Discussion
For the following reaction
2MnO4- + 5H2C2O4 + 6H+ \(\rightarrow\) 2Mn2+ + 8H2O + 10CO2
E°(MnO4-/Mn2+) = +1.51 V and E°(CO2/H2C2O4) = -0.49 V.
At 298 K, the equilibrium constant is
Answer: B
The equilibrium constant is 10338.
Ten electrons are transferred in the balanced equation (2 Mn from +7 to +2). The cell potential is the cathode (permanganate) minus the anode (oxalate) standard potential:
\[ \begin{aligned} E^{\circ}_{cell} &= 1.51 - (-0.49) = 2.00\ \text{V} \\ \log K &= \frac{nE^{\circ}}{0.0591} = \frac{10 \times 2.00}{0.0591} \\ &= 338 \end{aligned} \]
The other exponents do not follow from n = 10 and a cell potential of 2.00 V. A common slip is to subtract the two potentials (1.51 - 0.49 = 1.02 V), which gives log K of about 173, also not an option; 1038 is far too small for a ten-electron reaction with a 2 V driving force.
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