If the half cel reactions are given as
(i) `Fe^(2+)(Aq)+2e^(-)rarr Fe(s),E^(@)=-0.44 V`
(ii) `2H^(+)(sq)+1/2 O_(2)(g)+2e^(-)rarrH_(2)O(l) E^(@)=+1.23 V`
The `E^(@)` for the reaction
`Fe(s)+2H^(+)+1/2O_(2)(g)rarrFe^(2+)(g)rarr Fe^(2+)(aq)H_(2)O(l)` will be

Options

A. `+1.67 V`
B. `-1.67 V`
C. `+0.79 V`
D. `-0.79 V`

Answer

A

Explanation

`E_("cell")^(@)=E_("cathode")E_("anode")^(@)`
`=+1.23-(-0.44)=+1.67 V`