For the redox reaction
`xCr(OH)_(3)+YlO_(3)^(-)+zOH^(-)toxCrO_(4)^(2-)+yl^(-)+5H_(2)O`the sum of the stoichiometric coefficient `(x+y+z)` is equal to

Explanation

To solve the redox reaction given by the equation:

\[ x \text{Cr(OH)}_3 + y \text{IO}_3^- + z \text{OH}^- \rightarrow x \text{CrO}_4^{2-} + y \text{I}^- + 5 \text{H}_2\text{O} \]

we need to determine the stoichiometric coefficients \(x\), \(y\), and \(z\) and then find the sum \(x + y + z\).

### Step 1: Determine the oxidation states

1. **Chromium in Cr(OH)3**:
- The hydroxide ion (OH) has a charge of -1, and there are 3 OH groups.
- Let the oxidation state of Cr be \(x\).
- The equation becomes:
\[ x + 3(-1) = 0 \]
\[ x - 3 = 0 \]
\[ x = +3 \]
- Thus, the oxidation state of Cr in Cr(OH)3 is +3.

2. **Chromium in CrO4^{2-}**:
- Let the oxidation state of Cr in CrO4^{2-} be \(y\).
- Each oxygen has a charge of -2 and there are 4 oxygens:
- The equation becomes:
\[ y + 4(-2) = -2 \]
\[ y - 8 = -2 \]
\[ y = +6 \]
- Thus, the oxidation state of Cr in CrO4^{2-} is +6.

### Step 2: Determine the change in oxidation state

- The change in oxidation state for Cr is from +3 to +6, which means it loses 3 electrons (oxidation).

### Step 3: Analyze the iodine species

1. **Iodine in IO3^-**:
- Let the oxidation state of I in IO3^- be \(z\).
- The equation becomes:
\[ z + 3(-2) = -1 \]
\[ z - 6 = -1 \]
\[ z = +5 \]
- Thus, the oxidation state of I in IO3^- is +5.

2. **Iodine in I^-**:
- The oxidation state of I in I^- is -1.

### Step 4: Determine the change in oxidation state for iodine

- The change in oxidation state for I is from +5 to -1, which means it gains 6 electrons (reduction).

### Step 5: Balance the electrons

- To balance the electrons transferred:
- Chromium loses 3 electrons.
- Iodine gains 6 electrons.
- Therefore, we need to multiply the chromium reaction by 2 to balance the electrons:

\[ 2 \text{Cr(OH)}_3 + 1 \text{IO}_3^- + z \text{OH}^- \rightarrow 2 \text{CrO}_4^{2-} + 1 \text{I}^- + 5 \text{H}_2\text{O} \]

### Step 6: Determine the stoichiometric coefficients

- From the balanced equation:
- \(x = 2\) (for Cr(OH)3),
- \(y = 1\) (for IO3^-),
- \(z = 4\) (for OH^-).

### Step 7: Calculate the sum of stoichiometric coefficients

- Now, we can find the sum:
\[ x + y + z = 2 + 1 + 4 = 7 \]

### Final Answer

The sum of the stoichiometric coefficients \(x + y + z\) is equal to **7**.

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