`E^(@) ` of two reactions are given below :
`Cr^(3+) + 3e rarr Cr, E^(@) = -0.74 V`
`OCl^(-) + H_(2)O + 2e rarr Cl^(-) + 2OH^(-), E^(@) = 0.94 V`
What will be the `E^(@)` for ?
`3OCl^(-) + 2Cr + 3H_(2)O rarr 2Cr^(3+) + 3Cl^(-) + 6OH^(-)`

Options

A. `-1.68 V`
B. `1.68 V`
C. `-0.20 V`
D. `0.20 V`

Explanation

To determine the standard cell potential \( E^\circ \) for the reaction:

\[ 3 \text{OCl}^- + 2 \text{Cr} + 3 \text{H}_2\text{O} \rightarrow 2 \text{Cr}^{3+} + 3 \text{Cl}^- + 6 \text{OH}^- \]

we will use the provided half-reactions and their standard electrode potentials.

### Step 1: Identify the half-reactions and their potentials

The half-reactions given are:

1. \( \text{Cr}^{3+} + 3e^- \rightarrow \text{Cr}, \quad E^\circ = -0.74 \, \text{V} \) (Reduction)
2. \( \text{OCl}^- + \text{H}_2\text{O} + 2e^- \rightarrow \text{Cl}^- + 2\text{OH}^-, \quad E^\circ = 0.94 \, \text{V} \) (Reduction)

### Step 2: Balance the electrons in the half-reactions

To combine these half-reactions, we need to ensure that the number of electrons lost in the oxidation reaction equals the number of electrons gained in the reduction reaction.

- The first half-reaction involves 3 electrons (for Cr).
- The second half-reaction involves 2 electrons (for OCl).

To balance the electrons, we can multiply the first half-reaction by 2 and the second half-reaction by 3.

### Step 3: Write the balanced half-reactions

1. Multiply the first half-reaction by 2:
\[ 2 \text{Cr}^{3+} + 6e^- \rightarrow 2 \text{Cr}, \quad E^\circ = -0.74 \, \text{V} \]

2. Multiply the second half-reaction by 3:
\[ 3 \text{OCl}^- + 3 \text{H}_2\text{O} + 6e^- \rightarrow 3 \text{Cl}^- + 6 \text{OH}^-, \quad E^\circ = 0.94 \, \text{V} \]

### Step 4: Combine the half-reactions

Now, we can add the two half-reactions together:

\[ 3 \text{OCl}^- + 3 \text{H}_2\text{O} + 6e^- + 2 \text{Cr}^{3+} \rightarrow 3 \text{Cl}^- + 6 \text{OH}^- + 2 \text{Cr} + 6e^- \]

The electrons cancel out:

\[ 3 \text{OCl}^- + 2 \text{Cr} + 3 \text{H}_2\text{O} \rightarrow 2 \text{Cr}^{3+} + 3 \text{Cl}^- + 6 \text{OH}^- \]

### Step 5: Calculate the standard cell potential \( E^\circ \)

The standard cell potential \( E^\circ \) for the overall reaction is calculated using:

\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}} \]

In this case, the reduction potential for the second reaction is \( 0.94 \, \text{V} \) and the oxidation potential for the first reaction (which is negative) is \( -(-0.74) \, \text{V} = 0.74 \, \text{V} \).

Thus:

\[ E^\circ_{\text{cell}} = 0.94 \, \text{V} - (-0.74 \, \text{V}) = 0.94 \, \text{V} + 0.74 \, \text{V} = 1.68 \, \text{V} \]

### Final Answer

The standard cell potential \( E^\circ \) for the reaction is:

\[ E^\circ = 1.68 \, \text{V} \]