Question

Oxidation Half Reaction, Reduction Half Reaction, Final Oxidation Half Reaction, Final Reduction Half Reaction, Addition of Two Reactions, and Final/Balanced/Cancelled Redox Reaction for CH3CH=CH2+KMnO4 -> CH3CHOHCH2OH+MnO2 and C6H5-CH2CH2CH3+Na2Cr2O7 -> C6H5-COOH+CO2+CR^3+(in acid)

Answer 1

(- 1,-2,0 and -1) are oxidation state of carbon -1 state to 0 requries 1e^- -2 state to -1 requries 1e^-} total 2e^- are given rightarrow balance oxygen by adding H_2 O rightarrow balance hydrogen by H^+ to also equate change on both side oxidation half xn \r\n KMn O_4 goess to Mn O_2 KMn O_4 + 3e^- rightarrow KMn O_2 rightarrow balance oxygen KMn O_4 + 3e^- rightarrow Mn O_2 + 2 H_2 O rightarrow balance hydrogen KMn O_4 + 3e^- + 4H^+ rightarrow Mn O_2 + 2 H_2 O rightarrow balance change KMn O_4 + 3e^- + 4H^+ rightarrow Mn O_2 + 2 H_2 O + K^+ ......(2) reduction half cell x_n Adding 3 times eq(1) + 2 times eq(2) to cancell e^- on both sides Final equation material implication (2) C_6 H_5 - CH_2 CH_2 CH_3 + Na_2 Cr_2 o_7 rightarrow C_6 H_5 - COOH + CO_2 + Cr^3 + Now material implication C_6 H_5 - CH_2 CH_2 CH_3 rightarrow C_6 H_5 - COOH + 2CO_2 first c goes from -2 to + 3 rightarrow 5e^- second c goes from -2 to + 4 rightarrow 6e^- third c goes from -3 to + 4 rightarrow 7e^-/18e^-