# A level Redox reaction half equation calculation HELP

Can someone please explain simply how to solve this question, I know I need to find a ratio but the equation is already balanced so I’m confused where to get the ratio unless I’m wrong. I’m really bad with calculations and redox as a topic confuses me no matter how much I study it. Please no rude comments I don’t have a tutor/teacher or any peers to help me with this and I’m a self study student. How do I find the ratio to then calculate the next steps???

(edited 1 year ago)
Original post by Nairbear
Can someone please explain simply how to solve this question, I know I need to find a ratio but the equation is already balanced so I’m confused where to get the ratio unless I’m wrong. I’m really bad with calculations and redox as a topic confuses me no matter how much I study it. Please no rude comments I don’t have a tutor/teacher or any peers to help me with this and I’m a self study student. How do I find the ratio to then calculate the next steps???

With redox you must balance the electrons, so times the top equation by 5 and the bottom by 2 so the electrons cancel out…
Original post by Nairbear
Can someone please explain simply how to solve this question, I know I need to find a ratio but the equation is already balanced so I’m confused where to get the ratio unless I’m wrong. I’m really bad with calculations and redox as a topic confuses me no matter how much I study it. Please no rude comments I don’t have a tutor/teacher or any peers to help me with this and I’m a self study student. How do I find the ratio to then calculate the next steps???

You have to multiply the top equation by 5 and bottom equation by 2 so that the number of electrons are equal in both equations and can cancel out. You then cancel out the electrons and form a full equation. this makes the ratio between oxalate and manganate ions 2:5. Since the moles of oxalate is 0.02, you multiply by 2/5 to give moles of manganate ions and this makes 8x10-3 mol of manganate ions and then you use the equation volume = mol/concentration to find the volume needed of manganate ions. so youdo (8x10-3)/0.02 to give 0.4dm3 volume of manganate ions, and then to make it to cm3 you multiply by 1000 to give 400cm3. Hope this makes sense!
(edited 1 year ago)
Original post by T-A-27
You have to multiply the top equation by 5 and bottom equation by 2 so that the number of electrons are equal in both equations and can cancel out. You then cancel out the electrons and form a full equation. this makes the ratio between oxalate and manganate ions 2:5. Since the moles of oxalate is 0.02, you multiply by 2/5 to give moles of manganate ions and this makes 8x10-3 mol of manganate ions and then you use the equation volume = mol/concentration to find the volume needed of manganate ions. so youdo (8x10-3)/0.02 to give 0.4dm3 volume of manganate ions, and then to make it to cm3 you multiply by 1000 to give 400cm3. Hope this makes sense!

THANK YOU SO MUCH I completely forgot about balancing the electron part. Redox always messes me up I will try it again!!
(edited 1 year ago)
Original post by ashvinsingh
With redox you must balance the electrons, so times the top equation by 5 and the bottom by 2 so the electrons cancel out…

Thank you, I always mess up on redox and forgot this step. Will try again and hopefully get the answer.