In order to solve this, we need to know the standard cell potentials of the half reaction from the given overall reaction.
The half reactions with their standard cell potentials are:
<span>2ClO−3(aq) + 12H+(aq) + 10e- = Cl2(g) + 6H2O(l)
</span><span>E = +1.47
</span>
<span>Br(l) + 2e- = 2Br-
</span><span>E = +1.065
</span>
We solve for the standard emf by subtracting the standard emf of the oxidation from the reducation, so:
1.47 - 1.065 = 0.405 V
<span>The copper would go under oxidation since it will be losing two electrons. Copper starts out with an oxidation number of zero, but in order to balance the compound of CuO with the Oxygen having an oxidation number of -2, a positive 2 is required</span>
Answer:
Um english please I don’t know Jewish
Explanation:
Answer:
It does not matter where the sample of water came from or how it was prepared. Its composition, like that of every other compound, is fixed.
