Chemistry, asked by dpalei27, 11 months ago

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Answered by RvChaudharY50
37

Answer:

color change when Iodine is oxidized at the anode to form the brownish I2 on the electrode and yellow I3^(-1) in solution.

The reduction of H+ to form H2 (g) and OH− will turn the indicator blue at the cathode.

when filter paper dipped in potassium iodide brought near to HNO3 fumes, they form brown coloured iodine and nitrogen dioxide gas.

Answered by GETlost0hell
0

Answer:

Explanation:

color change when Iodine is oxidized at the anode to form the brownish I2 on the electrode and yellow I3^(-1) in solution.

The reduction of H+ to form H2 (g) and OH− will turn the indicator blue at the cathode.

when filter paper dipped in potassium iodide brought near to HNO3 fumes, they form brown coloured iodine and nitrogen dioxide gas.

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