abelkoontz Posted October 24, 2016 Report Share Posted October 24, 2016 I-(aq) + HSO4-(aq) -> I2(aq) + SO2(g) How do you solve this? I just learned this, but if they are both - charge on the left, how can one be oxidized and the other reduced if they are both 0 oxidation states on the right? WHAT AM I MISSING? They should be 0 on both sides right.. can someone solve this and explain it? I don't really get it. tHANKS Reply Link to post Share on other sites More sharing options...
kw0573 Posted October 24, 2016 Report Share Posted October 24, 2016 2I- + HSO4- + 3H+--> I2 + SO2 + 2H2O (Acidic condition balance) Now if we add 3 hydroxides to both sides 2I- + HSO4- + H2O --> I2 + SO2 + 3OH- The key is thus to first balance oxidation states, then balance oxygens, then balance H+ to get acid conditions, and if question asks, add equal number of hydroxides to both sides and find basic conditions. Reply Link to post Share on other sites More sharing options...
Msj Chem Posted October 24, 2016 Report Share Posted October 24, 2016 I have this (old) video that explains how to balance redox equations in acidic solutions. You might find it helpful. Reply Link to post Share on other sites More sharing options...
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