Metallurgical Thermodynamics · Electrochemistry · 2 marks · Multiple choice
A Zn–Fe galvanic cell has $E^0_{Zn^{2+}/Zn}=-0.76$ V and $E^0_{Fe^{2+}/Fe}=-0.44$ V, and spontaneously discharges at 0.5 V. Find the order of $[Fe^{2+}]/[Zn^{2+}]$. R = 8.314 J/(mol K), F = 96500 C/mol, T = 298 K.
Zinc oxidizes and iron ions reduce, so $E^0=0.32$ V. Nernst gives $E=E^0+(RT/2F)\ln([Fe^{2+}]/[Zn^{2+}])$. Hence the logarithm of the ratio is $2F(0.50-0.32)/(RT)=14.02$, giving approximately $1.23\times10^6$. Its order is $10^6$, C.
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