Find the equilibrium constant for the reaction cr(s) + zn2+(aq) → cr2+(aq) + zn(s) if the standard cell emf is −0.76 v at the cathode and −0.91 v at the anode. r t f = 0.025693 v

Given:

E⁰cathode = -0.76 V

E⁰anode = -0.91 V

To determine:

The equilibrium constant K for the reaction-

Cr(s) + Zn2+ (aq) ↔ Cr2+(aq) + Zn(s)

Explanation:

The relation between, E⁰cell and Keq is-

E⁰cell = (0.0592/n) log(Keq)

E⁰cell = E⁰cathode-E⁰anode = -0.76-(-0.91) = 0.15 V

n = # electrons in the reaction = 2

0.15 = (0.0592/2) * logKeq

logKeq = 5.07

Keq = 10⁵.⁰⁷ = 1.17*10⁵

Ans: the equilibrium constant is 1.17*10⁵

i think so it’s will be b yes i believe its