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Calculate the Equilibrium Constant for the Spontaneous Redox Reaction at 298

question 11

Multiple Choice

Calculate the equilibrium constant for the spontaneous redox reaction at 298 K that would occur using the following two half-reactions.
ClO₃^-(aq) + 6 H⁺(aq) + 5 e^- $\to$ $Calculate the equilibrium constant for the spontaneous redox reaction at 298 K that would occur using the following two half-reactions. ClO3-(aq) + 6 H+(aq) + 5 e- \to Cl2(g) + 3 H2O( ) E° = +1.47 V HClO(aq) + H+(aq) + e- \to Cl2(s) + H2O( ) E° = +1.63 V A) 5.04 \times 102 B) 1.98 \times 10-3 C) 2.32 \times 1052 D) 0.996 E) 3.26 \times 1013$ Cl₂(g) + 3 H₂O( $Calculate the equilibrium constant for the spontaneous redox reaction at 298 K that would occur using the following two half-reactions. ClO3-(aq) + 6 H+(aq) + 5 e- \to Cl2(g) + 3 H2O( ) E° = +1.47 V HClO(aq) + H+(aq) + e- \to Cl2(s) + H2O( ) E° = +1.63 V A) 5.04 \times 102 B) 1.98 \times 10-3 C) 2.32 \times 1052 D) 0.996 E) 3.26 \times 1013$ ) E° = +1.47 V
HClO(aq) + H⁺(aq) + e^- $\to$ $Calculate the equilibrium constant for the spontaneous redox reaction at 298 K that would occur using the following two half-reactions. ClO3-(aq) + 6 H+(aq) + 5 e- \to Cl2(g) + 3 H2O( ) E° = +1.47 V HClO(aq) + H+(aq) + e- \to Cl2(s) + H2O( ) E° = +1.63 V A) 5.04 \times 102 B) 1.98 \times 10-3 C) 2.32 \times 1052 D) 0.996 E) 3.26 \times 1013$ Cl₂(s) + H₂O( $Calculate the equilibrium constant for the spontaneous redox reaction at 298 K that would occur using the following two half-reactions. ClO3-(aq) + 6 H+(aq) + 5 e- \to Cl2(g) + 3 H2O( ) E° = +1.47 V HClO(aq) + H+(aq) + e- \to Cl2(s) + H2O( ) E° = +1.63 V A) 5.04 \times 102 B) 1.98 \times 10-3 C) 2.32 \times 1052 D) 0.996 E) 3.26 \times 1013$ )
E° = +1.63 V

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