The emf of the cell `Cr(s) abs(Cr^(3+)(1.0M))abs(Co^(2+)(1.0M))Co(s) [E^(o)" For " Cr^(3+)abs(Cr(s)=-0.74V & Co^(2+))Co(s) -0.28V]:`

Options

A. `-0.46` volt
B. `-1.02` volt
C. `+0.46` volt
D. `1.66` volt

Answer

C

Explanation

`E_(Cr^(3+) //Cr)^(@) = - 0.74 V , E_(Co^(2+) // Co)^(@) = -0.28 V`
The given cell reaction is
`Cr|Cr^(3+) (1.0M) || ) Co^(2+) (1.0M) ||Co`
`therefore` Cr is anode and Co is cathode
`E_(cell)^(@) = E_(C)^(@) - E_(A)^(@) = -0.28 - (-0.74)`
`= -0.28 + 0.74 = + 0.46 V `