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Question

Chemistry Question on Law Of Chemical Equilibrium And Equilibrium Constant

FeO42+2.2v  Fe3++0.70v  Fe2+0.45v  Fe0FeO^{2-}_4+\underrightarrow{2.2v}\;Fe^{3+}\underrightarrow{+0.70v}\;Fe^{2+}\underrightarrow{-0.45v}\;Fe^0
EFeO42+0  is  x×103VE^0_{FeO_4^{2+}}\; is \;x × 10^{-3} V. The Value of xx is __

Answer

The Answer is 1825