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Calculate the Equilibrium Constant for the Following Reaction at 25

question 22

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Calculate the equilibrium constant for the following reaction at 25 °C, 2 IO3-(aq) + 5 Hg( Calculate the equilibrium constant for the following reaction at 25 °C, 2 IO<sub>3</sub><sup>-</sup>(aq) + 5 Hg(   ) + 12 H<sup>+</sup>(aq) → I<sub>2</sub>(s) + 5 Hg<sup>2+</sup>(aq) + 6 H<sub>2</sub>O(   )  The standard reduction potentials are as follows: IO<sub>3</sub><sup>-</sup>(aq) + 6 H<sup>+</sup>(aq) + 5 e<sup>-</sup> → I<sub>2</sub>(s) + 3 H<sub>2</sub>O(   )  E° = +1.20 V Hg<sup>2+</sup>(aq) + 2 e<sup>-</sup> → Hg(   )  E° = +0.86 V A)  3 × 10<sup>-58</sup> B)  6 × 10<sup>5</sup> C)  3 × 10<sup>11</sup> D)  6 × 10<sup>28</sup> E)  3 × 10<sup>57</sup> ) + 12 H+(aq) → I2(s) + 5 Hg2+(aq) + 6 H2O( Calculate the equilibrium constant for the following reaction at 25 °C, 2 IO<sub>3</sub><sup>-</sup>(aq) + 5 Hg(   ) + 12 H<sup>+</sup>(aq) → I<sub>2</sub>(s) + 5 Hg<sup>2+</sup>(aq) + 6 H<sub>2</sub>O(   )  The standard reduction potentials are as follows: IO<sub>3</sub><sup>-</sup>(aq) + 6 H<sup>+</sup>(aq) + 5 e<sup>-</sup> → I<sub>2</sub>(s) + 3 H<sub>2</sub>O(   )  E° = +1.20 V Hg<sup>2+</sup>(aq) + 2 e<sup>-</sup> → Hg(   )  E° = +0.86 V A)  3 × 10<sup>-58</sup> B)  6 × 10<sup>5</sup> C)  3 × 10<sup>11</sup> D)  6 × 10<sup>28</sup> E)  3 × 10<sup>57</sup> )
The standard reduction potentials are as follows:
IO3-(aq) + 6 H+(aq) + 5 e- → I2(s) + 3 H2O( Calculate the equilibrium constant for the following reaction at 25 °C, 2 IO<sub>3</sub><sup>-</sup>(aq) + 5 Hg(   ) + 12 H<sup>+</sup>(aq) → I<sub>2</sub>(s) + 5 Hg<sup>2+</sup>(aq) + 6 H<sub>2</sub>O(   )  The standard reduction potentials are as follows: IO<sub>3</sub><sup>-</sup>(aq) + 6 H<sup>+</sup>(aq) + 5 e<sup>-</sup> → I<sub>2</sub>(s) + 3 H<sub>2</sub>O(   )  E° = +1.20 V Hg<sup>2+</sup>(aq) + 2 e<sup>-</sup> → Hg(   )  E° = +0.86 V A)  3 × 10<sup>-58</sup> B)  6 × 10<sup>5</sup> C)  3 × 10<sup>11</sup> D)  6 × 10<sup>28</sup> E)  3 × 10<sup>57</sup> )
E° = +1.20 V
Hg2+(aq) + 2 e- → Hg( Calculate the equilibrium constant for the following reaction at 25 °C, 2 IO<sub>3</sub><sup>-</sup>(aq) + 5 Hg(   ) + 12 H<sup>+</sup>(aq) → I<sub>2</sub>(s) + 5 Hg<sup>2+</sup>(aq) + 6 H<sub>2</sub>O(   )  The standard reduction potentials are as follows: IO<sub>3</sub><sup>-</sup>(aq) + 6 H<sup>+</sup>(aq) + 5 e<sup>-</sup> → I<sub>2</sub>(s) + 3 H<sub>2</sub>O(   )  E° = +1.20 V Hg<sup>2+</sup>(aq) + 2 e<sup>-</sup> → Hg(   )  E° = +0.86 V A)  3 × 10<sup>-58</sup> B)  6 × 10<sup>5</sup> C)  3 × 10<sup>11</sup> D)  6 × 10<sup>28</sup> E)  3 × 10<sup>57</sup> )
E° = +0.86 V


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