What is the Redox Reaction between aqueous solutions of Cu(NO3)2 and KI, to form

Question

What is the Redox Reaction between aqueous solutions of Cu(NO3)2 and KI, to form CuI, KNO3, and I2. The CuI precipitates from the reaction as a solid. Use the smallest whole number coefficients, including states of matter for each reactant and product. What is the Redox Reaction between aqueous solutions of Cu(NO3)2 and KI, to form CuI, KNO3, and I2. The CuI precipitates from the reaction as a solid. Use the smallest whole number coefficients, including states of matter for each reactant and product.

Explanation / Answer

n the reaction, the Cu2+ ion is reduced to the Cu+ ion, and the I- ion is oxidized to the I2 molecule, which has an oxidation state of 0. Write out the unbalanced equation, and see what you can make of it so far.

Cu(NO3)2(aq) + KI(aq) ? CuI(s) + I2(aq) + KNO3(aq)

To balance the electrons, you need to check the states of Cu and I. Cu goes from +2 to +1, which is equivalent to -1. I goes from -1 to 0, equivalent to +1. But you already have 2 I atoms on the right that have changed. For every I, you need a Cu, so you need 2 Cu per 2 I.

2Cu(NO3)2(aq) + ?KI(aq) ? 2CuI(s) + I2(aq) + ?KNO3(aq)

Now balance the rest.

2Cu(NO3)2(aq) + 4KI(aq) ? 2CuI(s) + I2(aq) + 4KNO3(aq)

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