The half cell in which the electrode gains electrons is where reduction occurs, and the half cell in which the electrode loses electrons is where oxidation occurs.
<h3><u>What is a Galvanic cell ?</u></h3>
Voltaic or galvanic cells are electrochemical devices that use spontaneous oxidation-reduction events to generate electricity. In order to balance the overall equation and highlight the actual chemical changes, it is frequently advantageous to divide the oxidation-reduction reactions into half-reactions while constructing the equations.
Two half-cells make up most electrochemical cells. The half-cells allow electricity to pass via an external wire by separating the oxidation half-reaction from the reduction half-reaction.
<h3><u>
Oxidation:</u></h3>
The anode is located in one half-cell, which is often shown on the left side of a figure. On the anode, oxidation takes place. In the opposite half-cell, the anode and cathode are linked.
<h3><u>Reduction:</u></h3>
The second half-cell, cathode, which is frequently displayed on a figure's right side. The cathode is where reduction happens. The circuit is completed and current can flow by adding a salt bridge.
To know more about processes in Galvanic cell, refer to:
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8.8 × 10-5 M is the [H3O+] concentration in 0.265 M HClO solution.
Explanation:
HClO is a weak acid and does not completely dissociate in water as ions.
the equation of dissociation can be written and ice table to be formed.
HClO +H2O ⇒ ClO- + H3O+
I 0.265 0 0
C -x +x +x
E 0.265-x +x +x
Now applying the equation of Ka, where Ka is given as 2.9 × 10-8.
Ka = ![\frac{[ClO-][H3O+]}{[HClO]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BClO-%5D%5BH3O%2B%5D%7D%7B%5BHClO%5D%7D)
2.9 × 10^-8 = ![\frac{[x] [x]}{[0.265-x]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bx%5D%20%5Bx%5D%7D%7B%5B0.265-x%5D%7D)
= 7.698 x
x = 8.8 × 10-5 M
The hydronium ion concentration is 8.8 × 10-5 M in 0.265 M solution of HClO.
Answer:
-100 kJ
Explanation:
We can solve this problem by applying the first law of thermodynamics, which states that:
where:
is the change in internal energy of a system
Q is the heat absorbed/released by the system (it is positive if absorbed by the system, negative if released by the system)
W is the work done by the system (it is positive if done by the system, negative if done on the system)
For the system in this problem we have:
W = +147 kJ is the work done by the system
Q = +47 kJ is the heat absorbed by the system
So , its change in internal energy is:
