Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
Protons and neutrons are located in the nucleus of the atom while the electrons move in the trajectory of the shell
<h3>Further explanation
</h3>
Isotopes are atoms whose no-atom has the same number of protons while still having a different number of neutrons.
So Isotopes are elements that have the same Atomic Number (Proton)
Isotopes of Helium : helium-3 and helium-4
protons = 2
electrons=protons=2
neutron=mass number-atomic number=3-2=1
protons = 2
electrons=protons=2
neutron=mass number-atomic number=3-2=1
protons = 2
electrons=protons=2
neutron=mass number-atomic number=4-2=2
Protons and neutron in the nucleus, electrons in the shell
Answer:
2H⁺(aq) + 2OH⁻(aq) --> 2H2O(l)
Explanation:
2HBr(aq)+Ba(OH)2(aq)⟶2H2O(l)+BaBr2(aq)
We break the compounds into ions. Only compounds in the aqueous form can be turned into ions.
The ionic equation is given as;
2H⁺(aq) + 2Br⁻(aq) + Ba²⁺(aq) + 2OH⁻(aq) --> 2H2O(l) + Ba²⁺(aq) + 2Br⁻(aq)
Upon eliminating the spectator ions; The net equation is given as;
2H⁺(aq) + 2OH⁻(aq) --> 2H2O(l)
<span>NaOH + HCl = NaCl + H2O</span>
