The oxidation state of 'Cr' in the final product formed by reaction between $\text{KI}$ and acidified $K_2Cr_2O_7$ is :
Show Hint
Remember the standard redox reactions of dichromate: in acidic medium, $Cr_2O_7^{2-}$ is always reduced to $Cr^{3+}$, corresponding to a change in oxidation state from $+6$ to $+3$.
The given question requires us to find the oxidation state of chromium (\(Cr\)) in the final product formed by the reaction between potassium iodide (\(\text{KI}\)) and acidified potassium dichromate (\(K_2Cr_2O_7\)).
Step-by-step Solution
Start by writing the balanced chemical equation for the reaction between \(\text{KI}\) and acidified \(K_2Cr_2O_7\): \(6\text{KI} + \text{K}_2\text{Cr}_2\text{O}_7 + 7\text{H}_2\text{SO}_4 \to 3\text{I}_2 + \text{Cr}_2\text{(SO}_4)_3 + 4\text{K}_2\text{SO}_4 + 7\text{H}_2\text{O}\)
In this reaction, \(\text{K}_2\text{Cr}_2\text{O}_7\) is reduced to \(\text{Cr}_2\text{(SO}_4)_3\).
Initially, the oxidation state of \(Cr\) in \(K_2Cr_2O_7\) is +6. To find this, consider the compound: \(K_2\text{Cr}_2\text{O}_7\) → 2 potassium (K), 2 chromium (Cr), and 7 oxygen (O).
The algebraic sum of the oxidation states in a neutral compound is zero: \(2(+1) + 2(x) + 7(-2) = 0\), where \(x\) is the oxidation state of Cr.
Simplifying the equation: \(2 + 2x - 14 = 0 \Rightarrow 2x - 12 = 0 \Rightarrow 2x = 12 \Rightarrow x = +6\). Therefore, the initial oxidation state of \(Cr\) is +6.
In the product, the chromium is part of \(\text{Cr}_2\text{(SO}_4)_3\).
Sulphate is a divalent anion (charge = -2). In \(\text{Cr}_2\text{(SO}_4)_3\), the sum of the charges must equal zero, thus: \(2(x) + 3(-2) = 0 \Rightarrow 2x - 6 = 0 \Rightarrow 2x = 6 \Rightarrow x = +3\). This implies that the oxidation state of Cr in the final compound is +3.
Therefore, the oxidation state of Chromium, \(Cr\), in the final product is +3.
Conclusion
Considering the above calculation and explanation, the correct answer is +3.
