0.59V
0.118V
1.18V
0.059V
Correct option is A. 0.59VH₂ → 2H⁺ + 2e⁻1 atm, 10⁻¹⁰; 0EH₂/H⁺ = 0 - 0.0592log(10⁻¹⁰)²EH₂/H⁺ = +0.59V