Can you help me balance an oxidation-reduction equation by the half reaction method?
Equation: CH3OH(aq)+CrO7 2-(aq)---->CH2O(aq)+Cr3+(aq)
Reduction half reaction:
14H+(aq) +Cr2O7 2-(aq)------>Cr3+ +7H20(l)
Thank you.
Answer:
half reactions:
Cr2O7 2- => 2Cr3+ (notice that Cr goes from +6 to +3, and there are two of them, so you need 6 electrons)
6e- 14H+ + Cr2O7 2- => 2Cr3+ + 7H2O
CH3OH => CH2O (C goes from -2 to 0, so 2 electrons are needed)
CH3OH => CH2O + 2e- + 2H+
so the two half reactions are:
6e- 14H+ + Cr2O7 2- => 2Cr3+ + 7H2O
CH3OH => CH2O + 2e- + 2H+
multiply the second by 3 and add the two:
14H+ + 3CH3OH + Cr2O7 2- => 2Cr3+ + 7H2O + 3CH2O + 6H+
answer:
8H+ + 3CH3OH + Cr2O7 2- => 2Cr3+ + 7H2O + 3CH2O
You need to equalize the total charge on each side of the half-reaction.
You need a two in front of the Cr3+, which then gives +6 on the right. On the left, you'd have +14 and -2 and then adding in the -6 from the six electrons will give +6 on the left-hand side.
HTH
14H+(aq) +Cr2O7 2-(aq)------>2Cr3+ +7H20(l)
Cr2O7 (2-)
each Cr is +7
-----> 2Cr3+
means each Cr grabbed
4 electrons
since theres (2-) on the Cr2O7 that accounts for 2 of the 8 extra electrons on the Cr3+, so you need to put in the pther 6 on the left to balance
Cr2O7-2(aq) + 14H+(aq) + 6e- --> 2Cr+3 + 7H2O(l)
CH3OH(aq) --> CH2O(aq) + 2H+(aq) + 2e- * 3
add
Cr2O7-2(aq) + 8H+(aq) + 3CH3OH(aq) -->
3CH2O(aq) + 2Cr+3(aq) + 7H2O(l)
add spectator ions
K2Cr2O7(aq) + 4H2SO4(aq) + 3CH3OH(aq) -->
3CH2O(aq) + Cr2(SO4)3(aq) + 7H2O(l) + K2SO4(aq)
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