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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12 Dahlias, 30 Tulips
12:30
12/2=6
30/6=5
12/6=2
=2:5
Answer:
B
Explanation:
Always remember blood first flows from atria (remember A comes before V) to ventricles and the right side always carries deoxgenated blood to the lungs which then arrives through the left side when oxygenated
Answer: Sugar
Explanation: Because after photosynthesis (Carbon dioxide, water, and sunlight) react, they make two products. Glucose (A sugar) and Oxygen.
Answer : The equilibrium concentration of NO is, 0.0092 M.
Solution :
First we have to calculate the concentration of NO.
The given equilibrium reaction is,
Initially conc. 0 0 0.1576
At eqm. (x) (x) (0.1576-2x)
The expression of 
will be,
![K_c=\frac{[NO]^2}{[N_2][O_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO%5D%5E2%7D%7B%5BN_2%5D%5BO_2%5D%7D)
By solving the term, we get:
Neglecting the 0.0839 value of x because it can not be more than initial value.
Thus, the value of 'x' will be, 0.0742 M
Now we have to calculate the equilibrium concentration of NO.
Equilibrium concentration of NO = (0.1576-2x) = [0.1576-2(0.0742)] = 0.0092 M
Therefore, the equilibrium concentration of NO is, 0.0092 M.
