Which of the following are correct :

Options

A. During electrolysis of `NaNO_3` (aq) using Pt electrodes `H_2` & `O_2` gases are liberated at cathode & anode respectively
B. at STP, Nemst equation is E =`E^@-0.059/n log _10 Q`
C. Higher is the `E_(OP)^@` of element greater is its reducing power
D. From Kohlrasch's Law `lambda_(mBa(OH)_2)^(oo) = lambda_(mBa^(2+))^(oo)+lambda_(mOH^(-))^(oo)`

Explanation

To determine which of the given statements are correct, we will analyze each option one by one based on the principles of electrochemistry and Kohlrausch's law.

### Step 1: Analyze Option A
**Statement**: During electrolysis of NaNO3 aqueous using platinum electrode, H2 and O2 gases are liberated at cathode and anode respectively.

- **Electrolysis of NaNO3**: In an aqueous solution of NaNO3, we have the following ions:
- Cations: Na⁺
- Anions: NO3⁻ and OH⁻ (from water)
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