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If E(Cd)^(0) = 0.408 V and E(Ag)^(0) = -...

If `E_(Cd)^(0) = 0.408 V and E_(Ag)^(0) = -799 V ` , the emf of the cell `Cd| Cd^(2+) || Ag^(+) | Ag` is

A

`-1.207` V

B

`+1.207` V

C

`-0.391 V `

D

`+0.391 V `

Text Solution

Verified by Experts

The correct Answer is:
B
E cell = `E_(Cd)^(0) - E_(Ag)^(0) = 0.408 - (-0.799) = + 1.207 V`
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