The following reaction takes place at 298 ~K in an electrochemical cell involving two metals A and B, A ²⁺( aq .)+ B ( s ) B ²⁺( aq )+ A ( s ) With [ A ²⁺ ]=4 10⁻³ M and [ B ²⁺ ]=2 10⁻³ M in the respective half-cells, the cell EMF is 1.091

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The following reaction takes place at 298 ~K in an electrochemical cell involving two metals A and B, A ²⁺( aq .)+ B ( s ) B ²⁺( aq )+ A ( s ) With [ A ²⁺ ]=4 10⁻³ M and [ B ²⁺ ]=2 10⁻³ M in the respective half-cells, the cell EMF is 1.091 ~V . The equilibrium constant of the reaction is closest to

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Correct answer: B. 2 10³⁷

E _ cell = E _ cell ^ - .0591 2 2 10⁻³ 4 10⁻³ 1.091= E ^ _ cell - .0591 2 (.5) E _ cell ^ =1.099 E _ cell ^ =- .0591 2 k k =- 1.099 2 .0591 =-37.22 k =2 10³⁷

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