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

4.0921 is the logarithm of the equilibrium constant.

Step-by-step explanation:

; E° = - 0.41 V

; E° = 0.80 V

Iron having negative value of reduction potential .So ,that means that it will loose electron easily and get oxidized.Hence, will be at anode.

=Reduction potential of cathode - Reduction potential of anode

; E° = - 0.41 V

; E° = 0.80 V

Net reaction:

n = 2

To calculate equilibrium constant, we use the relation between Gibbs free energy, which is:

and,

Equating these two equations, we get:

where,

n = number of electrons transferred = 2

F = Faraday's constant = 96500 C

= standard electrode potential of the cell = 1.21 V

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction =

Putting values in above equation, we get:

4.0921 is the logarithm of the equilibrium constant.

spinning a 1, 2, or 3. option d.

step-by-step explanation: when spinning a number graph up to 4, and aiming for any number less than 3, that automatically eliminates the number 4. now you have two numbers to spin on that are less than 3, but the number 3 is included in this question because it's no greater of an number than itself. i hope this you : )