Question

Determine the spontaneous reaction between Br2, I2, Br– and I– at 25 °C, Include physical states in your answer.

&

Balance this reduction-oxidation reaction. It occurs in an acidic solution.

I2 + PH3 I- + H3PO4

Answer 1

Solution.

1.

Write down the electrode potentials of bromine and iodine to determine who is the oxidizer and reducing agent.

$Br2(l) + 2e = 2Br^-(aq) \ \phi = +1.065 \ V$

$I2(s) + 2e = 2I^-(aq) \ \phi = +0.536 \ V$

In this situation, bromine is the oxidizer and iodine will be the reducing agent, so the reaction equation will look like this:

$Br2(l) + 2I^-(aq) = I2(s) + 2Br^-(aq)$

2.

$4I2 + PH3 + 4H2O = H3PO4 + 8H^+ + 8I^-$

Answer:

1.

$Br2(l) + 2I^-(aq) = I2(s) + 2Br^-(aq)$

2.

$4I2 + PH3 + 4H2O = H3PO4 + 8H^+ + 8I^-$