Question

For the given cell $P{t}_{D_2|D^{+}}||H^{+}|Pt|H_2$ if $E_{D_2|D^{\oplus }=0.003V}^{\ominus },$ what will be the ratio of $D^{\oplus }and{H}^{\oplus }at{25}^{\circ }C$ when he reaction: $D_2+2H^{\oplus }\to 2D^{\oplus }+H_2$ attains equilibrium :

• A. $1.34$
• B. $1.24$
• C. $1.124$
• D. $1.45$

Answer 1

The correct option is C $1.124$

$E_{cell}=E_{O{P}_{D_2}}=E_{R{P}_{H_2}}$
(OP = oxidation potential, RP = reduction potential)
$=E_{O{P}_{D_2|D^{\oplus }}}^{\ominus }-\frac{0.059}{2}log[D^{\oplus }{]}^2+E_{RP{H}^{\oplus }|H_2}+0.059log[H^{\oplus }{]}^2$
$0=0.003-\frac{0.059}{2}log\frac{[D^{\oplus }{]}^2}{[H^{\oplus }{]}^2}$
$\frac{\left[D^{\oplus }\right]}{\left[H^{\oplus }\right]}=1.124$