Chromium plating can involve the electrolysis of an electrolyte of an acidified mixture of chromic acid and chromium sulphate. If during electrolysis the article being plated increases in mass by `2.6g` and `0.6dm^(3)` of oxygen are evolved at an inert anode, the oxidation state of chromium ions being discharged must be `: (` assuming atomic weight of `Cr=52` and `1 mol e` of gas at room temperature and pressure occupies a volume at `24dm^(3))`

Options

A. `-1`
B. Zero
C. `+1`
D. `+2`

Explanation

To solve the problem of determining the oxidation state of chromium ions being discharged during the electrolysis of an acidified mixture of chromic acid and chromium sulfate, we can follow these steps:

### Step 1: Calculate the moles of oxygen evolved
Given that 0.6 dm³ of oxygen is evolved, we can use the molar volume of a gas at room temperature and pressure (RTP), which is 24 dm³/mol, to find the moles of oxygen.

\[ \text{Moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Molar volume at RTP}} = \frac{0.6 \, \text{dm}^3}{24 \, \text{dm}^3/\text{mol}} = 0.025 \, \text{mol} \]

### Step 2: Determine the charge associated with the oxygen evolution
The balanced half-reaction for the evolution of oxygen at the anode is:

\[ 4 \text{H}^+ + 4 \text{e}^- \rightarrow 2 \text{H}_2O + O_2 \]

From the reaction, we see that 4 moles of electrons are required to produce 1 mole of \(O_2\). Therefore, the moles of electrons (n) involved in the reaction can be calculated as follows:

\[ \text{Moles of electrons} = 4 \times \text{Moles of } O_2 = 4 \times 0.025 = 0.1 \, \text{mol} \]

### Step 3: Calculate the total charge (Q) using Faraday's constant
Using Faraday's constant (\(F \approx 96500 \, \text{C/mol}\)), we can calculate the total charge:

\[ Q = n \times F = 0.1 \, \text{mol} \times 96500 \, \text{C/mol} = 9650 \, \text{C} \]

### Step 4: Calculate the moles of chromium deposited
The mass increase of the article being plated is given as 2.6 g. To find the moles of chromium deposited, we use the molar mass of chromium (Cr = 52 g/mol):

\[ \text{Moles of Cr} = \frac{\text{Mass of Cr}}{\text{Molar mass of Cr}} = \frac{2.6 \, \text{g}}{52 \, \text{g/mol}} = 0.05 \, \text{mol} \]

### Step 5: Determine the charge associated with the deposition of chromium
Let \(x\) be the number of electrons lost by one mole of chromium during the plating process. The charge associated with the deposition of chromium can be expressed as:

\[ Q_{\text{Cr}} = \text{Moles of Cr} \times x \times F \]

Substituting the values we have:

\[ Q_{\text{Cr}} = 0.05 \, \text{mol} \times x \times 96500 \, \text{C/mol} \]

### Step 6: Set the charges equal
Since the charge for the deposition of chromium must equal the charge for the evolution of oxygen, we set the two equations equal:

\[ 0.05 \, x \times 96500 = 9650 \]

### Step 7: Solve for \(x\)
Solving for \(x\):

\[ 0.05 \, x \times 96500 = 9650 \implies x = \frac{9650}{0.05 \times 96500} = 2 \]

### Step 8: Determine the oxidation state of chromium
The oxidation state of chromium can be determined from the number of electrons lost. The maximum oxidation state of chromium is +6, and since 2 electrons are lost:

\[ \text{Oxidation state of Cr} = +6 - x = +6 - 2 = +4 \]

### Final Answer
The oxidation state of chromium ions being discharged must be +4.

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