (g) + 2MnO2 (s) + 8OH^(−) (aq)---------> 6Cl^(−) (aq) + 2MnO4^(−) (aq) + 4H2O (l)
E0cell = 0.80v
Answer:
Yes they woll react.
Explanation:
Ferous oxide and zinc will be product.
This is how I got to that answer. Since we don't know how many atoms there are in a mole, we use the number 6.02 x 10^-23. Now, just plug in what you have in the equation:
<span>1.75 moles ChCl3 x (6.02 x 10 ^-23) / 1 mole = 1.0535 x 10^-22 atoms. </span>
The answer to this is surface wave i think
