Question

i am stuck please help me. can you do a step by step guide with explanations flr i cna aplky this to other problems. thank you

Balance the following redox reaction in basic solution: sights Cr20,2- (aq) + Fe 2+ (aq) → Cr* (aq) + Fe+ (aq) Cr₂O cay - Cr? Feet cay Fest cay

Answer 1

The correct equation is Cr202- (aq) + 7 H2O + 6 Fe2+(aq)- 2Cr3+ (aq) +6 Fe3+ (aq) + 14 OH- Cr2O72- (aq) + Fe2(aq) - Cr3+ (aq) + Fe3+ (aq) 2 3 3 Oxidation numbers For the given redox reaction, Reduction half cell reaction is Cr2O72- (aq) Cr3+ (aq) decrease in oxidation number Oxidation half cell reaction is Fe2+(aq) - Fe3+ (aq) increase in oxidation number To write the balance equation following steps has to be followed Reduction half cell reaction Oxidation half cell reaction 1. Balancing otherthan 0 and H atoms Cr2O72- (aq) — 2Cr3+ (aq) Fe2+ (aq) - Fe3+ (aq) 2. Balancing O atoms 2. Cr2O72- (aq) - 2Cr3+ (aq) + 7 H20 (since there are 70 s present on the left side, 7 H2O should be added on the right hand side Fe2+(aq) — Fe3+ (aq) 3. Balancing H atoms 3. Cr2O72- (aq) + 14 H20 - 2Cr3+ (aq) + 7 H20 + 14 OH Fe2+(aq) — Fe3+ (aq) (since there are 14 H are present in right side 14 H2O should be added on the left hand side) To neutralize the charge 14 OH" were added on the right hand side 4. balancing the charges Fe2+(aq) — Fe3+ (aq) + 1 e Cr2O72- (aq) + 14 H20 + 6 e — 2Cr3+ (aq) + 7 H20 + 14 OH- Right hand side overall charge is -8 Charges were balanced by using electrons To neutralize the charge 6 e-are added on the left hand side. Now the total half cell reactions are Fe2+(aq) — Fe3+ (aq) + 1 e - ........ Cr2O72- (aq) + 14 H2O + 6 e - 2Cr3+ (aq) + 7 H2O + 14 OH To balance the number of electrons, multipling equation with 6 gives 6 Fe2+ (aq) — 6 Fe3+ (aq) + 6 e ........ 3 ........ By cancelling the common terms in equations 2 and 3 gives the final equation as shown here. 7 H20 Cr2072 (aq) + 14 H2O + € 6 Fe2(aq) - — 2Cr3+ (aq) + 7 H2O + 14 OH 6 Fe3+ (aq) + ........ ........ Cr2O72- (aq) + 7 H2O + 6 Fe2+(aq) — 2Cr3+ (aq) +6 Fe3+ (aq) + 14 OH-