What is the logarithm of the equilibrium constant,log K,at 25°C of the voltaic cell constructed from the following two half-reactions?
Fe²⁺ (aq) + 2e¯ Fe(s) ; E° = - 0.41V
Ag⁺(aq) + e^-
Ag(s) ; E° = 0.80 V

Question

What is the logarithm of the equilibrium constant,log K,at 25°C of the voltaic cell constructed from the following two half-reactions?
Fe²⁺ (aq) + 2e¯

What is the logarithm of the equilibrium constant,log K,at 25°C of the voltaic cell constructed from the following two half-reactions? Fe<sup>2+</sup> (aq) + 2e¯ Fe(s) ; E° = - 0.41V Ag<sup>+</sup>(aq) + e<sup>-</sup> <sup> </sup> Ag(s) ; E° = 0.80 V A) 6.7 B) 40.9 C) 20.3 D) 13.2 E) 67.6

Fe(s) ; E° = - 0.41V
Ag⁺(aq) + e^-
What is the logarithm of the equilibrium constant,log K,at 25°C of the voltaic cell constructed from the following two half-reactions? Fe2+ (aq) + 2e¯ Fe(s) ; E° = - 0.41V Ag+(aq) + e- Ag(s) ; E° = 0.80 V A) 6.7 B) 40.9 C) 20.3 D) 13.2 E) 67.6

Ag(s) ; E° = 0.80 V

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The Answer of What is the logarithm of the equilibrium constant,log K,at 25°C of the voltaic cell constructed from the following two half-reactions? Fe²⁺ (aq)+ 2e¯ Fe(s); E° = - 0.41V Ag⁺(aq)+ e^- Ag(s); E° = 0.80 V

What Is the Logarithm of the Equilibrium Constant,log K,at 25°C