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
Answer:
I think it is B
which I think is the correct answer
Hello young fellow friend I think the anwser is (C)
Answer:
90%
Explanation:
Percentage yield = ?
Theoretical yield = 50g
Actual yield = 45g
To calculate the percentage yield of a compound, we'll have to use the formula of percentage yield which is the ratio between the actual yield to theoretical multiplied by 100
Percentage yield = (actual yield / theoretical yield) × 100
Percentage yield = (45 / 50) × 100
Percentage yield = 0.9 × 100
Percentage yield = 90%
The percentage yield of the substance is 90%
Right answer is option d that o is oxidized.
