Answer:
There's a separate chemical reaction happening at the positive electrode, where incoming electrons recombine with ions taken out of the electrolyte, so completing the circuit. The electrons and ions flow because of the chemical reactions happening inside the battery—usually two of them going on simultaneously.
Explanation:
option (B) is right
Answer:
Equilibrium constant expression for 
:
![\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{[\mathrm{H_2CO_3}]}{\left[\mathrm{H^{+}\, (aq)}\right]^{2} \, \left[\mathrm{CO_3}^{2-}\right]}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20K%20%3D%20%5Cfrac%7B%5Cleft%28a_%7B%5Cmathrm%7BH_2CO_3%5C%2C%20%28aq%29%7D%7D%5Cright%29%7D%7B%5Cleft%28a_%7B%5Cmathrm%7BH%5E%7B%2B%7D%7D%7D%5Cright%29%5E2%5C%2C%20%5Cleft%28a_%7B%5Cmathrm%7B%7BCO_3%7D%5E%7B2-%7D%5C%2C%20%28aq%29%7D%7D%5Cright%29%7D%20%5Capprox%20%5Cfrac%7B%5B%5Cmathrm%7BH_2CO_3%7D%5D%7D%7B%5Cleft%5B%5Cmathrm%7BH%5E%7B%2B%7D%5C%2C%20%28aq%29%7D%5Cright%5D%5E%7B2%7D%20%5C%2C%20%5Cleft%5B%5Cmathrm%7BCO_3%7D%5E%7B2-%7D%5Cright%5D%7D)
.
Where
, 
, and 
denote the activities of the three species, and ![[\mathrm{H_2CO_3}]](https://tex.z-dn.net/?f=%5B%5Cmathrm%7BH_2CO_3%7D%5D)
, ![\left[\mathrm{H^{+}}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cmathrm%7BH%5E%7B%2B%7D%7D%5Cright%5D)
, and ![\left[\mathrm{CO_3}^{2-}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cmathrm%7BCO_3%7D%5E%7B2-%7D%5Cright%5D)
denote the concentrations of the three species.
Explanation:
<h3>Equilibrium Constant Expression</h3>
The equilibrium constant expression of a (reversible) reaction takes the form a fraction.
Multiply the activity of each product of this reaction to get the numerator.
is the only product of this reaction. Besides, its coefficient in the balanced reaction is one. Therefore, the numerator would simply be 
.
Similarly, multiply the activity of each reactant of this reaction to obtain the denominator. Note the coefficient "
" on the product side of this reaction. 
is equivalent to 
. The species 
appeared twice among the reactants. Therefore, its activity should also appear twice in the denominator:
.
That's where the exponent "" in this equilibrium constant expression came from.
Combine these two parts to obtain the equilibrium constant expression:
![\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \quad\begin{matrix}\leftarrow \text{from products} \\[0.5em] \leftarrow \text{from reactants}\end{matrix}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20K%20%3D%20%5Cfrac%7B%5Cleft%28a_%7B%5Cmathrm%7BH_2CO_3%5C%2C%20%28aq%29%7D%7D%5Cright%29%7D%7B%5Cleft%28a_%7B%5Cmathrm%7BH%5E%7B%2B%7D%7D%7D%5Cright%29%5E2%5C%2C%20%5Cleft%28a_%7B%5Cmathrm%7B%7BCO_3%7D%5E%7B2-%7D%5C%2C%20%28aq%29%7D%7D%5Cright%29%7D%20%5Cquad%5Cbegin%7Bmatrix%7D%5Cleftarrow%20%5Ctext%7Bfrom%20products%7D%20%5C%5C%5B0.5em%5D%20%5Cleftarrow%20%5Ctext%7Bfrom%20reactants%7D%5Cend%7Bmatrix%7D)
.
<h3 /><h3>Equilibrium Constant of Concentration</h3>
In dilute solutions, the equilibrium constant expression can be approximated with the concentrations of the aqueous "
" species. Note that all the three species here are indeed aqueous. Hence, this equilibrium constant expression can be approximated as:
![\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{\left[\mathrm{H_2CO_3\, (aq)}\right]}{\left[\mathrm{H^{+}\, (aq)}\right]^2\cdot \left[\mathrm{{CO_3}^{2-}\, (aq)}\right]}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20K%20%3D%20%5Cfrac%7B%5Cleft%28a_%7B%5Cmathrm%7BH_2CO_3%5C%2C%20%28aq%29%7D%7D%5Cright%29%7D%7B%5Cleft%28a_%7B%5Cmathrm%7BH%5E%7B%2B%7D%7D%7D%5Cright%29%5E2%5C%2C%20%5Cleft%28a_%7B%5Cmathrm%7B%7BCO_3%7D%5E%7B2-%7D%5C%2C%20%28aq%29%7D%7D%5Cright%29%7D%20%5Capprox%20%5Cfrac%7B%5Cleft%5B%5Cmathrm%7BH_2CO_3%5C%2C%20%28aq%29%7D%5Cright%5D%7D%7B%5Cleft%5B%5Cmathrm%7BH%5E%7B%2B%7D%5C%2C%20%28aq%29%7D%5Cright%5D%5E2%5Ccdot%20%5Cleft%5B%5Cmathrm%7B%7BCO_3%7D%5E%7B2-%7D%5C%2C%20%28aq%29%7D%5Cright%5D%7D)
.
So, Ba is divalent. This means that each Ba(OH)2 requires two I in order to produce one BaI2 molecule.
The two hydrogens from the 2 HI molecules will combine with 2 OH of Ba(OH)2 forming 2 water (H2O) molecules.
The equation which illustrates this reaction in as follows:
<span>2 HI (aq) + Ba(OH)2 (s) ........> BaI2 (aq) + 2 H2O (l)
</span>where:
aq refers to aqueous state
s refers to solid state and
l refers to liquid state.
<u><em>Answer:</em></u>
- <em>Respect </em>
- <em>Confidentiality </em>
<u><em /></u>
<u><em>Explanation:</em></u>
<em>*Hope this helps*</em>
<u><em /></u>
