Question

Consider the half reactions below and determine the overall reaction. H2C20_(aq) – 2 CO2(g) + 2 H+ (aq) + 2 e NO3 (aq) + 4 H*(aq) + 3 e — NO(g) + 2 H20 How many electrons are transferred in the oxidation half-reaction? Choose... How many electrons are transferred in the reduction half-reaction? Choose... How many electrons are transferred in the overall reaction? Choose... -

Answer 1

`1.The oxidation half equation can be written as-

$H_{2}C_{2}0_{4}(aq)\rightarrow 2CO_{2}(g)+2H^+(aq)+2e^-$

here is a transfer of 2 electrons occurs.

as it can be computed from that there is the charge on the left is 0, and the charge on the right is 2$\times 1$ = +2, and so the magnitude of the charge difference is (0-2) = 2

hence to balance we required 2 electron.

2.The reduction half -equation can be written as-

$NO^{3-}(aq)+4H^+(aq)+3e^-\rightarrow NO(g)+2H_{2}0$

Here is a transfer of 3 electrons occurs.

To have the same number of electrons in both half-reactions, multiply the oxidation reaction by 3 and multiply the reduction reaction by 2.

$3H_{2}C_{2}0_{4}(aq)\rightarrow 6CO_{2}(g)+6H^+(aq)+6e^-$

$2NO^{3-}(aq)+8H^+(aq)+6e^-\rightarrow 2NO(g)+4H_{2}0$

So $6e^-$ are transferred in the overall reaction.