Question

Consider the following cell reaction

$2\text{F}\text{e}\left(\text{s}\right)+{\left(\text{O}\right)}_2\left(\text{g}\right)+4{\left(\text{H}\right)}^{+}\left(\text{a}\text{q}\right)\to 2\text{F}{\left(\text{e}\right)}^{2+}\left(\text{a}\text{q}\right)+2{\left(\text{H}\right)}_2\text{O}\left(\text{l}\right)$

If ${\text{E}}_{\text{c}\text{e}\text{l}\text{l}}={\text{E}}_{\text{c}\text{e}\text{l}\text{l}}^{\text{o}}$ at $25^{\text{o}}\text{C}$ and $\left[\text{F}{\text{e}}^{2+}\right]=10^{-3}\text{M},$ ${\text{P}}_{{\text{O}}_2}$ $=0.01$ atm and $\text{p}\text{H}=\text{x}$

Value of $\text{x}$ is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A)

For ${\text{E}}_{cell}={\text{E}}_{cell}^o$
log $\text{Q}$ must be zero
$\text{Q}=1$
$Q=\frac{{\left[F{e}^{2+}\right]}^2}{{\left[H^{+}\right]}^4\left[P_{O_2]}\right]}\Rightarrow \frac{10^{-6}}{\left(10^{-4x}\right)(0.01)}=1$
$10^{-4}=10^{-4\text{x}}$
$\text{x}=1$