In a galvanic cell, the flow of electrons will be from the anode to cathode through the circuit .
Whether a cell is an electrolysis cell (non-spontaneous chemistry driven by forcing electricity from an external energy source) or a galvanic cell (spontaneous chemistry driving electricity), will determine the charge of the anode and the cathode. Depending on where the electrons encounter resistance and find it difficult to pass, a negative charge may emerge. Therefore, you cannot determine the direction of the current just on the charge on the electrode.
Oxidation and reduction always take place at the anode and cathode, respectively.
An element undergoes oxidation when it surrenders one or more electrons to become more positively charged. These electrons leave the chemicals in any type of cell and travel to the anode, where they enter the external circuit.
An element picks up an electron during reduction to become more negatively charged (less positive, lower oxidation state). These electrons are captured from the external circuit at the cathode in both types of cells.
Therefore, no matter what kind of cell you are dealing with, the oxidizing chemicals at the anode transfer the electrons to the external circuit; these electrons then move through the circuit from the anode to the cathode, where they are captured by the reducing chemicals. The electrons always go from the anode to the cathode via the external circuit.
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<span> 52.0ml of 0.35M CH3COOH : 0.052 L(0.35M) = .0182 mol of CH3COOH.
</span>
<span>31.0ml of 0.40M NaOH : .031 L(0.40M) = .0124 mol of NaOH.
</span>
<span>After the reaction, .0124 Mol CH3COO- is generated and .058 mol CH3COOH is left un-reacted. The concentration would be 12.4/V and 5.8/V, respectively. Therefore:
</span>
<span>pH = -log([H+]) = -log(Ka*[CH3COOH]/[CH3COO-]) </span>
<span>= -log(1.8x10^-5*5.8/12.4) = 5.07</span>
Explanation:
When we move across a period from left to right then there will occur an increase in electronegativity and also there will occur an increase in non-metallic character of the elements.
As calcium (Ca) is a group 2A element and rubidium (Rb) is a group 1A element. Hence, Rb being an alkali metal is more metallic in nature than calcium (alkaline earth metal).
Both magnesium (Mg) and radium (Ra) are group 2A elements. And, when we move down a group then as the size of element increases so, it becomes easy of the metal atom to lose an electron.
As a result, there occurs an increase in metallic character of the element. Hence, Radium (Ra) is more metallic in nature than magnesium (Mg).
Also, both bromine and iodine are group 17 elements. Since, both of them are non-metals and non-metallic character increases on moving down the group.
Therefore, bromine (Br) is more metallic than iodine.
Answer:
Explanation:
Hello there!
In this case, given the Henderson-Hasselbach equation, it is possible for us to compute the pH by firstly computing the concentration of the acid and the conjugate base; for this purpose we assume that the volume of the total solution is 0.025 L and the molar mass of the sodium base is 234 - 1 + 23 = 256 g/mol as one H is replaced by the Na:
And the concentrations are:
![[acid]=0.000855mol/0.025L=0.0342M](https://tex.z-dn.net/?f=%5Bacid%5D%3D0.000855mol%2F0.025L%3D0.0342M)
![[base]=0.000781mol/0.025L=0.0312M](https://tex.z-dn.net/?f=%5Bbase%5D%3D0.000781mol%2F0.025L%3D0.0312M)
Then, considering that the Ka of this acid is 2.5x10⁻⁵, we obtain for the pH:
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