Reduction refers to a chemical reaction in which electrons are gained by one of the atoms involved in the reaction and at the same time lowering of an oxidation state of that atom takes place.
In the given case, the reduction occurs at the cathode. In aqueous, V (vanadium) is present in +5 oxidation state that on reduction modifies to vanadium (aq) with +1 oxidation state.
The half reaction is:
V⁵⁺ (aq) + 4e⁻ → V¹⁺ (aq)
Answer: the equilibrium will be displaced to the right leading an increase on the quantities of y(g) and z(s).
Justification:
According to the rules of equilibrium, based on Le Chatellier's priciple, any change in a system in equilibrium will be tried to be compensated to restablish the equilibrium
The higher the amount, and so the concentration, of X(g), the more will the forward reaction proceed leading to an increase on the concentration of the products y(g) and z (s). Look that that will also be accompanied by a decreasing on the pressure, since 2 molecules of the gas X(g) are converted into 1 molecule of the gas y(g).
The answer is •c•
Hope this is correct I tried
<span>Germanium is the element that has 32 protons in its nucleus.</span>
Answer:
[H₃O⁺] = 4.3 × 10⁻¹² mol·L⁻¹; [OH⁻] = 2.4 × 10⁻³ mol·L⁻¹;
pH = 11.4; pOH = 2.6
Explanation:
The chemical equation is

For simplicity, let's re-write this as

1. Calculate [OH]⁻
(a) Set up an ICE table.
B + H₂O ⇌ BH⁺ + OH⁻
0.310 0 0
-x +x +x
0.310-x x x
![K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.100 - x} = 1.8 \times 10^{-5}](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Bb%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BBH%7D%5E%7B%2B%7D%5D%5Ctext%7B%5BOH%7D%5E%7B-%7D%5D%7D%7B%5Ctext%7B%5BB%5D%7D%7D%20%3D%201.8%20%5Ctimes%2010%5E%7B-5%7D%5C%5C%5C%5C%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.100%20-%20x%7D%20%3D%201.8%20%5Ctimes%2010%5E%7B-5%7D)
Check for negligibility:
(b) Solve for [OH⁻]
![\dfrac{x^{2}}{0.310} = 1.8 \times 10^{-5}\\\\x^{2} = 0.310 \times 1.8 \times 10^{-5}\\x^{2} = 5.58 \times 10^{-6}\\x = \sqrt{5.58 \times 10^{-6}}\\x = \text{[OH]}^{-} = \mathbf{2.4 \times 10^{-3}} \textbf{ mol/L}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.310%7D%20%3D%201.8%20%5Ctimes%2010%5E%7B-5%7D%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%200.310%20%5Ctimes%201.8%20%5Ctimes%2010%5E%7B-5%7D%5C%5Cx%5E%7B2%7D%20%3D%205.58%20%5Ctimes%2010%5E%7B-6%7D%5C%5Cx%20%3D%20%5Csqrt%7B5.58%20%5Ctimes%2010%5E%7B-6%7D%7D%5C%5Cx%20%3D%20%5Ctext%7B%5BOH%5D%7D%5E%7B-%7D%20%3D%20%5Cmathbf%7B2.4%20%5Ctimes%2010%5E%7B-3%7D%7D%20%5Ctextbf%7B%20mol%2FL%7D)
2. Calculate the pOH
![\text{pOH} = -\log \text{[OH}^{-}] = -\log(2.4 \times 10^{-3}) = \mathbf{2.6}](https://tex.z-dn.net/?f=%5Ctext%7BpOH%7D%20%3D%20-%5Clog%20%5Ctext%7B%5BOH%7D%5E%7B-%7D%5D%20%3D%20-%5Clog%282.4%20%5Ctimes%2010%5E%7B-3%7D%29%20%3D%20%5Cmathbf%7B2.6%7D)
3. Calculate the pH

4 Calculate [H₃O⁺]
