# Answers to problems in "Electrode Potentials"

## Question 5

The cell diagram is:

Pt | H_{2}(g) (P =
1 atm) | HCl (a = 1) | AgCl(s) | Ag(s) |

This is the Harned Cell. We first write down the potential determining equilibria

at the RHE: | AgCl(s) + e^{-} ⇌
Ag(s) + Cl^{-}(aq) |

at the LHE: | H^{+}(aq) + e^{-} ⇌
½H_{2}(g) |

Hence, the cell reaction is:

AgCl(s) + ½H_{2}(g)
⇌ Ag(s) +
H^{+}(aq) + Cl^{-}(aq) |

For one electron transfer processes, DG^{0}
= -FE^{0}_{cell}

Therefore, at 298K,

E^{0}_{cell} = 0.2366 - 4.856 x 10^{-4}(T
- 273) - 3.421 x 10^{-6}(T - 273)^{2}

= 0.2366 - 4.856 x 10^{-4}(298-273) - 3.421
x 10^{-6}(298 - 273)^{2}

= +0.225V

Therefore, ** DG ^{0}**
= -(96 490)(0.225) =

**-21.45kJ mol**

^{-1}

The reader is advised to review pp. 25-28. We quote
the result DS^{0} = F(∂E^{0}/
∂T)_{P}

Differentiating the expression given in the question
with respect to T (at constant P) using the chain rule (see Sivia,
D, "*Fundamentals of Mathematics for Chemists*";
Oxford Chemistry Primer, OUP (to be published)) gives,

^{0}/∂T)

_{P}= -4.856 x 10

^{-4 }- 2 x 3.421 x 10

^{-6}x (T - 273)

Therefore, at 298K, (∂E^{0}/∂T)_{P}
= -6.5665 x 10^{-4} V K^{-1}

Hence,** DS ^{0}
= -63.38J mol^{-1} K^{-1
}**

Since, DG^{0}
= DH^{0} - TDS^{0}

** DH ^{0}
= -40.33kJ mol^{-1}**