Chemistry

Problem

The reduction of

\text{NO}_3^-

occurs as

\text{NO}_3^- + 4\text{H}^+ + 3e^- \longrightarrow \text{NO} + 2\text{H}_2\text{O}; \quad E^\circ = 0.96 \text{ V}

The electrons are provided by Cd till that the solution originally having 0.1 M

\text{NO}_3^-

and 0.4 M

\text{H}^+

shows that 80% of

\text{NO}_3^-

ions are converted to NO showing 1 bar pressure. The reduction potential of remaining solution. a. 0.84 V b. 1.36 V c. 1.08 V d. 1.56 V