Balance the following redox reaction in basic solution.
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Question

Balance the following redox reaction in basic solution.

Cr₂O₇^2– (aq) + SO₃^2– (aq) → Cr³⁺ (aq) + SO₄^2– (aq)

Accepted Answer

Here we're going to balance this redox reaction in a basic environment. We're going to first break it up into its half reactions. We have chromium with chromium, sulfur with sulfur. That's going to give us our two half reactions initially. We have We have 2 chromiums here but only 1 here, so we put a 2. Then we have 1 sulfur and 1 sulfur, so we're fine. Next, balance out high, oxygens by adding water. So we have to add 7 waters here. Here we have 3 oxygens and here we we have 4, so we put 1 water here. Next we're going to balance out waters by adding h balance out hydrogens by adding h plus. So we have 7 times 2 is 14 h plus and then we have 2 hydrogens here so that's 2h+. Balance overall charge, so plus 14 minuteus 2 gives me plus 12 overall. And then here, this is 2 times plus 3 so that's plus 6. Here, this would be minus 2, overall for the charge. And over here, it's 2 times plus 1 is plus 2. Minus 2 is 0. So I'm gonna put 6 electrons here so that both sides are plus 6. And then I have to put 2 electrons here so that both sides are minus 2. Number of electrons are not the same, so I multiply this here by 3. So what I'm gonna get now is I'm gonna get cr 2072 minus plus 14h+6electrons gives me 2cr3+7waters. Then we have 3 s o three two minus plus 3 H two o gives me 3 s o four two minus plus 6 H plus plus 6 electrons. Cross out the intermediate, so this cancels out with this. This cancels out leaving us 8 here. Bring down everything. So this will represent our balanced equation in an acidic environment. So we're not quite done yet because we need to balance it in a basic environment. So we have that. So here, the 7 step is we look at how much o h minus, how much h plus is left and we add the same number as o h minus to both sides. And then, plus also 7 h2o. So, these here combine together to give me 8 waters. So we add everything. Now here you can have water on both sides, so all 7 of these cancel out with 7 from here, leaving us with one left. And that represents our balanced equation in a basic environment. We just lost 7 of these, so we have one left. So this would be our basic solution with its balance redox reaction. So remember, balancing it in this type of solution is almost the same as balancing it in an acidic environment except we have that last additional step where we add equal amounts of o h minus to both sides based on the amount of H plus remaining. When H+ and o h minus combine, they give us water molecules. We can't have water on both sides of the equation, so they're going to cancel out and we'll see what is remaining. When it comes to balancing redox reactions, remember the steps necessary in order to get your final answer.

Struggling with Analytical Chemistry?

Balance the following redox reaction in basic solution.
Cr₂O₇^2– (aq) + SO₃^2– (aq) → Cr³⁺ (aq) + SO₄^2– (aq)

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Balance the following redox reaction in basic solution.
Cr₂O₇^2– (aq) + SO₃^2– (aq) → Cr³⁺ (aq) + SO₄^2– (aq)