It would be C . The location of the star in the night sky
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:
Following are the complete balance of the given equation:
Explanation:
Given equation:
![Cu_2SO_4(aq) + Li_3PO_4(aq)](https://tex.z-dn.net/?f=Cu_2SO_4%28aq%29%20%2B%20Li_3PO_4%28aq%29)
![Cu_2So4\ (aq)+Li_3Po_4 \(aq) \longrightarrow Cu_3(Po_4)\ (aq)+Li_2So_4 \ (aq)](https://tex.z-dn.net/?f=Cu_2So4%5C%20%28aq%29%2BLi_3Po_4%20%5C%28aq%29%20%5Clongrightarrow%20Cu_3%28Po_4%29%5C%20%28aq%29%2BLi_2So_4%20%5C%20%28aq%29)
After Balancing the equation:
![3 Cu_2So4\ (aq)+ 2 Li_3Po_4\ (aq) \longrightarrow 2 Cu_3(Po_4)\ (aq)+ 3Li_2So_4 \ (aq)](https://tex.z-dn.net/?f=3%20Cu_2So4%5C%20%28aq%29%2B%202%20Li_3Po_4%5C%20%28aq%29%20%5Clongrightarrow%202%20Cu_3%28Po_4%29%5C%20%28aq%29%2B%203Li_2So_4%20%5C%20%28aq%29)
In the above equation, when the 3 mol Copper sulfate reacts with 2 mol lithium phosphate
, it will produce 2 mol Copper phosphate and 3 mol Lithium sulfate
.
<h3>
Answer:</h3>
= +0.43 V
<h3>
Explanation:</h3>
The two standard reduction potentials are;
Fe³⁺(aq) + e⁻ → Fe²⁺(aq) E₀ = +0.77 V
Cu²⁺(aq) + 2e⁻ → Cu(s) E₀ = +0.34 V
The overall reaction is;
Fe³⁺(aq) + Cu(s) → Fe²⁺(aq) + Cu²⁺(aq)
To calculate the standard cell potential of the overall reaction, we use the equation;
Ecell = E redu, cathode - E red, anode
From the reaction; Copper lost electrons to form copper ions (Cu²⁺) (Oxidation) while Fe³⁺ gained electrons to form Fe²⁺ (reduction)
Therefore;
E cell = + 0.77 v - (+0.34 v)
= + 0.43 V
Hence, the standard cell potential of the overall reaction is +0.43 V