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E(H^(+)//H(2))^(0)=0.00V, then E(D^(+)//...

`E_(H^(+)//H_(2))^(0)=0.00V,` then `E_(D^(+)//D_(2))^(0)` at `25^(@)C` will be

A

`0.00V`

B

more than zero `V`

C

less than zero `V`

D

car not be predicted

Text Solution

Verified by Experts

The correct Answer is:
3
D is more positive than `H`. Infact at `289^(')KE_(D^(+)//D_(2))^(@)=-0.01V`.
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