Question 11.5: The Nernst Equation Describe the cell based on the following...
The Nernst Equation
Describe the cell based on the following half-reactions:
VO_{2} ^{+} + 2 H^{+} + e^{-} → VO^{2+} + H_{2}O ξ° = 1.00 V (1)
Zn^{2+} + 2 e^{-} → Zn ξ° = - 0.76 V (2)
where
T = 25°C [VO_{2}^{+}] = 2.0 M [VO^{2+}] = 1.0 × 10^{-2} M
[H^{+}] = 0.50 M [Zn^{2+}] = 1.0 × 10^{-1} M
The balanced cell reaction is obtained by reversing reaction (2) and multiplying reaction (1) by 2:
2 × reaction (1)
2 VO_{2} ^{+} + 4 H^{+} + 2 e^{-} → 2 VO^{2+} + 2 H_{2}O ξ° (cathode)= 1.00 V
Reaction (2) reversed
Zn → Zn^{2+} + 2 e^{-} – ξ° (anode)= 0.76 V
Cell
reaction: 2 VO_{2} ^{+} (aq) + 4 H^{+} (aq) + Zn (s) → 2 VO^{2+} (aq) + 2 H_{2}O (l) + Zn^{2+} (aq) ξ° _{cell} = 1.76 V
Since the cell contains components at concentrations other than 1 M, we must use the Nernst equation, where n = 2 (since two electrons are transferred), to calculate the cell potential. At 25°C we can use the equation in the form
ξ = ξ° – \frac{0.0591}{n}\log (Q)
Thus
ξ = 1.76 – \frac{0.0591}{2}\log (\frac{[Zn^{2+}] [VO^{2+} ]²}{[VO_{2} ^{+} ]² [H^{+}] ^{4} } )
= 1.76 – \frac{0.0591}{2}\log (\frac{(1.0 × 10^{-1}) (1.0 × 10^{-2})²}{(2.0)² (0.50)^{4}} )
= 1.76 – \frac{0.0591}{2}\log (4 × 10^{-5}) = 1.76 + 0.13 = 1.89 V
The cell diagram is given in Fig. 11.9
