What is the chemical equation for:
CuI + NH4OH ---> ???
This equation should oxidise Cu+ to Cu2+.
I don't see anything there to oxidize Cu+1 to Cu+2. At least it isn't obvious to me.
To determine the chemical equation for the reaction between CuI (copper(I) iodide) and NH4OH (ammonium hydroxide) that oxidizes Cu+ to Cu2+, we need to consider the oxidation states of the elements involved.
First, let's determine the oxidation states of the elements in the reactants:
- Copper (Cu) is typically found in two oxidation states, +1 (Cu+) and +2 (Cu2+).
- Iodine (I) typically has an oxidation state of -1.
- Nitrogen (N) in NH4OH has an oxidation state of -3.
- Hydrogen (H) in NH4OH has an oxidation state of +1.
- Oxygen (O) in NH4OH has an oxidation state of -2.
Given this information, we can write the oxidation half-reaction:
Cu+ --> Cu2+
Since Cu is being oxidized from +1 to +2, we need to balance the charges by adding electrons to the reactant side:
Cu+ + e- --> Cu2+
Next, we need to write the reduction half-reaction.
NH4OH + e- --> NH3 + H2O
In this reaction, nitrogen is being reduced by gaining an electron. The ammonium ion (NH4+) is reduced to form ammonia (NH3) and water (H2O).
Finally, we can combine the oxidation and reduction half-reactions to obtain the balanced equation:
CuI + NH4OH --> Cu2+ + NH3 + H2O + I-
Note that the iodide ion (I-) is balanced with the copper ion (Cu2+).
Therefore, the chemical equation that oxidizes Cu+ to Cu2+ when CuI is reacted with NH4OH is:
CuI + NH4OH --> Cu2+ + NH3 + H2O + I-