Write a balanced equation for each of the following redox reactions in an acidic solution using

Write a balanced equation for each of the following redox...

Key Concepts

Balancing in Acidic Solutions

When balancing redox reactions in an acidic solution, additional steps involve using water (H2O) to balance oxygen atoms and hydrogen ions (H+) to balance hydrogen atoms. Excess electrons, H2O, and H+ are introduced appropriately to ensure that both mass and charge are balanced in the final equation, reflecting the conditions present in an acidic medium.

Half-Reaction Method

The half-reaction method is a systematic approach used to balance redox equations by dividing the overall reaction into two separate half-reactions: one for oxidation and one for reduction. Each half-reaction is balanced independently for atoms and charge, and then the half-reactions are combined while ensuring the electrons exchanged cancel out. This method helps in managing the complexity of electron transfer in redox processes.

Redox Reactions

Redox reactions involve the transfer of electrons between chemical species, leading to changes in oxidation states. These reactions encompass both oxidation (loss of electrons) and reduction (gain of electrons) processes that occur simultaneously. The fundamental idea is that the total number of electrons lost in oxidation must equal the number gained in reduction, ensuring charge conservation.

Transcript

00:01 To balance two redox reactions shown in the screen which take place in accruous acidic conditions so let's start with the first reaction oxidation of tin 2 plus to tin 4 plus by iodate anion so let's start with oxidation reaction first tin 2 plus loses two electrons and becomes tin 4 plus and this reaction doesn't need any additional balancing so additionally iodate is reduced to iodine and we have to balance this reaction in the acididia so first let's balance iodine and now we have to add three water molecules to the right side of the equation to balance oh sorry six water molecules to the right side of the equation to balance oxygen so now there are six oxygen atoms on each side so now we have to balance hydrogen and charge so on the right there are 12 hydrogens therefore we need to add 12 hydrogens to the left side of the equation now let's calculate the total charge on the left so it's 12 minus 2 so it's 10 and on the rate side the net charge is zero so therefore we have to add 10 electrons to the last side of the equation so now it's balanced and basically 10 electrons means that each i .dn.

02:04 5 plus gets 5 electrons and is reduced to iret in 0 which makes sense so now we can balance oxidation reaction which is simply 2 oh which we've already done and now we can calculate the overall net reaction so overall it's a 10 electron reaction so therefore we have to multiply 10 by 5 and second reaction by 1 so now let's write down the total reaction so 5tin 2 plus reacts with 2 iodate anions 12 protons and it forms 15 plus 1 iodine and 6 water molecules so let's check for the charge balance is plus 20 on the right and on the left it's 10 minus 2 plus 12 which makes 20 so there are 12 hydrogens on the left 12 hydrogens on the right 15 atoms on the left 15 atoms on the right 2 iodine atoms on the right to iodine atoms on the left so and 6 oxygens on each side so therefore it's balanced now let's move on to the next equation which is oxidation of arsenic 03 -3 minus to arsenic 04 3 minus so this is oxidation because oxidation number of arsenic changes from plus 3 to plus 5 and it's oxidized by bromine which is reduced in turn now let's balance the equation in the acetic media so first we'll start with oxidation reaction so in the oxidation reaction arsenic 03 -3 minus is oxidized to arsenic 04 3 minus so we have to balance oxygen first.

05:00 So therefore we add one water molecule to the left side of the equation.

05:07 So now we have four oxygen atoms on each side.

05:12 Now we need to balance hydrogen.

05:14 So we add two protons to the right side of the equation.

05:20 So and now we need to balance charge.

05:22 The net charge on the left side of the equation is negative 3...

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