Please explain this to me
2I- --> I2 + 2e
Cr2O72- ==> Cr3+
6e + 14H + Cr2O72- ==> 2Cr3+ + 7H20
I don't understand how the left side gets 6 electrons.
Cr on the left is +6 and there are two of them for a total of +12. On the right 2*+3 = +6. So both Cr atoms went from 12 on the left to 6 on the right which is a gain of 6e.
I can see that now thank you very much
To understand how the left side of the equation gets 6 electrons, let's break down the oxidation states of the elements involved.
In the first reaction:
2I- --> I2 + 2e
The iodide ion (I-) has an oxidation state of -1. In iodine (I2), the oxidation state is 0. This means that each iodine atom gains one electron in the reaction, resulting in the formation of I2, and 2 electrons are released.
Therefore, the left side of the equation has a total of 2 electrons.
In the second reaction:
Cr2O72- --> 2Cr3+ + 3H2O
The dichromate ion (Cr2O72-) has an overall oxidation state of +6. In chromium (Cr3+), the oxidation state is +3. This implies that each chromium ion gains 3 electrons in the reaction.
Since the reaction involves two chromium ions (Cr3+), the total number of electrons gained is 2 x 3 = 6 electrons.
Therefore, the left side of the equation has 6 electrons, representing the reduction occurring in the reaction.
To understand how the left side of the chemical equation gets 6 electrons, we need to examine the charges and oxidation states of the elements involved.
In the first equation:
2I- --> I2 + 2e
"I2" refers to molecular iodine, which has zero charge. "2e" represents 2 electrons gained during the reaction.
The left side has two iodide ions (I-) with a charge of -1 each. Together, they have a total charge of -2.
For the charges to balance in a chemical equation, the charges on the left and right sides must be equal. Therefore, we need to find a way for the left side to have a charge of -2.
Since two electrons (2e) have a total charge of -2, we can add them to the left side of the equation to balance the charges:
2I- + 2e --> I2 + 2e
Now, the left side has a charge of -2 and the right side also has a charge of -2.
In the second equation:
Cr2O72- ==> Cr3+
The "Cr2O72-" represents the dichromate ion, which has a charge of -2. On the right side, "Cr3+" represents chromium ions with a charge of +3 each.
To balance the charges, we need to determine how many electrons are involved. Since the charge on the dichromate ion is -2 and each chromium ion has a charge of +3, a total of 6 electrons must be involved in the reaction.
6e + 14H+ + Cr2O72- --> 2Cr3+ + 7H2O
By adding 6 electrons to the left side of the equation, we balance the charges. Now, the left side has a charge of -2 (6(-) and 14(+) cancel each other out), and the right side also has a charge of -2 (2(+3) and 7(0) cancel each other out).
So, in summary, the left side of the equation has 6 electrons because we balance the charges by adding those electrons to account for the difference in charges between the reactants and products.