In the reaction below, when balanced (question marks filled in with whole number
ID: 724003 • Letter: I
Question
In the reaction below, when balanced (question marks filled in with whole numbers), the coefficient for S(s) is 3. I need an explanation as to why that is, and what would be the method for figuring that out. Thanks!?(Cr2O7)2- (aq) + ?H2S(aq) --> ?Cr3+(aq) + ?S(s)
Explanation / Answer
first u have to find the oxidation state of Cr and S in both the reactants and products and then u have to balance the charge and molecules with the help of H2O and H+ in case acidic aqueous solution and H2O and OH- in case of basic solutions H2S = S + 2 H+ + 2e- ---------------- (1) Cr2O72- + 14 H+ + 6e-= 2 Cr3+ + 7 H2O -------------(2) now multiply eq (1) by 3 and add both the eqs 3H2S = 3S + 6 H+ + 6e- Cr2O72- + 14 H+ + 6e-= 2 Cr3+ + 7 H2O adding the reactants side and products side separately we get 3H2S + Cr2O72- + 14 H+ + 6e- ----> 3S + 6 H+ + 6e- + 2 Cr3+ + 7 H2O Now we can cancel the similar items on both the sides of reaction like H+ and e- 6e- at the reactants side are cancelled by 6e- at the product side. similarly Out of 14 H+ ions at the reactants side 6 are subtracted by the 6H+ ions present at the product side. So final eq is 3 H2S + Cr2O72- + 8 H+ = 3 S + 2 Cr3+ + 7 H2O