# Answers to problems in "Electrode Potentials"

** **

**QUESTION 12**

**(i)
**The half-cell reaction for the anthraquinone
monosulphonate redox-couple is

1/2 A(aq)
+ e^{- }+ H^{+ }⇌ 1/2 AH_{2}
(aq)

which implies the following electrochemical cell

Pt|H_{2}(g) (P=1 atm) | H^{+}(aq) (a=1)
||A(aq)(a_{A}), H^{+}(aq) (a_{H}^{+}), AH_{2}
(aq) (a_{AH2}) |Pt

The corresponding Nernst equation is as follows noting that the hydrogen electrode above is standard

E = E^{0}_{A/AH2} + RT/F ln[(a_{A}^{1/2} a_{H}^{+})/(
a_{AH2}^{1/2})]

Since anthraquinone and anthrahydroquinone are neutral molecules and uncharged their activities can, to a good degree of accuracy, by approximated by their concentrations. The Nernst equation is simplified to

E = E^{0}_{A/AH2} + RT/F ln[ ([A]^{1/2} a_{H}^{+})/(
[AH_{2}]^{1/2}) ]

**(ii)
**The Nernst equation from (i) can be written as

E = E^{0}_{A/AH2} + RT/F ln[ ([A]^{1/2}
a_{H}^{+})/(
[AH_{2}]^{1/2}) ] + RF/T ln(a_{H}^{+})

Substituting
log_{10}N = lnN/ln(10) into the above gives

E = E^{0}_{A/AH2} + RT/F ln[ ([AH_{2}]^{1/2})/(
[A]^{1/2}) ] + ln(10)RF/T log_{10}(a_{H}^{+})

Noting
that pH = -log_{10}(a_{H}^{+}), the Nernst equation can
be written in terms of pH

E = E^{0}_{A/AH2} + RT/F ln[ ([AH_{2}]^{1/2})/(
[A]^{1/2}) ] + ln(10)RF/T pH

Differentiating the above equation with respect to pH gives

dE/d(pH) = -ln(10)RT/F

At 25°C, T = 298K.

dE/d(pH) = -ln(10) x 298 x 8.31 / 96485.3 = -59mV per pH unit

** **

**(iii)
**The generalised redox process

B + ne^{- }+ mH^{+ }⇌ BH_{m}

can be written as an one electron half-cell reaction

1/n B + e^{- }+ m/n H^{+ }⇌ 1/n
BH_{m}

The Nernst equation for the generalised redox process is therefore

E = E^{0}_{B/BHm}+ RT/F ln[ (a_{B}^{1/n}
a_{H+}^{m/n})/( a_{BHm}^{1/n})
]

E = E^{0}_{B/BHm}+ RT/F ln[ (a_{B}^{1/n})/(
a_{BHm}^{1/n}) ] + RF/T ln(a_{H+}^{m/n})

E = E^{0}_{B/BHm}+ RT/F ln[ (a_{B}^{1/n})/(
a_{BHm}^{1/n}) ] + mRF/nT ln(a_{H+})

E = E^{0}_{B/BHm}+ RT/F ln[ (a_{B}^{1/n})/(
a_{BHm}^{1/n}) ] + mRF/nT log_{10}(a_{H+})

Substituting
in pH = -log_{10}(a_{H}^{+}) gives

E = E^{0}_{B/BHm}+ RT/F ln[ (a_{B}^{1/n})/(
a_{BHm}^{1/n}) ] + mRF/nT pH

Differentiating the above equation with respect to pH gives

dE/d(pH) = - mRT/nF ln(10)

**(iv)**
Acid dissociation constant, K_{A}, for the dissociation reaction HA⇌ H^{+}
+ A^{-} is
given by

K_{A }= a_{H}^{+}
a_{A}^{-} / a_{HA}

_{ }

**(v)
**For A/AH_{2} system (e.g.
anthraquinone/anthrahydroquinone), m = 2, n = 2

slope = dE/d(pH) = 2RT/2nF ln(10) = -59mV per pH unit**
**

**
** pKa_{1} (AH^{-}/AH_{2})
= -log(2x10^{-8})=7.7; pKa_{2} (A^{2-}/AH^{-})
= -log(1.2x10^{-11})=10.9

Accordingly,
the reduced species (anthrahydroquinone) exists predominantly as AH_{2}
for pH less than 7.7, as AH^{-} in the pH range 7.7 to 10.9 and as A^{2-}
for pH > 10.9. Applying the results derived above leads to the following E –
pH behaviour using the value n = 2 throughout but m = 0, 1 or 2 as appropriate.

**(vi,
vii)**

** **