Question

When the following equation is balanced properly under acidic conditions, what are the coefficients of the species shown? Cr3+ + H2S— Cr2+ + S reactant, product, neither) with a coefficient of Water appears in the balanced equation as a (Enter 0 for neither.) How many electrons are transferred in this reaction?

Answer 1

According to the Electron Theory of Redox Reaction, reduction is the taking of electron by the reactant and Oxidation is the liberating of electron by the reactant.

Here also we are using this concept in determining the half reaction of the Redox reactions. As in the reduction process, reactants take the electron, the Oxidation number of the reactants decreases in that case. Similarly, in the Oxidation reaction, reactants liberate the electron, so the Oxidation number of the reactants increases.

The electron transfered in this reaction are 2. It is calculated from the balanced equation of the Redox Reaction. We eliminate the 2 electrons from the both side of the equation. This is the total electron transfered in this reaction.

The condition of the medon reaction is acidie environment. So, we have to balance the equation with Ht ions due to the acidic condition. Now, there are two half-reactions in one redon reactions. Cotte - G2 Creduction) electrons are taken by the reactants and the onidation number of Cr3+ decreases from +3 to +2. So ist in a reduction process. HzS – S +245 +2é Conidation) electron are liberated by the reactant and the onidation number of His inescases from 2 to 0. So it in an onidation в осе». Now, the lealanced equation, 268134 + 2€ + H2S ->2C%2*+ S +2* +2€ On 2Cr3+ + H2S – 2002+ +5+2+ And water doesn't affeas in this equation So the coefficient of water in Zero.

The sredox reaction is done under acidic condition, so the balancing of the equation is done by Ht ion only. The two half reactions of the recon brocess Sn't +25 - Sn (reduction) 0. As election is taken by the reactant and the oxidation number of so decreases from +2 to O, So this is a reduction process. Similarly, 2560* +3460 - 56,05 + 6H+ + 4e As elections are liberated by the reactant and the oridation number of 56 also increases from +3 to +5 here. So thin is an oxidation Beaction. Now, to balance these two half equation, 0x2 +®, we get 25m2+ + e +2560++ 34,0 → 250 +56,05+6+4h 05 25m2+ + 2 S60* + 3H20 → 25m + Sb2O5 +6H+ Water appeares in the balanced equation as preactant with a coefficient of 3.
The total electron transfered in this Redox Reaction is 4. It is calculated from the balanced equation of the reaction. We eliminate the 4 e^{​​​​​​-} from the both side of the reaction. This is the total electron transfered in this Redox Reaction.