Consider the given data.
`{:(,"Half-cell reaction","Standard reduction"),(,,"potential"E^(@)("volts")),(1.,Cr_(2)O_(7)^(2-)+14H^(+)+6e hArr 2Cr^(3+)+7H_(2)O,1.33),(2.,Cr^(3+)+3ehArr Cr(s),-0.74),(3.,Cu^(+)+ehArrCu(s),0.51),(4.,Cu^(2+)+2ehArrCu(s),0.34):}`
The numerical value of the standard cell potential for the reaction, `Cr(s) +3Cu^(+2)(aq) hArr 3Cu^(+)+Cr^(+3)(aq)` is:

Explanation

To calculate the standard cell potential for the reaction \( \text{Cr(s)} + 3\text{Cu}^{2+}(aq) \rightleftharpoons 3\text{Cu}^{+}(aq) + \text{Cr}^{3+}(aq) \), we will follow these steps:

### Step 1: Identify Relevant Half-Reactions
From the given data, we have the following half-reactions:

1. \( \text{Cr}^{3+} + 3e^- \rightleftharpoons \text{Cr(s)} \) with \( E^\circ = -0.74 \, \text{V} \)
2. \( \text{Cu}^{2+} + 2e^- \rightleftharpoons \text{Cu(s)} \) with \( E^\circ = 0.34 \, \text{V} \)
3. \( \text{Cu}^{+} + e^- \rightleftharpoons \text{Cu(s)} \) with \( E^\circ = 0.51 \, \text{V} \)
...