Question

Consider the following cell reaction

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

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
$\text{Q} \, = \, 1$
$Q =\frac{\left[ Fe ^{2+}\right]^{2}}{\left[ H ^{+}\right]^{4}\left[ P _{\left. O _{2}\right]}\right]} \Rightarrow \frac{10^{-6}}{\left(10^{-4 x }\right)(0.01)}=1$
$10^{- 4}=10^{- 4 \text{x}}$
$\text{x}=1$