Cr(s) | Cr3+(aq) Pb2+(aq) | Pb(s)
Chromium
Cr2O72- + 6Fe2++ 14H+ --> 2Cr3+ + 6Fe3+ + 7H2O
Chromium (II) chloride = CrCl2Chromium (III) chloride = CrCl3 Chromium (IV) chloride = CrCl4
The two relevant half-equations are: [1] SO2 + 2H2O --> (SO4)2- + 4H+ + 2e [2] (Cr2O7)2- + 14H+ + 6e --> 2Cr3+ + 7H2O Multiply [1] by 3; add together; cancel down H+ and H2O. Giving: 3SO2 + (Cr2O7)2- + 2H+ --> 3(SO4)2- + 2Cr3+ + H2O (The potassium ions are spectator ions) Edit: Can the answerer please give the complete equation? The ionic one is pretty "obvious"; I would prefer the full one. Thanks
oxidation number of S is +6.It oxidize into SO4 ^-2
Cr2O72-(aq) + 3SO2 (g) + 2H+(aq)= 2Cr3+ (aq) + 3SO4 2- (aq) H2O (l)
From orange to (yellowish) green: It is a powerfull oxidant, used to titrate Fe2+ from Mohr's salt to the oxidised form Fe3+ by the following: Cr2O72−(aq) + 14H+ + 6e− → 2Cr3+(aq) + 7H2O
First:Comparing the electropotential values of both redox couples (being equal, see below 1. and 2.)AND considering that Cl2 gas might escape the mixture giving toxic fumes,it does NOT look a good titration method.2Cl- --> Cl2,g + 2e- (delta)Eo= -1.36VCr2O72- + 14H+ + 6e- --> 2Cr3+ + 7H2O (delta)Eo= +1.36VSecond:There is no exact answer possible because the titration end point can never be established, you'll never know when the last Cl2 is escaped, and there is no sharp (sudden) change in colour, and the reaction at the endpoint is very slow.Third: The answer is 60.9 mLThe equivalent point (theorethical end point) can be calculated as follows:mEq reductor (HCl) = mEq oxidator (K2Cr2O7)0.325 (mEq/mL HCl) * VHCl (mL) = 0.526 (mEq/mL K2Cr2O7) * 37.6 (mL K2Cr2O7) So VHCl (mL) = 0.526 (mEq/mL K2Cr2O7) * 37.6 (mL K2Cr2O7) / 0.325 (mEq/mL HCl) = 60.854 = 60.9 mL
net ionic equation only has the ions that become a precipitate: 2Cr3+(aq)+3CO32-(aq)--->Cr2(CO3)3(s) spectator ions are the ones that do not make a precipitate: SO42-,NH4+ no precipitate(combines to form another aqueous solution)=no reaction. hope this helps
chromium (III) acetate at least this is what masteing chemistry told me after I guessed wrong three times I however am not to sure because my chem. teacher "doctor" Ott did not bother teaching me how to figure this out before she assigned it for homewor