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) rar 3Cu^(+)+Cr^(+3)(aq)` is:

Options

A. `-1.08` volts
B. `-0.40` volts
C. `1.08` volts
D. `2.50` volts .

Answer

C

Explanation

The half -cell reactions of the given cell will be
`Cr to Cr^(3+) + 3e^(-) , E^(@) = + 0.74 V`
`Cu^(2+) + 2e^(-) to Cu , E^(@) = + 0.34` V
`therefore E_(cell) = 0.74 + 0.34 = 1.08 V`