# Question 6e2b5

Mar 5, 2016

Depends on the species at hand,

#### Explanation:

Species like

$Z {n}^{+ 2} + 2 {e}^{-} \rightarrow Z {n}^{0}$
${S}_{2} {O}_{8}^{-} 2 + 2 {e}^{-} \rightarrow 2 S {O}_{4}^{-} 2$

Aren't really affected by the pH, although species like

8 H^+ +MnO_4^− + 5 e^(−) rarr Mn^(2+) + 4 H_2O#
$2 C {r}^{3 +} + 21 {H}_{2} O \rightarrow C {r}_{2} {O}_{7}^{-} 2 + 14 {H}_{3} {O}^{+} + 6 {e}^{-}$
$2 {H}^{+} + 2 {e}^{-} \rightarrow {H}_{2}$

Are influenced by the pH, since the pH will influence the reduction potential, which will influence the $\Delta E$ of the system.

In general, if we have species that

$a A + b {H}^{+} + n {e}^{-} \rightarrow c B + k {H}_{2} O$

(Having a semirreaction that won't necessarily be real, just for explanation purposes, assuming of course this is a balanced equation as well as remembering we always write these equations as reductions and not oxidations)

The reduction potential (at standard temperature and pressure) will be given by

$E = {E}^{0} - \frac{59.15}{n} \cdot \log \left({\left[B\right]}^{c} / \left({\left[{H}^{+}\right]}^{b} {\left[A\right]}^{a}\right)\right)$

(For units in milivolts, if you'd rather work with volts, just use $0.05915$)

From there all you need to do to see how the pH wil affect the reaction is plug in ${10}^{- p H}$ in the place of $\left[{H}^{+}\right]$ or $\left[{H}_{3} {O}^{+}\right]$ depending on the author and go on your merry way.