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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} ...

KIKI and CuSO4CuS{O_4} solution when mixed give:
A. CuI2+K2SO4Cu{I_2} + {K_2}S{O_4}
B. Cu2I2+K2SO4C{u_2}{I_2} + {K_2}S{O_4}
C. Cu2I2+K2SO4+I2C{u_2}{I_2} + {K_2}S{O_4} + {I_2}
D. CuI2+K2SO4+I2Cu{I_2} + {K_2}S{O_4} + {I_2}

Explanation

Solution

Hint: Try to recall that when KIKI and CuSO4CuS{O_4} are mixed together then they undergo a redox reaction and not precipitation or double decomposition because it involves oxidizing agents like I{I^ - } and reducing agent like Cu2+C{u^{2 + }}. 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{I^ - }) is a good reducing agent and cupric ion (Cu2+C{u^{2 + }}) is a good oxidizing agent.
So, when potassium iodide,KIKI is added to the copper sulphate, CuSO4CuS{O_4} solution, then they will undergo redox reaction in which iodide ion reduces cupric ion, Cu2+C{u^{2 + }} to cuprous ion, Cu2+C{u^{2 + }} (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+2e2Cu+12C{u^{ + 2}} + 2{e^ - } \to 2C{u^{ + 1}}
Oxidation half equation: I1I2+e{I^{ - 1}} \to {I_2} + {e^ - }.
To balance these both half equations, double the oxidation half equation (to equalise electrons) and then add both reduction and half equations together.
(I1I2+e)×2({I^{ - 1}} \to {I_2} + {e^ - }) \times 2
2Cu+2+2e2Cu+12C{u^{ + 2}} + 2{e^ - } \to 2C{u^{ + 1}}
-------------------Add
2Cu+2+I12Cu++I22C{u^{ + 2}} + {I^{ - 1}} \to 2C{u^ + } + {I_2}
Now, add in the spectator ions:
2CuSO4+2KICu2I2+K2SO4+I22CuS{O_4} + 2KI \to C{u_2}{I_2} + {K_2}S{O_4} + {I_2}
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 (KI3K{I_3}).
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.