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^-"
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