SO 4 -2 + I -1 --->I 2 + S -2 Solution since this is in acidic solution, balance

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since this is in acidic solution, balance the atoms first followingthese steps. 1)balance atoms other than H and O 2) then balance O atoms, by adding H2O on the other sidewith appropriate coefficients 3) balance H atoms, by adding H+ on other side w/appropriate coefficients. now break this into 2 half equations SO4-2 ---->S2- I-1 -----> I2 -follow the 3 steps above, we can see that the Sulfur arebalanced already, now we need 4 Oxygens on the right side, so weadd 4H2O. Doing so gives the right side 8 H, so we mustbalance that by adding 8 H+ on the left. SO4-2 + 8H+---->S2- + 4H2O -for the 2nd half reactionm, we balance the I by putting thecoefficient 2 infront of I-1. Since there are no O or Hwe dont need to worry about them. 2I-1 ----->I2 - now for the reaction SO4-2 +8H+ ---->S2- + 4H2O thecharge on the left side is (-2+8) = +6 and the charge on the rightis (-2+0) = -2. So we must make the bigger charge equal tothe smaller one, by adding 8electrons SO4-2 + 8H+ +8e----->S2- + 4H2O -For the rxn. 2I-1 ----->I2 thecharge on the left is -2 while the right is 0 so we add 2e to theright 2I-1 ----->I2+2e- SO4-2 + 8H+ +8e----->S2- + 4H2O 2I-1 ----->I2+2e- (4) there is addition of 8e in the 1st rxn and 2e in the 2nd. thecommon denominator is 8 so we multiply the 2nd rxn by 4 to get8e. 4I-1 ----->4I2+8e- SO4-2 + 8H+ +8e----->S2- + 4H2O (now u can add theproducts and reacants together) 4I- + SO4-2 +8H+ -----> 4I2 + S2- +4H2O

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