Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / 4 / b

Oxidize zinc at the anode and reduce nitrate to nitric oxide at the inert platinum cathode. Multiplying the oxidation half-reaction by three and the reduction half-reaction by two cancels six electrons:

3Zn (s) + 2N O_{3}^{-} (aq) + 8 H^{+} (aq) ⟶ 3Z n^{2 +} (aq) + 2NO (g) + 4 H_{2} O (l) .

(1)

The standard cell potential is the cathode reduction potential minus the anode reduction potential:

E^{\circ} = 0.957 - (- 0.762) = 1.719 V .

(2)

The written direction is spontaneous under standard conditions, since

Δ_{r} G^{\circ} = - 6 F E^{\circ} < 0

. Multiplying a half-reaction changes its extensive Gibbs energy, but does not multiply its intensive electrode potential.

Solution

 0  0

The Nernst equation for the six-electron cell reaction is

E = E^{\circ} - \frac{RT}{6 F} ln Q, Q = \frac{a _{Z n^{2 +}}^{3} a _{NO}^{2}}{a _{N O_{3}^{-}}^{2} a _{H^{+}}^{8}} .

(1)

Here pure solid zinc and liquid water have unit thermodynamic activity. For ideal nitric oxide gas,

a_{NO} = p_{NO} / p^{\circ}

with

p^{\circ} = 1 bar

; for ideal dissolved species,

a_{i} = [i] / c^{\circ}

. The reaction quotient is dimensionless. The platinum supplies an electronically conducting surface and is absent from the net chemical equation.

Solution

 0  0

Bulk electroneutrality gives $[N O_{3}^{-}] = [H^{+}] + 2 [Z n^{2 +}]$ . Each dissolved zinc ion contributes two positive elementary charges, whereas each nitrate contributes one negative charge.

This assumes complete nitric acid dissociation, only

H^{+}

and

Z n^{2 +}

as significant cations, and only nitrate as a significant anion. It neglects

O H^{-}

at acidic pH, other dissolved salts, zinc complexes and additional ionic nitrogen products. It is a bulk charge balance; a thin charged electrode interface does not invalidate bulk electroneutrality.

Solution

 0  0

At pH

3

[H^{+}] = 1 0^{- 3} M

, so the nitrate concentration is

0.001 + 2 (0.1) = 0.201 M

. With ideal thermodynamic activities,

Q = \frac{( 0.1 ) ^{3} ( 0.1 ) ^{2}}{( 0.201 ) ^{2} ( 1 0 ^{- 3} ) ^{8}} = 2.47519 \times 1 0^{20} .

(1)

Hence

E = 1.719 - \frac{8.3145 ( 298 )}{6 ( 96485 )} ln (2.47519 \times 1 0^{20}) = 1.5180 V .

(2)

The activity of nitric oxide is

0.1

because the supplied pressure is

0.1 bar

and the gas standard is

1 bar

. This numerical result additionally approximates zinc and nitrate thermodynamic activities by their standard-concentration ratios; without their activity coefficients, a more precise nonideal value is not specified.

This is the Gibbs energy of reaction at the stated composition and gas pressure, not the standard Gibbs energy of reaction. Its negative sign confirms thermodynamic favourability in the written direction under those conditions.