# Half-Reaction Balance equation?

Using the half-reaction method, balance the two equations below:

1)

OCl− (aq) + I−(aq) → I2(aq) + Cl−(aq)

2)

I2(aq) + S2O3

2−(aq) → I−(aq) + S4O6

2−(aq)

Thanks!

Relevance

OCl- +2H+ + 2e- = Cl- + H2O

2 I- = I2 + 2e-

OCl- + 2 H+ + 2 I- = Cl- + I2 + H2O

I2 + 2e- = 2 I-

2S2O32- = S4O62- + 2e-

I2 + 2 S2O32- = 2 I- + S4O62-

• NO2^- ion will become NO (artwork out Oxidation States of N) N(III) will become N(II) ie. is decreased by ability of electron benefit N^5+ + e --------> N^2+ (artwork out sort of electrons in touch) NO2^- + e --------> NO (stability O by ability of including H2O on suitable facet) NO2^- + e --------> NO + H2O (stability H by ability of including H^+ on suitable facet) 2H^+ + NO2^- + e --------> NO + H2O (examine fee balances: left (2+) + (-) + (-) = 0: perfect=0 that's a million/2 equation for help of the nitrite ion In^+ will become In^3+ ie. is oxidised by ability of electron loss In^+ -------> In^3+ + 2e that's a million/2 equation for oxidation of the In^+ combine the two a million/2 equations to get rid of the electrons 2 x help equation + oxidation equation 4H^+ + 2NO2^- + 2e + In^+ --------> 2NO + 2H2O + In^3+ + 2e 4H^+ + 2NO2^- + In^+ --------> 2NO + 2H2O + In^3+ this could frequently be sufficient, yet placed into sort asked for 2H^+ + 2HNO2 + In^+ --------> 2NO + 2H2O + In^3+ (NB. The 2H^+ shows the elect for an acidic ecosystem)