"E_{cell}=E_{0cell}+\\frac{0.059}{2}lg\\frac{a_{Cd^{2+}}}{a_{Fe^{2+}}}"
"a_{Cd^{+2}}=a_{Fe^{+2}}\u00b710^{\\frac{2\u00b7(E_{cell}-E_{0cell})}{0.059}}=0.1\u00b710^{\\frac{2\u00b7(-0.02-0.04)}{0.059}}=9.2\u00b710^{-4}M"
Answer "9.2\u00b710^{-4}M"
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