Study the given figure carefully.
Which of the following reactions explains the above change most appropriately?

Z n S O_{4} + C u \to C u S O_{4} + Z n

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C u S O_{4} + F e \to F e S O_{4} + C u

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F e S O_{4} + C u \to C u S O_{4} + F e

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C u S O_{4} + Z n \to Z n S O_{4} + C u

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Solution

The correct option is B $C u S O_{4} + F e \to F e S O_{4} + C u$
The blue coloured solution is $C u S O_{4}$ solution. Iron $(F e)$ is more reactive than copper $(C u)$ . So, $F e$ displaces $C u$ from the $C u S O_{4}$ solution and forms $F e S O_{4}$ solution which is green in color. Brown-red deposition in a test tube is of copper. Such types of reactions are called displacement reactions.

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Similar questions

C u S O_{4} + F e \to F e S O_{4} + C u

F e S O_{4} + Z n \to Z n S O_{4} + F e

C u S O_{4} + F e \to F e S O_{4} + C u

F e S O_{4} + Z n \to Z n S O_{4} + F e

Z n + C u S O_{4} \to Z n S O_{4} + C u

E_{Z n / Z n^{2 +}}^{\circ} = 0.765 v o l t; E_{C u^{2 +} / C u}^{\circ} = 0.347 v o l t

C u S O_{4} + Z n \to C u + Z n S O_{4}

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