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.

62.5 paisa

step-by-step explanation:

4000*70 = 280000

280000-(2000*90) = 100000

4000-2400 = 1600

100000/1600 = 62.5

( give brainliest)

answer: your gay

step-by-step explanation:

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