balance using half-reduction method CH3OH+MnO4-=CO2+MnO2 (acidic solution) Solut

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a) Cu -> Cu 2+ + 2e 3e + 4H+ + NO3 - -> NO + 2H2O 3Cu + 8H+ + 2NO3 - -> 3Cu 2+ + 2NO + 4H2O b) instead of telling you all the answers, i think it's better to teach you how to fish yourself. firstly, take on one reactant and its product as a half-equation, for example: Cr2O7(aq) ---> Cr3+( aq) firstly, balance no of atoms other than O and H Cr2O7 2-(aq) ---> 2 Cr3+( aq) then, since there are 7O on the left, add H2O on the right to balance it. Cr2O7 2- (aq) ---> Cr3+( aq) + 7H2O (l) now there are 14H on the right, add H+ on the left to balance it. 14H+ + Cr2O7 2- (aq) ---> 2Cr3+( aq) + 7H2O (l) now all the atoms are balanced, but the charges are not. add e to balance that. 6 electrons are needed on the left. 6e + 14H+ (aq) + Cr2O7 2- (aq) ---> 2Cr3+( aq) + 7H2O (l) so this is one completed half reaction, for the other half, 2Cl - (aq) -> Cl2 (g) + 2e- then coming to combining these 2 half-equations together, we balance the electrons, eqn 1 has 6 on the left, whereas eqn2 has 2 on the right, so we need to multiply by 3 to eqn2, thus giving us 6Cl-(aq) +14H+(aq)+Cr2O7 2- (aq)---> 2Cr3+( aq) + 7H2O (l)+ 3Cl2 (g)

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