Question

Oxidation potential of 0.05M h2so4

Answer 1

$\begin{array}{c c c c c} H_2SO_4 & \longrightarrow & 2H^+ & + & SO_4^{2-}\\ 0.05M && 0 && 0\\ 0 && 0.1M && 0.05M\\ \end{array}\\ E_{ox}=E_{ox}^0-\dfrac{RT}{zF}\ln{Q}$
For the reaction,
$\dfrac{1}{2}H_2(g)\longrightarrow H^+(aq)+e^-\\ E_{ox}=E_{ox}^0-0.059\log{[H^+]}\\ =0-0.059\log{0.1}\\ =0.059V$