The electrode potentials are :
`O_2 + 4 e^- + 4H^(+) to 2 H_(2) O`
`Ce^(4+) + e^(-) to Ce^(3+)`
Will `Ce^(4+)` oxidize `H_(2)O` to `O_(2)` in acidic solution ?

Explanation

`Ce^(4+)` can oxidize water as : `4 Ce^(4+) + 2 H_(2) O to O_(2) + 4 Ce^(3+) + 4H^(+)`
Let us calculate its `E^(@)` .
`Ce^(4+) + e^- to Ce^(3+)`
`O_(2) + 4e^(-) + 4 H^+ to 2 H_(2) O`
Reversing eqn. (iii) `2 H_(2) O to O_(2) + 4e^(-) + 4 H^(+)`
Multiply eqn. (ii) by 4 `4 Ce^(4+) + 4 e^(-) to 4 Ce^(3+)`
Adding eqn. (iv) and (v) `2 H_(2) O + 4 Ce^(4+) to O_2 + 4 Ce^(3+) + 4 H^(+)`
Since `E^(@)` is positive , reaction is spontaneous and so `Ce^(4+)` will oxidize `H_(2)O` to `O_(2)`.