To calculate the corrosion potential of iron in a 0.1 M solution of sulfuric acid at 25C, we need to consider the two possible redox reactions involving iron:1 Fe Fe + 2e oxidation half-reaction 2 Fe Fe + 3e oxidation half-reaction We will use the Nernst equation to calculate the potential for each reaction:E = E - RT/nF * lnQwhere E is the electrode potential, E is the standard electrode potential, R is the gas constant 8.314 J/molK , T is the temperature in Kelvin 25C = 298.15 K , n is the number of electrons transferred, F is the Faraday constant 96485 C/mol , and Q is the reaction quotient.For the first reaction, the standard electrode potential E Fe/Fe = -0.44 V. The reaction quotient Q is given by the concentration of Fe ions, which we will assume to be equal to the concentration of sulfuric acid 0.1 M .E Fe/Fe = -0.44 - 8.314 * 298.15 / 2 * 96485 * ln 0.1 E Fe/Fe -0.44 - 0.0129 * ln 0.1 E Fe/Fe -0.44 + 0.0299E Fe/Fe -0.4101 VFor the second reaction, the standard electrode potential E Fe/Fe = -0.04 V. The reaction quotient Q is given by the concentration of Fe ions, which we will assume to be equal to the concentration of sulfuric acid 0.1 M .E Fe/Fe = -0.04 - 8.314 * 298.15 / 3 * 96485 * ln 0.1 E Fe/Fe -0.04 - 0.0086 * ln 0.1 E Fe/Fe -0.04 + 0.0199E Fe/Fe -0.0201 VNow, we need to determine which reaction is more likely to occur. Since the potential for the Fe/Fe reaction is less negative than the potential for the Fe/Fe reaction, the Fe/Fe reaction is more likely to occur.Therefore, the corrosion potential of iron in a 0.1 M solution of sulfuric acid at 25C is approximately -0.0201 V.