Question
Question: \(KI\) and \(CuS{O_4}\) solution when mixed give: A. \(Cu{I_2} + {K_2}S{O_4}\) B. \(C{u_2}{I_2} ...
KI and CuSO4 solution when mixed give:
A. CuI2+K2SO4
B. Cu2I2+K2SO4
C. Cu2I2+K2SO4+I2
D. CuI2+K2SO4+I2
Solution
Hint: Try to recall that when KI and CuSO4 are mixed together then they undergo a redox reaction and not precipitation or double decomposition because it involves oxidizing agents like I− and reducing agent like Cu2+. Now, by using this you can easily find the correct option for the given question.
Complete step by step solution:
It is already known to you that iodide ion (I−) is a good reducing agent and cupric ion (Cu2+) is a good oxidizing agent.
So, when potassium iodide,KI is added to the copper sulphate, CuSO4 solution, then they will undergo redox reaction in which iodide ion reduces cupric ion, Cu2+ to cuprous ion, Cu2+ (blue to white precipitate)and it themselves get oxidized to iodine gas.
The best way to look at the above reaction is via the half equations:
Reduction half equation: 2Cu+2+2e−→2Cu+1
Oxidation half equation: I−1→I2+e−.
To balance these both half equations, double the oxidation half equation (to equalise electrons) and then add both reduction and half equations together.
(I−1→I2+e−)×2
2Cu+2+2e−→2Cu+1
-------------------Add
2Cu+2+I−1→2Cu++I2
Now, add in the spectator ions:
2CuSO4+2KI→Cu2I2+K2SO4+I2
So, the brownish color of solution is due to the liberation of iodine gas.
Therefore, from above we can easily conclude that option C is the correct option to the given option.
Note: It should be remembered to you that the iodine formed in the above reaction reacts with potassium ions in the reaction mixture to give potassium triiodide (KI3).
Also, you should remember that the liberated iodine gas can be titrated against a standardized solution of sodium thiosulphate with starch as an indicator to determine the amount of copper sulphate present in the solution.